Equations whose unknown is a function, the order of an equation, what counts as a solution, general and particular solutions, initial-value problems and how many conditions each order needs, and verifying a proposed solution by substitution.
Subject: Calculus II · 65 slides · symbolic lesson
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Title
Calculus II · Section 4.1
Equations whose unknown is a whole function
Objectives
Every equation you have solved so far had a number for its answer. In this chapter the answer is a function, and the equation tells you about its derivatives. This first section is about reading such equations and checking answers to them.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 352-361 — learning objectives 4.1.1 to 4.1.5
Up to now, solving an equation meant finding a number, or a few numbers, that make it true. A differential equation is different: its unknown is a whole function, and the equation describes how that function changes. The answer is a function, and very often a whole family of functions.
This first section of Chapter 4 does not yet teach clever solving methods. It teaches you to read these equations: to say how complicated one is, to say what counts as an answer, and to check a proposed answer with complete confidence. It also shows you the simplest kind you can already solve, where the derivative is given outright and you only need to antidifferentiate.
The last part explains why these equations matter. Physical laws are almost always statements about rates of change, so the moment you write a law down, you have written a differential equation.
Warm-up
Discussion prompt
From Calculus I: find every function whose derivative is 3x². Then find the one whose graph passes through the point (1, 5).
Write your answer before revealing it. You have done this kind of problem before, in the antiderivatives section of Calculus I, perhaps without calling it a differential equation.
Notice the two stages. First, every antiderivative of three x squared: x cubed plus any constant, a whole family of functions. Second, one extra fact, the point on the graph, picks out a single member of the family, the one with constant 4.
Hold on to that shape. It is the skeleton of everything in this lesson: a differential equation gives a family, and extra information chooses the member. The new ideas are about equations where the derivative is not handed to you on its own, and about how much extra information you need.
Section
Part 1
Concept
\[ y' = 3x^2 \]
Read it as an instruction: start with some function y of x, differentiate it, and the result must be three x squared. The unknown is not a number but the whole function.
Figure (svg): The curve y equals x cubed from x equals minus 1.6 to 1.6, with short yellow tangent segments at x equals minus 1, 0.5 and 1.2 whose slopes are 3, 0.75 and 4.32, each equal to three x squared.
\[ y = x^3 \;\Longrightarrow\; y' = 3x^2 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 352 — the opening example
Read the equation as an instruction about slopes. Whatever function y is, its derivative at each x must be three x squared. So the question is not which number works, but which function works.
The figure shows the curve y equals x cubed with three tangent segments drawn in. At x equal to minus 1 the slope is 3; at one half it is three quarters; at 1.2 it is 4.32. In each case the slope equals three times the square of x, exactly what the equation demands. And it does so at every x at once, not just at the three that are drawn.
That is what makes x cubed a solution. It is not the only one: adding a constant slides the curve up or down without changing any slope, so x cubed plus 4, or minus 7, works just as well. You will see shortly that this is always the case for this kind of equation.
Concept
differential equation — An equation involving an unknown function y equal to f of x and one or more of its derivatives.
solution — A function that satisfies the differential equation when it and its derivatives are substituted into the equation.
| equation | a solution |
|---|---|
| y′ = 2x | y = x² |
| y′ + 3y = 6x + 11 | y = e⁻³ˣ + 2x + 3 |
| y″ − 3y′ + 2y = 24e⁻²ˣ | y = 3eˣ − 4e²ˣ + 2e⁻²ˣ |
Solutions are not unique: x squared plus 4 also solves the first equation, because the derivative of a constant is zero.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 352-353 — Definition and Table 4.1
These two definitions are the foundation of the chapter, so read them slowly. A differential equation involves an unknown function and at least one of its derivatives. A solution is a function that makes the equation true when you substitute it and its derivatives in.
The table shows three examples from the textbook, in increasing difficulty. In the first, the derivative is simply equal to a function of x. In the second, the unknown y appears alongside its derivative, so you cannot just integrate. The third involves the second derivative as well. You will verify the second and third rows yourself in a moment.
The final sentence matters more than it looks. A differential equation usually has infinitely many solutions, because constants disappear when you differentiate. So when you are asked to verify a solution, you are checking that a particular function is one member of that family, not that it is the only answer.
Notation
Annotate
On: \( y' + 3y = 6x + 11 \quad\text{for every } x \)
Step through the annotations one at a time. The key thing to absorb is that y is a function, so y prime and three y are also functions, and the equation says that one particular combination of them equals another function, six x plus eleven.
Because y appears undifferentiated on the left, you cannot solve this by integrating the right side. Integrating would give you information about y prime alone, but the equation mixes y prime with y. Methods for this kind of equation come in Section 4.5. For now, the job is only to check a candidate.
The last annotation is the one people forget. The equation must hold for every x in an interval. That is a far stronger requirement than holding at one x, and it is what makes verification a matter of simplifying expressions rather than plugging in numbers.
Worked example
Verify that the function below is a solution of the differential equation.
\[ y = e^{-3x} + 2x + 3, \qquad y' + 3y = 6x + 11 \]
Differentiate the candidate
Why: The chain rule on the exponential brings down minus 3.
\[ y' = -3e^{-3x} + 2 \]
Substitute into the left side
Why: Put y prime and three times y side by side.
\[ y' + 3y = \left(-3e^{-3x} + 2\right) + 3\left(e^{-3x} + 2x + 3\right) \]
Distribute the 3
Why: Remove the parentheses.
\[ = -3e^{-3x} + 2 + 3e^{-3x} + 6x + 9 \]
Combine like terms
Why: The exponentials cancel exactly.
\[ = 6x + 11 \]
Check against the right side
Why: The left side has become the right side for every x, so the candidate is a solution; a spot value at x equal to 1 agrees too.
\[ x = 1: \quad (-3e^{-3} + 2) + 3(e^{-3} + 5) = 17 = 6(1) + 11 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 353 — Example 4.1
This is the model verification, and every one you do will follow the same four moves. Differentiate the candidate. Substitute the candidate and its derivative into the left side. Simplify. Compare with the right side.
The chain rule gives the derivative of e to the minus three x as minus three times the same exponential. When you add three times the candidate, the two exponential terms are minus three and plus three times the same thing, so they cancel exactly. What survives is two plus six x plus nine, which is six x plus eleven.
The comparison is between expressions, and it holds for every x. The final spot check at x equal to 1 is not the proof; it is a quick safeguard against an algebra slip. If the numbers had disagreed, you would know something went wrong somewhere above.
One small warning: the printed textbook solution has a typo, writing e to the minus two x in the middle lines. The answer is right, but trust the exponent in the candidate, minus three x.
Picture it
Figure (svg): Two panels. Left: for y equals e to the minus 3x plus 2x plus 3, the left side y prime plus 3y (dashed) lies exactly on the right side 6x plus 11 for every x. Right: for the candidate x squared and the equation y prime equals 2y, the left side 2x and the right side 2x squared are different curves that meet only at x equals 0 and x equals 1.
On the left the two sides of Example 4.1 are drawn as functions of x, and they are one and the same line. On the right, x squared is tested in the equation y prime equals 2y: the two sides are different curves that happen to meet twice.
Each panel treats the two sides of an equation as two functions of x and draws both. On the left, the two sides of Example 4.1: the dashed yellow line lies exactly on the thick blue one, because after simplifying they are the same function. That is what being a solution looks like.
On the right, the candidate x squared is tested in the equation y prime equals two y. The left side is two x, a straight line; the right side is two x squared, a parabola. They meet at the two marked points, x equal to 0 and x equal to 1, and differ everywhere else.
This is why plugging in one convenient number is not a check. If you happened to choose x equal to 0 or 1, the two sides would agree and you would wrongly accept the candidate. Only comparing the expressions, which is comparing the whole curves, settles it.
Worked example
\[ y = 2e^{3x} - 2x - 2, \qquad y' - 3y = 6x + 4 \]
Differentiate
Why: The exponential brings down a 3.
\[ y' = 6e^{3x} - 2 \]
Compute minus three times y
Why: Distribute the minus 3 across all three terms.
\[ -3y = -6e^{3x} + 6x + 6 \]
Add the two
Why: The exponentials cancel.
\[ y' - 3y = 6e^{3x} - 2 - 6e^{3x} + 6x + 6 \]
Simplify
Why: Collect the constants.
\[ y' - 3y = 6x + 4 \]
Check at a point as well
Why: At x equal to 0: y is 0 and y prime is 4, so the left side is 4, as is the right.
\[ y(0) = 0, \; y'(0) = 4: \quad 4 - 3(0) = 4 = 6(0) + 4 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 353 — Checkpoint 4.1
Try this one yourself before revealing the steps; it is the same four moves as Example 4.1. The only new feature is the minus sign in front of three y, which has to be distributed across all three terms of the candidate.
The exponential again cancels: six e to the three x from the derivative against minus six e to the three x from the minus three y. What is left is minus two, plus six x, plus six, which collects to six x plus four, the right side.
Notice how the constants in the candidate were chosen. The minus two x and minus two are exactly what is needed to produce six x plus four; change either of them and the check fails. When you meet the methods that produce such candidates later in the chapter, this is the pattern they generate.
Trap
Testing x squared in the equation y prime equals 2y:
\[ x = 0: \quad y' = 0, \; 2y = 0 \]
\[ \Longrightarrow\; y = x^2 \text{ solves it?} \]
Wrong. One matching value proves nothing.
Compare the two sides as functions. They agree only where 2x equals 2x squared, at x equal to 0 and 1, and nowhere else.
\[ 2x \ne 2x^2 \text{ for } x \ne 0, 1 \]
A point check can only reject a candidate, never confirm one.
This mistake usually happens under time pressure. You want to check quickly, so you plug in zero, both sides come out zero, and you declare victory. But one agreeing value is weak evidence, because two different functions can easily agree at one point.
The fix is to compare the two sides as functions. For x squared in the equation y prime equals two y, the left side is two x and the right side is two x squared. They are equal only at zero and at one. Everywhere else they differ, so x squared is not a solution.
Remember the asymmetry. A single point where the sides disagree is enough to reject a candidate, and that can be a useful shortcut. A single point where they agree proves nothing at all.
Worked example
The third row of Table 4.1 involves a second derivative. Verify it.
\[ y = 3e^{x} - 4e^{2x} + 2e^{-2x}, \qquad y'' - 3y' + 2y = 24e^{-2x} \]
First derivative
Why: Differentiate each exponential.
\[ y' = 3e^{x} - 8e^{2x} - 4e^{-2x} \]
Second derivative
Why: Differentiate again.
\[ y'' = 3e^{x} - 16e^{2x} + 8e^{-2x} \]
Collect the e to the x terms
Why: Coefficients 1, minus 3 and 2 multiply 3, 3 and 3.
\[ e^{x}: \quad 3 - 9 + 6 = 0 \]
Collect the e to the 2x terms
Why: Minus 16, then minus 3 times minus 8, then 2 times minus 4.
\[ e^{2x}: \quad -16 + 24 - 8 = 0 \]
Collect the e to the minus 2x terms
Why: 8, then minus 3 times minus 4, then 2 times 2.
\[ e^{-2x}: \quad 8 + 12 + 4 = 24 \]
Check the total against the right side
Why: Only the last family survives, with coefficient 24: exactly the right side.
\[ y'' - 3y' + 2y = 0 + 0 + 24e^{-2x} = 24e^{-2x} \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 353 — Table 4.1
Second-order equations are verified the same way, with one extra differentiation. The organising trick here is to group by exponential. The candidate is built from three different exponentials, and since they are genuinely different functions, each group must balance separately.
For the e to the x terms, every derivative is still three e to the x, so the coefficients are one, minus three and two applied to three: three minus nine plus six, which is zero. For e to the two x, the second derivative gives minus sixteen, minus three times minus eight gives twenty-four, and two times minus four gives minus eight: zero again.
Only e to the minus two x survives: eight, plus twelve, plus four, which is twenty-four. That is exactly the right side. Grouping like this turns a long, error-prone expansion into three small sums, each easy to check.
Concept
Figure (svg): The graph of y equals one over one minus x, with a dashed vertical asymptote at x equals 1. The left branch, drawn in blue, rises from near 0 toward infinity as x approaches 1; the right branch, in orange, comes up from minus infinity toward 0.
A solution has to be differentiable wherever it is claimed to work. One over one minus x fails to exist at x equal to 1, so the formula gives one solution on the left of 1 and a separate one on the right.
\[ y = \frac{1}{1-x} \quad \text{on } (-\infty, 1) \text{ or on } (1, \infty) \]
Look at the graph before the algebra. The formula one over one minus x has a vertical asymptote at x equal to 1, where the denominator is zero. The function does not exist there, so it certainly has no derivative there.
A solution must be differentiable wherever it is claimed to satisfy the equation, and it must do so on an interval, an unbroken stretch of the number line. So this formula actually describes two different solutions: the blue branch on the numbers less than 1, and the orange branch on the numbers greater than 1.
This is the first time in the chapter that the domain matters, and it will not be the last. When a solution blows up at a finite value, as this one does, the interval on which it lives is part of the answer.
Worked example
Verify the candidate. The equation squares the unknown, so it is not of the kind in Table 4.1.
\[ y = \frac{1}{1-x}, \qquad y' = y^2 \]
Rewrite as a power
Why: Easier to differentiate.
\[ y = (1-x)^{-1} \]
Differentiate with the chain rule
Why: The inner derivative is minus 1, which cancels the minus from the power.
\[ y' = -(1-x)^{-2}\cdot(-1) = (1-x)^{-2} \]
Square the candidate
Why: The right side of the equation.
\[ y^2 = \left(\frac{1}{1-x}\right)^2 = (1-x)^{-2} \]
Compare
Why: The same expression, at every x except 1.
\[ y' = (1-x)^{-2} = y^2, \quad x \ne 1 \]
Check with numbers
Why: At x equal to 0 both sides are 1; at x equal to 3, y is minus one half and both sides are one quarter.
\[ x = 3: \quad y' = (-2)^{-2} = \tfrac14 = \left(-\tfrac12\right)^2 = y^2 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 362 — Exercise 11
The equation here squares the unknown, which puts it outside the family of Table 4.1, where y and its derivatives only appear to the first power. The verification method does not care: differentiate, substitute, compare.
Rewriting the candidate as one minus x to the power minus one makes the chain rule clean. The outer derivative brings down minus one and lowers the power to minus two; the inner derivative is minus one, and the two minus signs cancel. Squaring the candidate gives the same power, so the two sides agree wherever the candidate is defined.
The numerical check at x equal to 3 is on the right-hand branch of the graph, where the candidate is negative. Squaring a negative number gives a positive one, and the derivative there is positive too: both are one quarter. The equation holds on each branch separately.
Intuition
To verify a candidate you only differentiate and simplify, and differentiation always works. To find a solution you must run differentiation backwards, and for most equations there is no formula that does that.
\[ \text{verify: } y \;\mapsto\; y', y'' \;\mapsto\; \text{substitute} \]
So verification is always your first tool. Whether the answer came from a textbook, a classmate or a computer, you can test it in a few lines, and every worked example in this chapter ends that way.
This slide is about why verification deserves so much attention. Differentiation is a mechanical process: given any formula built from the usual functions, the rules always produce its derivative. So checking a candidate is always possible, even if it takes a few lines.
Solving is the reverse process, and reverse processes are hard. Most differential equations cannot be solved with a formula at all, and the rest need specific methods that you will meet in Sections 4.3 and 4.5. Section 4.2 shows how to approximate solutions when no formula is available.
So get into the habit now. Whenever you produce a solution, whether by hand, from a table or from software, finish by substituting it back. It costs a minute and catches most mistakes, which is why every worked example in this chapter ends with a check.
Sorting
Sort into buckets
Differentiate each candidate and compare with twice the candidate.
Sort all six before checking. For each candidate, compute its derivative and compare it with twice the candidate, as whole expressions.
The exponential e to the two x works because its derivative brings down a factor of two. Multiplying by five changes nothing essential: the derivative is ten e to the two x, which is still twice the candidate. The zero function also works, since its derivative is zero, which is twice zero. All three belong to one family, a constant times e to the two x.
The failures are instructive. Adding one to the exponential breaks it, because the derivative ignores the one but twice the candidate does not; the sides differ by two everywhere. The candidate e to the three x has the wrong growth rate. And x squared fails for the reason you saw in the trap. Notice that for this equation, unlike y prime equals two x, adding a constant does not produce another solution.
Section
Part 2
Concept
order — The highest order of any derivative of the unknown function that appears in the equation.
\[ y', \quad y'', \quad y''', \quad y^{(4)}, \quad y^{(5)}, \;\ldots \]
Past the third derivative, primes become unreadable, so the order is written as a number in parentheses. The order counts how many times the unknown has been differentiated, nothing else: coefficients, powers and products do not change it.
| equation | order |
|---|---|
| y′ = 2x | 1 |
| y″ − 3y′ + 2y = 24e⁻²ˣ | 2 |
| y‴ = 0 | 3 |
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 353 — Definition of order
The order is the most basic way to classify a differential equation, and it is found by looking, not by calculating. Scan the equation for derivatives of the unknown, and the order is the highest one you find.
The notation changes after the third derivative. Four primes are hard to read and easy to miscount, so the fourth derivative is written as y with a small four in parentheses, and so on upward. The parentheses matter: without them the four would look like a power.
Everything else in the equation is irrelevant to the order. A coefficient like x squared in front of a derivative, a power on a derivative, a product of two derivatives, a function of x on the right side: none of these changes how many times y has been differentiated. You will see shortly why this number matters so much.
Notation
Annotate
On: \( x^2 y''' - 3x y'' + x y' - 3y = \sin x \)
Step through the annotations. This equation looks intimidating because every term has a coefficient that depends on x, but the order only asks about the derivatives themselves.
The third derivative is present, and nothing higher is, so the order is three. The lower derivatives, the function itself, the coefficients and the sine on the right side are all part of the equation, but none of them can raise the order.
A useful habit is to cover up everything except the derivatives of y, reading only the list y triple prime, y double prime, y prime and y. The largest in that list is the order. This works however messy the rest of the equation becomes.
Worked example
\[ \text{(a) } y' - 4y = x^2 - 3x + 4 \]
Highest derivative in (a)
Why: Only a first derivative appears.
\[ y' \;\Longrightarrow\; \text{order } 1 \]
\[ \text{(b) } x^2y''' - 3xy'' + xy' - 3y = \sin x \]
Highest derivative in (b)
Why: A third derivative is present.
\[ y''' \;\Longrightarrow\; \text{order } 3 \]
\[ \text{(c) } 4xy^{(4)} - 6x^2y'' + 12x^4y = x^3 - 3x^2 + 4x - 12 \]
Highest derivative in (c)
Why: The superscript four in parentheses is a fourth derivative.
\[ y^{(4)} \;\Longrightarrow\; \text{order } 4 \]
\[ \text{Checkpoint: } (x^4 - 3x)y^{(5)} - (3x^2+1)y' + 3y = \sin x\cos x \]
Highest derivative in the checkpoint
Why: A fifth derivative; its messy coefficient is irrelevant.
\[ y^{(5)} \;\Longrightarrow\; \text{order } 5 \]
Check by listing every derivative present
Why: Take the largest from each list; none of the powers of x can raise it.
\[ \{1, 0\}, \; \{3,2,1,0\}, \; \{4,2,0\}, \; \{5,1,0\} \;\to\; 1, 3, 4, 5 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 354 — Example 4.2 and Checkpoint 4.2
These four are routine once you know what to look for, and the steps show the single idea applied four times: find the highest derivative. Part (a) has only a first derivative. Part (b) reaches a third. Part (c) uses the parenthesised four, a fourth derivative.
The checkpoint is designed to distract you with an elaborate coefficient, x to the fourth minus three x, multiplying the fifth derivative. The coefficient is irrelevant. The fifth derivative is present, so the order is five.
The final line is a check you can always carry out: list the orders of every derivative that appears, including zero for the function itself, and take the largest. Writing the list out forces you to look at every term, which protects against missing a high derivative buried in the middle of a long equation.
Trap
Exercise 2, classified in a hurry:
\[ (y')^2 = y' + 2y \]
\[ \Longrightarrow\; \text{order } 2? \]
Wrong. The 2 is an exponent.
Squaring the first derivative is arithmetic on y prime; it does not differentiate again. No second derivative appears, so the order is 1.
\[ (y')^2 = y'\cdot y' \;\Longrightarrow\; \text{order } 1 \]
Compare y double prime, which is a second derivative.
The mistake comes from the notation. A small two above and to the right looks like it could mean either a power or a second derivative, and in a hurry the two blur together.
Here the two is outside the parentheses, so it is a power: the first derivative multiplied by itself. No differentiation beyond the first has happened, so the order is one. If the equation had contained y double prime, or the second derivative written with d squared y over d x squared, the order would have been two.
Why does this matter beyond getting a classification question right? Because the order tells you how many conditions a problem needs. Misreading this equation as second order would lead you to look for two initial values when one is enough.
Matching
Match the pairs
Why: Look only for the highest derivative. The square on y in (a) and the seventh power in (d) are powers, not orders; the product y″y′ in (c) does not matter because y‴ is already there.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 362 — Exercises 1, 3 and 4
Match all four before checking. The right-hand column has one of each order from one to four, so each equation has exactly one partner.
The distractions are deliberate. The first equation squares y, but squaring y is not differentiating it, so the order is one. The second has y double prime on the right side, which counts just as much as on the left: order two. The third multiplies two derivatives together, but it already contains the third derivative, so the product is irrelevant: order three.
The fourth raises the first derivative to the seventh power. That power is enormous but meaningless for the order; the fourth derivative on the left is what sets it. When you feel unsure, cover up the powers and coefficients and list the derivatives.
Concept
Take the simplest second-order equation, the free fall of a ball, and undo it one derivative at a time.
\[ s'' = -9.8 \]
\[ s' = -9.8t + C_1 \]
\[ s = -4.9t^2 + C_1t + C_2 \]
Each antidifferentiation brings its own constant, so an equation of order n has a general solution with n arbitrary constants, and needs n pieces of information to pin them down.
Figure (svg): Four parabolas s equals minus 4.9 t squared plus v0 t plus 3, for launch speeds v0 of 0, 5, 10 and 15, all starting at the same point (0, 3) and spreading apart; the v0 equals 10 curve is highlighted.
Here is why the order deserves its own name. Take the free-fall equation, second derivative of height equal to minus 9.8, and undo it one step at a time. The first antidifferentiation gives the velocity, with a constant C sub one. The second gives the height, with a new constant C sub two.
Two antidifferentiations, two constants. The same count holds in general: an equation of order n has a general solution with n arbitrary constants, and so n pieces of information are needed to fix them all.
The figure shows what goes wrong with too little information. Every curve starts at height 3, so each satisfies the condition on the starting height. But they have different launch speeds, and each is a perfectly good solution of the equation. Only a second condition, the starting velocity, can pick out the highlighted throw.
Section
Part 3
Concept
Any constant C works, because a constant differentiates to zero. The converse is the real claim: nothing else works. Suppose y is any solution and subtract x squared.
\[ \frac{d}{dx}\left(y - x^2\right) = y' - 2x = 2x - 2x = 0 \]
\[ y - x^2 = C \;\Longrightarrow\; y = x^2 + C \]
A function with zero derivative on an interval is constant, by the Mean Value Theorem. So this one-parameter family is every solution: the general solution.
Figure (svg): Parabolas y equals x squared plus C for C equal to minus 4, minus 2, 0, 2 and 4, stacked vertically, plus the highlighted parabola y equals x squared plus 3 passing through the point (2, 7).
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 354 — Figure 4.2
The first half of the claim is easy: x squared plus any constant has derivative two x. The interesting half is that there are no other solutions, and the argument is short enough to remember.
Take any solution y and look at y minus x squared. Its derivative is y prime minus two x, which is zero because y is a solution. A function whose derivative is zero on an interval must be constant there, a consequence of the Mean Value Theorem. So y minus x squared is some constant C, and y is x squared plus C.
That is why the formula with an arbitrary constant deserves the name general solution: it captures every solution, not just some. The figure reproduces Figure 4.2 from the textbook, with the even values of C from minus four to four, plus the highlighted member with C equal to 3 that the next example will single out. Remember that C can be any real number, not only an integer.
Picture it
Figure (svg): A grid of short line segments whose slope at each point equals 2x: flat along the y-axis, tilting down on the left and up on the right. Three parabolas y equals x squared plus C, for C equal to minus 2, 0 and 2, follow the segments exactly.
Each little segment has the slope the equation demands at that point, two times x. A solution is any curve that runs along the segments everywhere; the parabolas do, and one passes through every point of the plane.
Each short segment in the grid is drawn with the slope the equation assigns to that point, two times x. Along the vertical axis the segments are flat, because x is zero there. To the right they tilt up, more steeply as x grows; to the left they tilt down.
The equation says nothing about height, which is why every segment in a vertical column has the same slope. A solution is a curve that follows the segments everywhere, and the three parabolas drawn do exactly that. They are the same curve shifted up or down.
This picture is called a slope field or direction field, and it is the main tool of the next section. For equations you cannot solve with a formula, it still shows you what the solutions look like.
Prediction
\[ y = x^2 + C_1 \quad\text{and}\quad y = x^2 + C_2, \quad C_1 \ne C_2 \]
Predict first
Can two different solutions of y′ = 2x ever meet at a point?
Correct: Never: their gap is always C₁ − C₂
Why: Subtract the two formulas: the x squared cancels and the difference is the constant C₁ − C₂ at every x, which is not zero. Curves that are always the same nonzero distance apart never touch, which is why one point is enough to single out one curve.
\[ (x^2 + C_1) - (x^2 + C_2) = C_1 - C_2 \ne 0 \]
Commit to one option before revealing. It is tempting to think the parabolas might cross near the bottom, where they bend, but the algebra settles it in one line.
Subtract one member from another. The x squared terms cancel, leaving the difference of the two constants, the same at every x. Two curves that are always the same vertical distance apart can never meet. The figure of the family shows exactly this: the parabolas are stacked like copies of one another.
This is what makes initial conditions work. Because the curves never meet, each point of the plane lies on exactly one of them. So naming a single point is enough to pick out a single solution, which is exactly what the next example does.
Worked example
Find the particular solution of the differential equation that passes through the given point.
\[ y' = 2x, \qquad (2, 7) \]
Write the general solution
Why: Every solution has this form.
\[ y = x^2 + C \]
Substitute the point
Why: x equal to 2 and y equal to 7.
\[ 7 = 2^2 + C \]
Simplify
Why: Two squared is four.
\[ 7 = 4 + C \]
Solve for C
Why: Subtract 4.
\[ C = 3 \]
Write the particular solution
Why: Put C back.
\[ y = x^2 + 3 \]
Check both requirements
Why: The derivative is 2x, so it solves the equation, and at x equal to 2 the height is 7, so it passes through the point.
\[ y' = 2x \;\checkmark, \qquad y(2) = 4 + 3 = 7 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 355 — Example 4.3
This is the simplest possible way to choose one member of a family. The general solution has one unknown constant, and a point on the graph gives one equation for it.
Substitute the coordinates of the point, x equal to 2 and y equal to 7, into the general solution. That gives seven equals four plus C, so C is 3, and the particular solution is x squared plus 3. In the family picture, this is the highlighted parabola passing through the marked point.
The check asks two separate questions, and both must be answered. Does the function still solve the equation? Yes, its derivative is two x. Does it pass through the point? Yes, at x equal to 2 it takes the value 7. A function that passes the first test but not the second is the wrong member of the right family.
Worked example
\[ y' = 4x + 3, \quad (1, 7), \quad \text{general: } y = 2x^2 + 3x + C \]
Confirm the general solution
Why: Differentiate it once.
\[ \frac{d}{dx}\left(2x^2 + 3x + C\right) = 4x + 3 \]
Substitute the point
Why: x equal to 1, y equal to 7.
\[ 7 = 2(1)^2 + 3(1) + C \]
Simplify
Why: Two plus three.
\[ 7 = 5 + C \]
Solve for C
Why: Subtract 5.
\[ C = 2 \;\Longrightarrow\; y = 2x^2 + 3x + 2 \]
Check the point
Why: At x equal to 1 the height is 7.
\[ y(1) = 2 + 3 + 2 = 7 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 355 — Checkpoint 4.3
This time the general solution is given to you, but it is still worth confirming it before using it. One differentiation shows that two x squared plus three x plus C has derivative four x plus three, exactly the equation.
Then the same routine: substitute the point, simplify to seven equals five plus C, and solve to get C equal to 2. The particular solution is two x squared plus three x plus two.
The check at the point uses the final answer, not the general formula. That is deliberate: it tests the whole chain of steps, including the moment where C was put back into the formula. A slip there, such as writing the constant as 7 instead of 2, would be caught immediately.
Worked example
Verify the general solution, then find the particular solution through (0, 12).
\[ y' = 3x^2y, \qquad y = Ce^{x^3} \]
Differentiate the general solution
Why: Chain rule: the derivative of x cubed is 3x squared.
\[ y' = Ce^{x^3}\cdot 3x^2 \]
Recognise y inside it
Why: The factor C times e to the x cubed is y itself.
\[ y' = 3x^2\left(Ce^{x^3}\right) = 3x^2y \;\checkmark \]
Substitute the point
Why: x equal to 0, y equal to 12; e to the 0 is 1.
\[ 12 = Ce^{0} = C \]
Write the particular solution
Why: C is 12.
\[ y = 12e^{x^3} \]
Figure (svg): Curves y equals C times e to the x cubed for C equal to minus 6, minus 3, 3 and 6, plus the highlighted curve with C equal to 12 through the point (0, 12). All of them flatten toward the x-axis on the left and never cross; C equals 0 is the x-axis itself.
Check the point and the equation
Why: At x equal to 0 the height is 12, and the derivative is 36x squared times the exponential, which is 3x squared times y.
\[ y(0) = 12, \quad y' = 36x^2e^{x^3} = 3x^2\left(12e^{x^3}\right) \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 362 — Exercise 20
In every family so far the constant has been added, which slides the curves up and down. Here the constant multiplies, which stretches the curve vertically instead. The family still works the same way: one constant, one condition.
The verification uses the chain rule. The derivative of e to the x cubed is three x squared times e to the x cubed, and the constant C just comes along. Once you spot that C times the exponential is y itself, the derivative becomes three x squared times y, which is the equation.
The point has x equal to 0, where the exponential equals 1, so C is simply the y value, 12. In the figure the highlighted curve passes through that point. Notice that the curves with positive and negative C lie on opposite sides of the horizontal axis and never cross it; the zero solution, the axis itself, separates them.
Discrimination
Sort into buckets
Sort each function by whether it is a whole family or one member.
Sort the six before checking. The test is simple: does the formula still contain a constant you are free to choose? If so, it describes infinitely many functions and is a general solution. If every constant has been fixed, it is one particular function.
Four of these still have free constants, including the free-fall height with two of them, C sub one and C sub two. The other two, x squared minus 3 and twelve e to the x cubed, have all their constants pinned down; each came from choosing one member of its family.
Do not be misled by numbers. A particular solution can contain plenty of numbers, like minus 4.9 or 12, and still be completely determined. What matters is whether any letter remains that could be changed while the function stays a solution.
Counterexample
Discussion prompt
Someone claims: if two functions both solve y′ = 2x, they must be the same function. Give a counterexample, and say what extra information would make the claim true.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 353 — the remark after Table 4.1
Write your counterexample before revealing. Any two members of the family y equals x squared plus C will do, because their derivatives are identical.
The more important part of the question is the second half. The claim fails because a differential equation on its own does not carry enough information to determine a single function. Something extra must be supplied, and the most common extra information is the value of the function at a single starting point.
That combination, an equation plus starting information, is so important that it has a name, which the next part of the lesson introduces: an initial-value problem.
Section
Part 4
Concept
initial-value problem — A differential equation together with one or more initial values: the value of the unknown (and, if needed, of its derivatives) at one chosen input.
\[ y' = 2x, \qquad y(3) = 7 \]
The equation supplies the family; the initial value picks the member. The name comes from physics, where the input is usually time t and the condition is known at the start, t equal to 0.
\[ y = x^2 + C, \quad 7 = 9 + C \;\Longrightarrow\; y = x^2 - 2 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 355 — Initial-value problems
An initial-value problem bundles together the two stages you have been using all along: a differential equation, which gives a family of solutions, and an initial value, which picks one member. The phrase initial value refers to an ordered pair, an input and the value the function must take there.
In the example the equation is y prime equals two x, whose general solution you already know, and the initial value says that at x equal to 3 the function equals 7. Substituting gives seven equals nine plus C, so C is minus 2.
The word initial comes from physics. The independent variable is often time t, and the information you usually have is the state of the system at the start, time zero. That is why so many initial values in this chapter are given at t equal to 0.
Concept
An equation of order n has n constants in its general solution, so it needs n initial values, all given at the same input.
\[ y'' - 3y' + 2y = 4e^{x} \]
\[ y(0) = 2, \qquad y'(0) = -1 \]
| order | constants | initial values needed |
|---|---|---|
| 1 | 1 | y(t₀) |
| 2 | 2 | y(t₀) and y′(t₀) |
| n | n | y, y′, …, y⁽ⁿ⁻¹⁾ at t₀ |
Too few and a whole family is left over. Too many and the extra condition usually contradicts the others, leaving no solution at all.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 355 — the general rule for initial values
The number of initial values you need matches the order, because each initial value can fix one constant, and the order tells you how many constants there are. A first-order equation needs one; a second-order equation needs two.
For higher orders, the extra conditions are on the derivatives, and they are all given at the same input. In the second-order example from the textbook, the conditions are the value of y at zero and the value of y prime at zero. Giving the value of y at two different points is a different kind of problem, called a boundary-value problem, which behaves quite differently.
The final sentence is worth thinking through. With too few conditions, the constants cannot all be determined, and a family of solutions remains. With too many, the extra conditions typically contradict each other. For instance, y prime equals two x with the value 1 at zero and the value 5 at one has no solution, because the first condition forces x squared plus 1, which equals 2 at one, not 5.
Notation
Annotate
On: \( y' + 2y = 3e^{t}, \qquad y(0) = 3 \)
Step through the annotations in order. The problem has two parts written side by side, the differential equation and the initial value, and a solution must satisfy both.
The equation is first order, so its general solution has one constant, and the textbook tells you that it is two e to the minus two t plus C times e to the t. You are not expected to derive that yet; Section 4.5 shows how. What you can do already is check it and use it.
The last annotation is the one that costs marks. It is easy to verify that a function satisfies the differential equation and forget the initial value, or the other way round. A complete check has two lines, one for each part.
Worked example
Verify that the function solves the initial-value problem.
\[ y = 2e^{-2t} + e^{t}; \qquad y' + 2y = 3e^{t}, \quad y(0) = 3 \]
Differentiate
Why: The first exponential brings down minus 2.
\[ y' = -4e^{-2t} + e^{t} \]
Substitute into the left side
Why: Add twice the candidate.
\[ y' + 2y = \left(-4e^{-2t} + e^{t}\right) + 2\left(2e^{-2t} + e^{t}\right) \]
Distribute
Why: Remove the parentheses.
\[ = -4e^{-2t} + e^{t} + 4e^{-2t} + 2e^{t} \]
Combine
Why: The e to the minus 2t terms cancel.
\[ = 3e^{t} \;\checkmark \]
Check the initial value too
Why: Both exponentials equal 1 at t equal to 0.
\[ y(0) = 2e^{0} + e^{0} = 2 + 1 = 3 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 356 — Example 4.4
This is Example 4.1 with an extra requirement. First comes the familiar verification: differentiate, substitute, simplify. The derivative of two e to the minus two t is minus four e to the minus two t, and when you add twice the candidate, those exponential terms cancel, leaving three e to the t.
Then the second part, which is quick but essential: evaluate the candidate at t equal to 0. Both exponentials equal 1 there, so the value is two plus one, which is 3. The initial value is satisfied.
Only when both checks succeed can you say that the function solves the initial-value problem. A function that passes only the first is a solution of the equation, but it belongs to a different initial-value problem.
Picture it
Figure (svg): Curves y equals 2 e to the minus 2t plus C e to the t for C equal to minus 1, minus 0.5, 0, 0.5 and 1.5, with the curve C equal to 1 highlighted and passing through the point (0, 3).
The family is two e to the minus 2t plus C e to the t. Every curve drawn solves the differential equation; the initial value y of 0 equal to 3 is satisfied by exactly one of them.
\[ y(0) = 2 + C = 3 \;\Longrightarrow\; C = 1 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 356-357 — Figure 4.3
This is the picture behind Example 4.4. Each curve in the figure is a member of the family two e to the minus two t plus C e to the t, for a different value of C. Every one of them satisfies the differential equation.
At t equal to 0 each curve passes through the height two plus C, so the curves cross the vertical axis at different heights. Exactly one of them passes through 3, the highlighted one, with C equal to 1. That is the solution to the initial-value problem.
Notice how differently the curves behave for large t. With positive C they shoot upward; with negative C they plunge down; with C equal to zero they decay toward zero. A small change in the starting value changes the long-term behaviour completely, which is worth remembering whenever a model's starting value is only approximately known.
Tweak it
Parameter explorer
The curve is y = 2e^(−2t) + Ce^t. Slide C and watch where the curve crosses t = 0. Which C gives y(0) = 3, and what does C do to the curve for large t?
\[ y = 2e^{-2t} + {C}e^{t} \]
Start with C at zero. The curve is then just two e to the minus two t, which starts at height 2 and decays toward zero. Now increase C slowly and watch the point where the curve crosses the vertical axis rise.
At C equal to 1 the curve crosses at height 3: that is the particular solution from Example 4.4. Every other setting of C gives a different curve that still satisfies the differential equation, but starts at the wrong height.
Now push C negative and watch the right-hand end of the curve. The term C e to the t grows so quickly that it eventually dominates, and the curve heads downward without bound. The sign of C decides the long-term fate of the solution, and the initial value decides the sign of C.
Worked example
\[ y = 3e^{2t} + 4\sin t; \qquad y' - 2y = 4\cos t - 8\sin t, \quad y(0) = 3 \]
Differentiate
Why: Chain rule on the exponential; sine differentiates to cosine.
\[ y' = 6e^{2t} + 4\cos t \]
Compute minus twice y
Why: Distribute the minus 2.
\[ -2y = -6e^{2t} - 8\sin t \]
Add
Why: The exponentials cancel.
\[ y' - 2y = 4\cos t - 8\sin t \;\checkmark \]
Check the initial value
Why: The sine of 0 is 0.
\[ y(0) = 3e^{0} + 4\sin 0 = 3 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 356 — Checkpoint 4.4
Try this yourself before looking at the steps. It has the same two-part structure as Example 4.4, with a trigonometric term alongside the exponential.
The derivative is six e to the two t plus four cos t. Subtracting twice the candidate cancels the exponentials and leaves four cos t minus eight sin t, exactly the right side. The sine from the candidate and the cosine from its derivative both survive, which is why the right side needs both.
At t equal to 0 the sine is zero and the exponential is one, so the value is 3, matching the initial condition. Both parts check, so the function solves the initial-value problem.
Step zero
\[ y' = 3e^{x} + x^2 - 4, \qquad y(0) = 5 \]
Discussion prompt
Before any integrating: what is the order, how many constants will the general solution have, is one condition enough, and why can you simply integrate both sides here?
Write down your answers to all four questions before revealing. This is the thirty seconds of thinking that stops you starting on the wrong method.
The equation is first order, so the general solution will have one constant, and the single initial value is exactly enough to determine it. The key structural fact is that y prime stands alone on the left and the right side is a function of x only, with no y in it. That means you can integrate both sides directly with respect to x.
Contrast this with Example 4.4, where y appeared next to its derivative. There, integrating the right side would not help, because the left side is not just a derivative. Recognising which situation you are in is the first step of any differential equation problem.
Worked example
\[ y' = 3e^{x} + x^2 - 4, \qquad y(0) = 5 \]
Antidifferentiate both sides
Why: Each side brings its own constant.
\[ y + C_1 = 3e^{x} + \tfrac13x^3 - 4x + C_2 \]
Merge the constants
Why: C sub 2 minus C sub 1 is just one constant; call it C.
\[ y = 3e^{x} + \tfrac13x^3 - 4x + C \]
Substitute the initial value
Why: x equal to 0, y equal to 5.
\[ 5 = 3e^{0} + \tfrac13(0)^3 - 4(0) + C \]
Simplify and solve
Why: e to the 0 is 1.
\[ 5 = 3 + C \;\Longrightarrow\; C = 2 \]
Write the solution
Why: Put C back.
\[ y = 3e^{x} + \tfrac13x^3 - 4x + 2 \]
Figure (svg): Curves y equals 3 e to the x plus x cubed over 3 minus 4x plus C for C equal to minus 2, 0, 4 and 6, with the member C equal to 2 highlighted through the point (0, 5).
Check the equation and the value
Why: Differentiate back; then evaluate at 0.
\[ y' = 3e^{x} + x^2 - 4 \;\checkmark, \quad y(0) = 3 + 2 = 5 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 357-358 — Example 4.5
Now you solve rather than verify. Integrating both sides with respect to x gives y on the left and an antiderivative on the right, and each integration produces its own constant. The textbook writes them as C sub one and C sub two to be honest about that.
Moving C sub one to the right gives C sub two minus C sub one, and a difference of two arbitrary constants is just one arbitrary constant. So the general solution has one constant, as the order promised. In practice, you can write a single constant on the right from the start.
Substituting the initial value is where care matters: the exponential term is 3 at x equal to 0, not zero, so C is 2, not 5. The figure shows the family with the member through height 5 highlighted, and the final check differentiates back to the original equation and confirms the starting value.
Worked example
\[ y' = x^2 - 4x + 3 - 6e^{x}, \qquad y(0) = 8 \]
Antidifferentiate
Why: Power rule on each power of x; the exponential is its own antiderivative.
\[ y = \tfrac13x^3 - 2x^2 + 3x - 6e^{x} + C \]
Substitute the initial value
Why: Every power of x vanishes at 0.
\[ 8 = 0 - 0 + 0 - 6e^{0} + C \]
Solve for C
Why: Minus 6 plus C equals 8.
\[ C = 14 \]
Write the solution
Why: Put C back.
\[ y = \tfrac13x^3 - 2x^2 + 3x - 6e^{x} + 14 \]
Check the equation and the value
Why: Differentiate back, then evaluate at 0.
\[ y' = x^2 - 4x + 3 - 6e^{x} \;\checkmark, \quad y(0) = -6 + 14 = 8 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 358 — Checkpoint 4.5
Do this one before revealing the steps; it is the same method as Example 4.5. The right side is a sum of powers of x and an exponential, so antidifferentiate term by term.
Minus four x becomes minus two x squared and three becomes three x. The exponential term, minus six e to the x, is its own antiderivative. At x equal to 0 all the powers vanish but the exponential does not: it contributes minus 6. So eight equals minus six plus C, and C is 14.
The check differentiates the answer back to the original right side and confirms that the value at zero is 8. If you got C equal to 8, the next slide is for you.
Trap
Checkpoint 4.5 done too fast:
\[ y(0) = 8 \;\Longrightarrow\; C = 8 \]
\[ y = \tfrac13x^3 - 2x^2 + 3x - 6e^{x} + 8 \]
Wrong. Evaluate it at 0.
The other terms need not vanish at 0: here the exponential contributes minus 6.
\[ y(0) = -6 + 8 = 2 \ne 8 \]
Always substitute and solve for C. It equals the starting value only when the rest of the formula is zero there.
\[ -6 + C = 8 \;\Longrightarrow\; C = 14 \]
This shortcut is tempting because it often works. In Example 4.3's family, and in the baseball problems coming up, every term except C is zero at the starting point, so C does equal the starting value.
But it is only a coincidence of those examples. Here the exponential term is minus 6 at zero, so setting C equal to 8 produces a function whose starting value is 2. The correct move is always to substitute the initial value into the general solution and solve for C, which gives 14.
The quickest protection is the second half of the check. Evaluate your final answer at the starting point. If it does not return the initial value, the constant is wrong.
Section
Part 5
Concept
A ball moves under gravity alone. Newton's second law says force equals mass times acceleration, and acceleration is the derivative of velocity.
\[ F = ma = m\,v'(t) \]
Gravity pulls down with force minus m times g, with g about 9.8 metres per second squared. Set the two forces equal.
\[ m\,v'(t) = -mg \]
\[ v'(t) = -g \]
The mass cancels, so every ball obeys the same equation. What the law gives you is a rate of change; the velocity itself is what you solve for.
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 358-359 — Figures 4.4 and 4.5
This slide shows where differential equations come from. Newton's second law relates force to acceleration, and acceleration is the rate of change of velocity. So as soon as you know the forces, you have an equation for the derivative of velocity.
For a ball moving under gravity alone, ignoring air resistance, the only force is its weight, mass times g, pointing down. The textbook takes up as positive, so the force is minus m g. Setting the two expressions for force equal gives m times v prime equals minus m g, and dividing by m leaves v prime equals minus g.
The mass has cancelled. Every object, heavy or light, obeys the same equation when gravity is the only force, which is Galileo's famous observation. The law itself tells you only how fast the velocity changes; to find the velocity, you still need an initial value.
Worked example
A ball is thrown upward at 10 metres per second from 3 metres up. Find its velocity, and its velocity after 2 seconds.
\[ v'(t) = -9.8, \qquad v(0) = 10 \]
Antidifferentiate
Why: A constant integrates to a line.
\[ v(t) = -9.8t + C \]
Use the initial velocity
Why: At t equal to 0 the line's height is C.
\[ 10 = -9.8(0) + C \;\Longrightarrow\; C = 10 \]
Write the velocity
Why: Put C back.
\[ v(t) = -9.8t + 10 \]
Evaluate at t equal to 2
Why: Multiply, then add.
\[ v(2) = -19.6 + 10 = -9.6 \text{ m/s} \]
Figure (svg): The line v equals minus 9.8 t plus 10 from t equals 0 to 2.3. The region between the line and the t-axis is green where v is positive, up to t about 1.02, and red where v is negative after that. The point (2, minus 9.6) is marked.
Check and interpret
Why: The derivative is minus 9.8 and the start is 10, so both parts hold; the negative sign means that by 2 seconds the ball is falling at 9.6 metres per second.
\[ v' = -9.8 \;\checkmark, \quad v(0) = 10 \;\checkmark, \quad v(2) < 0 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 359-360 — Example 4.6
Translate the words first. Thrown upward at 10 metres per second means the initial velocity is plus 10, and gravity gives v prime equal to minus 9.8. The starting height, 3 metres, and the mass play no part in this velocity problem; they are there for the next example.
Antidifferentiating a constant gives a straight line, minus 9.8 t plus C, and at t equal to 0 only C survives, so C is 10. Evaluating at 2 seconds gives minus 9.6 metres per second.
The figure shows why the sign is the whole story. The velocity starts at 10 and drops by 9.8 every second. It is positive, shaded green, while the ball rises, crosses zero at the top of the flight a little after one second, and is negative, shaded red, while it falls. At 2 seconds the ball is on its way down at 9.6 metres per second.
Concept
Velocity is the derivative of height, so once v is known, height s solves another first-order problem, with the starting height as its condition.
\[ s'(t) = v(t), \qquad s(0) = s_0 \]
Chaining the two problems is the same as solving one second-order problem with two conditions: the starting velocity and the starting height.
\[ s'' = -g, \qquad s'(0) = v_0, \quad s(0) = s_0 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 360 — the position initial-value problem
Once you know the velocity, height is the next question. Velocity is the derivative of height, so height satisfies its own first-order initial-value problem, with the starting height as its initial value.
Look at what happens when you chain the two problems. The first used the starting velocity; the second uses the starting height. Together they solve a single second-order problem, the second derivative of height equals minus g, with two initial values given at the same moment.
That is the general rule from the previous part in action: second order, two constants, two conditions. You will solve it both ways, first in two stages as the textbook does, then in one go.
Worked example
The same ball: find its height, and its height after 2 seconds.
\[ s'(t) = -9.8t + 10, \qquad s(0) = 3 \]
Antidifferentiate
Why: Power rule on each term.
\[ s(t) = -4.9t^2 + 10t + C \]
Use the initial height
Why: At t equal to 0 only C survives.
\[ 3 = -4.9(0)^2 + 10(0) + C \;\Longrightarrow\; C = 3 \]
Write the height
Why: Put C back.
\[ s(t) = -4.9t^2 + 10t + 3 \]
Evaluate at t equal to 2
Why: Two squared is 4.
\[ s(2) = -4.9(4) + 20 + 3 = 3.4 \text{ m} \]
Figure (svg): The height s equals minus 4.9 t squared plus 10 t plus 3 from t equals 0 until the ball lands near t equals 2.31. Vertical arrows at t equals 0, 0.5, 1.5 and 2 show the velocity: pointing up and long at the start, down afterwards. The top is near (1.02, 8.10) and the point (2, 3.4) is marked.
Check by differentiating back
Why: The derivative returns the velocity, and the start is 3 metres; the mass, 0.15 kilograms, was never used.
\[ s'(t) = -9.8t + 10 = v(t) \;\checkmark, \quad s(0) = 3 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 360-361 — Example 4.7
The velocity from Example 4.6 is now the right side of the equation, and the initial value is the starting height of 3 metres. Antidifferentiating gives minus 4.9 t squared plus 10 t plus C, and at t equal to 0 only C survives, so C is 3.
At 2 seconds the height is 3.4 metres: the ball has gone up to its peak and come most of the way back down. The figure shows the whole flight, with arrows for the velocity at several times. The long upward arrow at the start shrinks to nothing at the top, around 8.1 metres, and then grows downward.
The check is twofold again: differentiate back to recover the velocity, and confirm the starting height. And notice what never appeared anywhere in the calculation: the mass of 0.15 kilograms. It cancelled at the very first step, when the law was turned into an equation.
Worked example
A rock falls from rest from a height of 100 metres. Find its velocity; then, going one step further, when and how fast it lands.
\[ v'(t) = -9.8, \qquad v(0) = 0 \]
Antidifferentiate and use v of 0 equal to 0
Why: From rest means the constant is 0.
\[ v(t) = -9.8t + C, \quad 0 = C \;\Longrightarrow\; v(t) = -9.8t \]
Solve for the height
Why: Antidifferentiate again; the starting height is 100.
\[ s(t) = -4.9t^2 + C_2, \quad s(0) = 100 \;\Longrightarrow\; s(t) = 100 - 4.9t^2 \]
Set the height to zero
Why: Landing means s equals 0.
\[ 4.9t^2 = 100 \;\Longrightarrow\; t = \sqrt{100/4.9} \approx 4.518 \text{ s} \]
Find the impact velocity
Why: Evaluate v at the landing time.
\[ v(4.518) \approx -44.27 \text{ m/s} \]
Figure (svg): The height s equals 100 minus 4.9 t squared, starting flat at 100 metres at t equals 0 and curving down to hit the ground at t about 4.52 seconds, where a dot is drawn.
Check with energy
Why: A fall from height h reaches speed root 2gh, the same number without any calculus.
\[ \sqrt{2(9.8)(100)} = \sqrt{1960} \approx 44.27 \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 360 — Checkpoint 4.6
From rest means the initial velocity is zero, which makes the constant zero and the velocity simply minus 9.8 t. That is the checkpoint's answer; the rest of the slide goes one step further, as a physics problem would.
Antidifferentiating again with a starting height of 100 gives the height function, and setting it to zero finds the landing time, about 4.52 seconds. The velocity at that moment is about minus 44.27 metres per second.
The check comes from a completely different direction: conservation of energy says an object dropped from height h hits the ground at the square root of two g h. That formula gives the same 44.27, without any differential equation. Two independent routes to the same number is about as strong a check as you can get.
Worked example
Solve the ball's motion in one go, straight from acceleration, with both conditions.
\[ s'' = -9.8, \qquad s'(0) = 10, \quad s(0) = 3 \]
Antidifferentiate once
Why: First constant.
\[ s' = -9.8t + C_1 \]
Apply the velocity condition
Why: It belongs to s prime, so it fixes C sub 1.
\[ 10 = C_1 \;\Longrightarrow\; s' = -9.8t + 10 \]
Antidifferentiate again
Why: Second constant.
\[ s = -4.9t^2 + 10t + C_2 \]
Apply the height condition
Why: It belongs to s, so it fixes C sub 2.
\[ 3 = C_2 \;\Longrightarrow\; s = -4.9t^2 + 10t + 3 \]
Check against Example 4.7
Why: The same function, found in one problem; two conditions for two constants.
\[ s'' = -9.8, \quad s'(0) = 10, \quad s(0) = 3 \;\checkmark \]
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, pp. 359-361 — Examples 4.6 and 4.7 combined
Here the baseball is solved directly from the acceleration, as a second-order initial-value problem with two conditions. Antidifferentiate once to get the velocity, with constant C sub one; antidifferentiate again to get the height, with constant C sub two.
The important thing is which condition fixes which constant. C sub one appears in the velocity formula, and it equals the velocity at time zero, so the velocity condition fixes it. C sub two appears when you integrate again, and it equals the height at time zero, so the height condition fixes it.
The result is the same height function as Example 4.7, as it must be. Two routes, two stages or one problem, and each uses exactly two conditions, one per constant.
Error analysis
Annotate
On: \( s' = -9.8t + C_1, \quad s(0) = 3 \;\Longrightarrow\; C_1 = 3 \)
Read the line and try to find the problem before revealing the annotations. The integration is correct; the mistake is in which condition was used.
The constant C sub one sits in the formula for velocity, and at time zero it equals the starting velocity. The given condition, s of zero equals 3, is about height. Using it to fix C sub one silently changes the problem: the ball is now thrown at 3 metres per second instead of 10.
The cost is not small. The peak of the flight drops from about 8.1 metres to about 3.46 metres, a completely different throw. The protection is to write each condition next to the function it describes and use it only after you have the formula for that function.
Intuition
Nature is usually easier to describe by how things change than by what they are. Newton's law of cooling says a hot drink cools at a rate proportional to its difference from room temperature.
\[ T'(t) = -0.1\left(T - 20\right), \qquad T(0) = 90 \]
\[ T = 20 + 70e^{-0.1t}: \quad T' = -7e^{-0.1t} = -0.1\left(70e^{-0.1t}\right) \;\checkmark \]
Figure (svg): Four curves T equals 20 plus (T0 minus 20) e to the minus 0.1 t for starting temperatures 90, 60, 20 and 5 degrees, over 40 minutes. The hot ones fall and the cold one rises, all approaching the dashed line T equals 20.
You can check the candidate without knowing how to solve the equation. Solving it is Section 4.3's job.
Physical laws are usually statements about how things change, because change is what can be observed and measured locally. Newton's law of cooling says the temperature of a hot drink changes at a rate proportional to how far it is above room temperature. Written down, that is a differential equation.
You cannot solve it yet with the methods of this section, because T appears on the right side. But you can check a proposed solution, and the second line does exactly that. The derivative of 20 plus seventy e to the minus 0.1 t is minus seven e to the minus 0.1 t, which equals minus 0.1 times the difference from room temperature. It starts at 90, so it solves the initial-value problem.
The figure shows the same law with four different starting temperatures. Hot drinks cool toward 20, a cold one warms toward 20, and one starting at 20 stays there. After ten minutes, the drink that started at 90 is at about 45.75 degrees. One law, many solutions, and the initial value picks the one you need.
Estimation
\[ v' = -g, \quad v(0) = 25 \text{ m/s}: \quad g_{\text{Earth}} = 9.8, \; g_{\text{Mars}} = 3.711 \]
Predict first
A ball is thrown up from the ground at 25 m/s. On Mars gravity is 3.711 instead of 9.8. How many times longer is it in the air?
Correct: About 2.6 times
Why: Height is 25t − g t²/2, which returns to zero at t = 50/g. The time aloft is inversely proportional to g, so the ratio is 9.8/3.711 ≈ 2.64: 5.10 s on Earth against 13.47 s on Mars. The mass cancels, but g does not.
\[ t_{\text{air}} = \frac{2v_0}{g}: \quad \frac{50}{9.8} \approx 5.10, \quad \frac{50}{3.711} \approx 13.47 \]
Pick an answer before revealing. The equation is the same on both planets, v prime equals minus g; only the number g changes.
With a throw from the ground, height is 25 t minus half of g t squared, which returns to zero at t equal to 50 over g. So the time in the air is inversely proportional to g. On Earth that is about 5.10 seconds; on Mars, with g equal to 3.711, it is about 13.47 seconds, a ratio of about 2.64.
The tempting wrong answer is that it is the same time because mass cancels. Mass does cancel, but gravity does not, and gravity is what differs between the planets. A differential equation with a parameter describes a whole range of situations at once.
Real world
Figure (svg): Two parabolas for a ball thrown up at 25 metres per second. On Earth it peaks at about 31.9 metres and lands after 5.10 seconds; on Mars it peaks at about 84.2 metres and lands after 13.47 seconds.
Discussion prompt
Exercises 56 and 57: throw a 1 kg ball straight up at 25 m/s on Earth and on Mars. How much longer is it in the air on Mars, and how much higher does it go?
OpenStax Calculus Volume 2, §4.1 Basics of Differential Equations §4.1, p. 363 — Exercises 56 and 57
Attempt both parts before revealing. The equation for height is the same for both planets with a different g, so you can solve it once with g as a letter and then substitute each value.
The ball is at the top when the velocity is zero, at t equal to 25 over g, and the height there is 25 squared over twice g. On Earth that is about 31.89 metres, on Mars about 84.21 metres. So the Mars throw goes about 52.32 metres higher and stays up about 8.37 seconds longer.
The figure shows both flights on the same axes. They start with the same slope, because the initial velocity is the same, and separate as the different accelerations take effect. Notice that both the time and the height scale by the same factor, about 2.64, because both are inversely proportional to g.
Section
Part 6
Pattern
Figure (svg): A flow diagram with two lanes. Top lane, a candidate is given: differentiate it, substitute into both sides, and it is a solution only if the sides agree for every x. Bottom lane, only the equation is given: antidifferentiate to get a general solution with one constant per order, then use one condition per constant to get the particular solution, then check.
This is the whole lesson as a procedure. The top lane of the diagram is for when someone hands you a candidate: differentiate, substitute, simplify, compare as functions. The bottom lane is for when you only have the equation.
In the bottom lane, the order comes first, because it tells you how many constants to expect and how many conditions you will need. Then antidifferentiate, merging constants as you go, apply each condition to the function it describes, and solve for the constants.
Both lanes end in the same place: substitute back into the equation, and evaluate at the starting point. Solving without checking is the most common way to lose marks on these problems, and the check is mechanical.
Comparison
Comparison matrix
| equation | order | initial values needed | general solution |
|---|---|---|---|
| y′ = 2x | 1 | 1 | x² + C |
| y′ = 3x²y | 1 | 1 | Ce^(x³) |
| s″ = −9.8 | 2 | 2 | −4.9t² + C₁t + C₂ |
| y‴ = 0 | 3 | 3 | C₁x² + C₂x + C₃ |
Fill in every blank before checking. Each row asks the same three questions: what is the order, how many initial values does that require, and what does the general solution look like.
The first two rows are first order, so each needs one initial value, even though one family is shifted by adding C and the other stretched by multiplying by C. The free-fall row is second order, needs two conditions, and its general solution has the two constants you found by antidifferentiating twice.
The last row, the third derivative equal to zero, extends the pattern. Antidifferentiating three times gives a quadratic with three constants, so three initial values are needed, typically the value, the slope and the second derivative at one point.
Ranking
Put in order
Order the steps for solving y′ = f(x), y(x₀) = y₀.
Why: Counting comes first so you know how many conditions to expect. The constants must be merged before substituting, or you will solve for two unknowns from one equation. The check always comes last and tests both parts.
Drag the steps into order before checking. Most of the order is forced, but two places deserve thought.
Merging the integration constants must come before substituting the initial value. If you keep C sub one and C sub two separate, one initial value gives one equation in two unknowns, which looks unsolvable even though only their difference matters.
The check must come last and must include both parts: differentiate the answer back to the original equation, and evaluate it at the starting point. Only a function that passes both is the solution.
Check
Check your understanding
Which function is a solution of y′ = 3y?
Answer: B
Why: Differentiating 5e^(3x) gives 15e^(3x), which is 3 times 5e^(3x). The left side equals the right side for every x, so it is a solution.
Differentiate each candidate and compare with three times the candidate, as whole expressions. Only one of them survives.
The tempting wrong answers each fail in a different way. Adding a constant to the exponential breaks it, because the derivative loses the constant but three times the candidate does not. Putting the 3 in front instead of in the exponent gives the wrong growth rate. And x cubed agrees with the equation only at two points, which is not agreement at all.
Check
Check your understanding
What is the order of (y″)³ + x y′ = x⁵?
Answer: A
Why: The highest derivative present is the second derivative y″. Cubing it is a power, not a further differentiation, so the order is 2.
Find the highest derivative that appears, and ignore everything else. The second derivative is there, and nothing higher, so the order is two.
Each wrong answer comes from reading a different number in the equation as the order: the power on the derivative, the power of x on the right side, or a product of the two. None of these counts how many times y has been differentiated.
Check
Check your understanding
Solve y′ = 2x + 1 with y(0) = 4.
Answer: A
Why: Antidifferentiating gives x² + x + C, and at x = 0 that is C, so C = 4. Check: the derivative is 2x + 1 and y(0) = 4.
Antidifferentiate, then substitute the initial value. The antiderivative of two x plus one is x squared plus x plus C, and at x equal to 0 that is just C, so C is 4.
Each wrong answer skips or misplaces one of the steps. One forgets the initial value entirely; one never antidifferentiates; one puts the constant in the wrong place. The two-part check, differentiate back and evaluate at zero, catches every one of them.
Explain it to yourself
Discussion prompt
Using the thrown ball, explain in two or three sentences why a second-order equation needs two initial values, and what would go wrong with only the starting height.
Write your two or three sentences before revealing. A good answer names the two constants and says what each one means physically.
Undoing the second derivative takes two antidifferentiations, and each brings its own constant. For the ball, one constant turns out to be the launch speed and the other the starting height. Knowing only the starting height fixes one of them and leaves the other free, so every possible launch speed still gives a solution.
That is exactly the fan of curves you saw earlier, all starting from the same point. The second condition, the starting velocity, is what collapses the fan to a single throw.
Exit ticket
\[ y' = 6x^2 - 2, \qquad y(1) = 3 \]
Discussion prompt
Give the order, solve the initial-value problem, and check your answer in both parts.
This brings the lesson together in one problem. Identify the order, antidifferentiate, use the initial value, and check both parts.
The equation is first order with the derivative given outright, so integrate: two x cubed minus two x plus C. Now the initial value is at x equal to 1, not 0, so the other terms do not vanish; here they happen to cancel, two minus two, which leaves C equal to 3.
The check is the two lines you have written many times now: the derivative returns six x squared minus two, and the value at 1 is 3. If both hold, you are done.
Recap
| idea | what it means | how you use it |
|---|---|---|
| differential equation | an equation for an unknown function and its derivatives | read it as a rule the derivatives must obey |
| solution | a function that satisfies it for every x in an interval | verify by differentiating and substituting |
| order | the highest derivative present | predicts the number of constants |
| general / particular | the whole family / one member | one condition fixes one constant |
| initial-value problem | equation + as many initial values as the order | solve, then check both parts |
\[ \text{order } n \;\Longrightarrow\; n \text{ constants} \;\Longrightarrow\; n \text{ initial values} \]
Next, Section 4.2 draws the slopes a first-order equation prescribes, and follows them numerically when no formula can be found.
Stewart, Calculus: Early Transcendentals 8e, §9.1 Modeling with Differential Equations §9.1, pp. 586-590 — the same material in Stewart
A differential equation is a rule the derivatives of an unknown function must obey, and a solution is a function that obeys it for every input on an interval. You verify a candidate by differentiating, substituting and comparing expressions, never by testing one point.
The order is the highest derivative present. It tells you how many constants the general solution has, and therefore how many initial values an initial-value problem needs. The general solution is the whole family; the initial values pick the particular member, and the final check tests both the equation and the initial values.
So far you can solve only equations where the derivative is given outright. The next sections widen that: Section 4.2 draws slope fields and approximates solutions numerically, and Sections 4.3 to 4.5 develop exact methods for the most important types, including the cooling law.
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