4.5 First-order Linear Equations

1. First-order Linear Equations

Title

Calculus II · Section 4.5

One multiplication turns the left side into a single derivative

2. What this lesson gives you

Objectives

Section 4.3 solved equations you could separate. Many of the most useful equations in science cannot be separated, because the unknown appears added to something else rather than multiplied. This lesson solves a whole class of them with one idea.

• Recognise a linear first-order equation and write it in standard form
• Derive the integrating factor from the product rule, and use it to solve
• Solve initial-value problems, and split a solution into transient and steady state
• Model and solve falling with air resistance, electrical circuits and mixing tanks

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 408-421 — learning objectives 4.5.1 to 4.5.3

Separation of variables is powerful, but it only works when the right side of the equation is a product of a piece in x and a piece in y. The moment the unknown is added to something, as in y prime equals x plus y, separation stops working, and a huge number of real models have exactly that shape.

This lesson handles all of those equations at once, provided the unknown appears only to the first power. The idea is a single multiplication, chosen so that the left side of the equation becomes the derivative of one product. After that, one integration finishes the job.

The last two parts put the method to work on a ball with air resistance, on electrical circuits, and on a mixing tank. In every one of them you will see the same structure: a part of the solution that fades away and a part that stays.

3. Before anything new: try to separate it

Warm-up

\[ \frac{dy}{dx} = x + y \]

Give this a real attempt before revealing. Try dividing by y, or subtracting y, and watch what happens: any move that gets the y away from the x on the right leaves a y somewhere on the wrong side.

The underlying reason is structural. Separation needs the right side to factor into a function of x times a function of y, so that you can divide by the y factor. A sum of x and y has no such factorisation, and no algebraic trick creates one.

Keep this equation in mind. You will solve it in Part 3 with the integrating factor, and it will take five short lines. The answer contains an exponential that separation could never have produced.

4. Linear equations and standard form

Section

Part 1

5. Adding air resistance to a thrown ball

Concept

Earlier the velocity of a ball thrown upward satisfied one line, because gravity was the only force. Now let air push back with a force proportional to the velocity.

\[ \frac{dv}{dt} = -32 \quad\longrightarrow\quad m\frac{dv}{dt} = -kv - mg, \quad v(0) = v_0 \]

The new equation has v on both sides, added to a constant, so it will not separate as neatly. It is the model problem of this section: equation 4.13.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 408 — equation 4.13 and Figure 4.24

Start with the physics, because it explains where these equations come from. Newton's second law says mass times acceleration equals the total force. Without air, the only force is gravity, and the equation has no v on the right, so you just integrate.

Air resistance changes that. For small, slow objects the drag force is roughly proportional to the speed, with some constant k, and it always points against the motion. Look at the two panels: while the ball rises, drag points down along with gravity; while it falls, drag points up, against gravity.

Writing the drag as minus k times v handles both panels with one formula, because v itself changes sign when the ball turns over. The result has v on the right-hand side added to a constant. That is a first-order linear equation, and by the end of this lesson you will solve it completely.

6. What makes a first-order equation linear

Concept

Linear first-order equation — An equation that can be written as a coefficient function times y prime, plus a coefficient function times y, equals a function of x. The coefficients may be anything in x; y and its derivative appear only to the first power, never multiplied together or inside another function.

\[ a(x)\,y' + b(x)\,y = c(x) \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 409 — equation 4.14 and the examples

The word linear here is about the unknown function, not about x. The coefficients a, b and c can be as complicated as you like in x: sines, logarithms, polynomials. What matters is that y and y prime each appear to the first power, on their own, multiplied only by functions of x.

Read down the table. On the left, every equation has y prime times something in x plus y times something in x equal to something in x. On the right, each equation breaks the rule in a specific way: a power of y prime, a cube of y, or y trapped inside a sine.

When you test an equation, look at every place y or y prime appears and ask whether it is to the first power and not inside any other function. One failure anywhere and the equation is nonlinear.

7. Reading equation 4.14 piece by piece

Notation

Step through the annotations and match each one to a part of the equation. The general form has three slots, each filled by a function of x alone: one multiplying y prime, one multiplying y, and one standing by itself on the right.

The right-hand side deserves a name because it has a physical meaning in almost every model. In the ball equation it is the weight, in a circuit it is the battery or generator, in a mixing tank it is the salt pouring in. It drives the system, which is why it is called the forcing term.

Notice also what the form does not allow: no product of y with y prime, no y inside an exponential, no square. Those are exactly the things that make an equation hard, and excluding them is what makes a general method possible.

8. Linear and separable are different questions

Intuition

Separable asks whether the right side factors into a function of x times a function of y. Linear asks whether y enters only to the first power. Neither contains the other.

When an equation is both, either method works. The new method is needed exactly for the second row.

It is tempting to think of linear and separable as two levels of difficulty, one inside the other. They are not. They are two independent tests, and the table shows every combination.

The first row, y prime equals x times y, passes both tests, so you may use either method; separation is usually quicker. The second row is the warm-up equation, linear but not separable, and it is the reason this lesson exists. The third row is the logistic-type equation from Section 4.4, separable but not linear because of the y squared hidden in y times 2 minus y. The last row passes neither test, and needs methods beyond this course.

So when you meet a first-order equation, run both tests. Whichever one passes tells you which tool to pick up.

9. Linear or not?

Sorting

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 420 — Exercises 208 to 212

For each equation, find every appearance of the unknown and its derivative, and ask whether it is to the first power and multiplied only by something in the independent variable.

Three of these pass. The first has y multiplied by x squared, which is fine. The second has y multiplied by t. The fourth has no y at all on the right, which is the special case where the coefficient of y is zero; it is linear, and in fact you could just integrate it directly.

The other three fail in three different ways. A y squared is a power. An e to the y puts the unknown inside another function. And y times y prime multiplies the unknown by its own derivative, which the linear form never allows even though neither factor is squared.

10. Standard form: make the coefficient of y′ equal to 1

Concept

Divide the whole equation by the coefficient of y prime, then give the two new coefficients names.

\[ (3x^2 - 4)y' + (x-3)y = \sin x \]

\[ y' + \frac{x-3}{3x^2-4}\,y = \frac{\sin x}{3x^2-4} \]

In general, divide equation 4.14 by a(x):

\[ y' + \underbrace{\frac{b(x)}{a(x)}}_{p(x)}\,y = \underbrace{\frac{c(x)}{a(x)}}_{q(x)} \quad\Longrightarrow\quad y' + p(x)\,y = q(x) \]

Division by a(x) is only allowed where a(x) is not zero, so standard form already tells you which intervals a solution can live on.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 409 — equations 4.15 and 4.16

Standard form is the arrangement the whole method is built on: y prime alone with coefficient one, then p of x times y, then everything else on the right as q of x. Getting there is usually one division.

The worked line shows the book's example. Dividing every term by 3x squared minus 4 leaves y prime with coefficient one, and the new coefficients are fractions. That is completely normal: p and q are allowed to be messy functions.

The last point is easy to miss. You can only divide by a of x where it is not zero, so standard form quietly restricts where your solution is valid. In the example, 3x squared minus 4 is zero at two values of x, and a solution built from standard form lives on an interval that avoids them.

11. Example 4.15(a) and (c): rearrange, then divide

Worked example

Put each equation in standard form and identify p and q.

\[ \text{(a) } y' = 3x - 4y \qquad \text{(c) } y = 3y' - 4x^2 + 5 \]

\[ y' + 4y = 3x \;\Longrightarrow\; p(x) = 4, \; q(x) = 3x \]

\[ 3y' - y = 4x^2 - 5 \]

\[ y' - \frac13 y = \frac43 x^2 - \frac53 \]

\[ p(x) = -\frac13, \quad q(x) = \frac43 x^2 - \frac53 \]

\[ 5 = 3y' + 5 \Rightarrow y' = 0; \qquad y' = \tfrac13(5) - \tfrac53 = 0 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 410 — Example 4.15a and c

Both parts follow the same two moves: collect the y and y prime terms on the left with everything else on the right, then divide by whatever multiplies y prime.

In part (a), y prime already has coefficient one, so only the rearranging is needed. Adding 4y to both sides moves the y term left, and its coefficient becomes plus 4. In part (c), after rearranging you have 3 y prime, so you must divide every term by 3, including the right side. The coefficient of y becomes minus one third, and that minus sign belongs to p.

The check is worth adopting as a habit. Pick any point, work out the slope the original equation gives there, and then the slope the standard form gives. They must agree, because the two forms are the same equation. At x equal to 0 and y equal to 5 both give a slope of 0.

12. Example 4.15(b): clear the fraction first

Worked example

\[ \frac{3xy'}{4y-3} = 2 \qquad (x > 0) \]

\[ 3xy' = 2(4y - 3) \]

\[ 3xy' = 8y - 6 \]

\[ 3xy' - 8y = -6 \]

\[ y' - \frac{8}{3x}\,y = -\frac{6}{3x} = -\frac{2}{x} \]

\[ p(x) = -\frac{8}{3x}, \quad q(x) = -\frac{2}{x} \]

\[ y' = \frac{2(4-3)}{3} = \frac23, \qquad y' = \frac83 - 2 = \frac23 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 410 — Example 4.15b (the printed q(x) is corrected)

This one has the unknown in a denominator, so the first job is to clear it. Multiplying both sides by 4y minus 3 is safe as long as that expression is not zero, and it turns the equation into something with no fractions in y.

After expanding and moving the 8y across, you divide by 3x. That is allowed only because the problem says x is positive; at x equal to 0 the original equation would read 0 equals 2, which is impossible.

Watch the last division carefully. Minus 6 divided by 3x is minus 2 over x. The printed textbook gives minus 2 over 3x here, which is a misprint, and the check shows it: at x and y both equal to 1, the original equation gives a slope of two thirds, and so does the standard form with minus 2 over x. With the misprinted value you would get two, which is wrong.

13. Checkpoint 4.15: finish the standard form

Fill the middle

\[ \frac{(x+3)y'}{2x - 3y - 4} = 5 \]

Multiply out to get (x + 3) times y prime equal to 10x minus 15y minus 20, move the y term left, and divide by x + 3. Fill in the two numbers.

\[ y' + \frac{\square}{x+3}\,y = \frac{10x - \square}{x+3} \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 410 — Checkpoint 4.15

Work it through on paper before typing the two numbers. The only danger is losing track of a sign or a factor when you clear the denominator.

Multiplying both sides by 2x minus 3y minus 4 gives x plus 3 times y prime equal to 5 times that expression, which is 10x minus 15y minus 20. The y term is on the wrong side, so add 15y to both sides. Now the left side is x plus 3 times y prime plus 15y.

Dividing every term by x plus 3 gives p of x equal to 15 over x plus 3, and q of x equal to 10x minus 20 over x plus 3. Notice that the 15 came from the right-hand side with a minus sign and changed sign when it crossed over.

14. Trap: reading p off the wrong side

Trap

15. The integrating factor

Section

Part 2

16. The left side is almost a product rule

Concept

Multiply the standard form by a function mu of x, not yet chosen.

\[ \mu(x)y' + \mu(x)p(x)\,y = \mu(x)q(x) \]

Compare the left side with the product rule for mu times y:

\[ \frac{d}{dx}\left[\mu(x)\,y\right] = \mu(x)y' + \mu'(x)\,y \]

The first terms already agree. The second terms agree if mu is chosen so that its derivative is mu times p.

\[ \mu'(x) = \mu(x)\,p(x) \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 411 — equation 4.19 and the product rule

Here is the central idea of the lesson. Multiply the standard form by some function mu of x that you have not yet chosen. The left side becomes mu times y prime plus mu times p times y.

Now compare that with the product rule for the product mu times y. Its derivative is mu times y prime plus mu prime times y. The first terms already match exactly. The second terms match if mu prime equals mu times p.

So the plan is: choose mu so that its derivative is mu times p. Then the entire left side is the derivative of mu times y, a single derivative, and an equation of the form derivative of something equals a known function can be solved by integrating once. The whole method is this one recognition.

17. Solving for the factor: a separable equation

Concept

The condition on mu is itself a first-order equation, and a separable one, so Section 4.3 solves it.

\[ \frac{\mu'(x)}{\mu(x)} = p(x) \]

\[ \int \frac{\mu'(x)}{\mu(x)}\,dx = \int p(x)\,dx \]

\[ \ln|\mu(x)| = \int p(x)\,dx + C \]

\[ |\mu(x)| = e^{C}\,e^{\int p(x)\,dx} \]

\[ \mu(x) = C_2\,e^{\int p(x)\,dx} \]

Any nonzero constant works, so take the simplest one.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 411 — the derivation of the integrating factor

The condition that mu prime equals mu times p is itself a differential equation, and a pleasant one: it separates. Divide by mu and you have mu prime over mu equal to p of x.

The left side is the derivative of the natural log of the absolute value of mu, so integrating both sides gives the log of mu equal to the integral of p plus a constant. Exponentiating removes the log, and the constant turns into a multiplying constant.

The book keeps that constant, calling it C sub 2, to show that there is a whole family of factors that work. You only need one of them, so the natural choice is C sub 2 equal to one. The next slides show why that choice is completely harmless, and why the same freedom does not apply to the constant that appears later.

18. Reading the integrating factor

Notation

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 412 — step 2 of the problem-solving strategy

Go through the annotations one at a time. The formula is short, but each piece hides a way to go wrong.

The most important annotation is the second one: p means the coefficient of y after you have divided through by the coefficient of y prime. If the equation came to you as x times y prime plus 3y, then p is 3 over x, not 3. Reading p from the wrong form is the single most common way to get the wrong factor.

The last annotation is a useful shortcut. When the integral of p is a multiple of a logarithm, the exponential undoes it and the factor becomes a power of x. Three over x gives x cubed; one over x minus 2 gives x minus 2. Factors are exponentials only when p is not of that logarithmic kind.

19. Why this constant may be dropped

Intuition

Suppose you kept the constant and used a factor of 5 times e to the 3x. Every term of the equation is multiplied by it.

\[ 5e^{3x}y' + 15e^{3x}y = 5e^{3x}q(x) \]

The 5 is a common factor of every term, so dividing it out gives back exactly the equation you get with e to the 3x alone. Only one factor is needed, so choose the constant to be 1.

\[ \frac{d}{dx}\left[5e^{3x}y\right] = 5e^{3x}q \iff \frac{d}{dx}\left[e^{3x}y\right] = e^{3x}q \]

The constant from the LAST integration is different: it produces the whole family of solutions, and dropping it loses all but one of them.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 412 — the note on the integrating constant

This slide answers a question that bothers careful students: why are you allowed to throw away a constant of integration here, when you have been told never to do that?

Look at what the constant does. A factor of 5 times e to the 3x multiplies every single term of the equation by 5 as well as by e to the 3x. The 5 then divides straight back out, and you are left with exactly the equation you would have had with e to the 3x alone. So every choice of constant leads to the same solutions, and you might as well pick 1.

The constant that appears when you integrate the collapsed equation is completely different. It is added, not multiplied, and it is what produces a whole family of solutions. Throw it away and you keep only one solution out of infinitely many.

20. Multiply, collapse, integrate, divide

Concept

With mu chosen, the whole method is four lines.

\[ \mu(x)y' + \mu(x)p(x)\,y = \mu(x)q(x) \]

\[ \frac{d}{dx}\left[\mu(x)\,y\right] = \mu(x)q(x) \]

\[ \mu(x)\,y = \int \mu(x)q(x)\,dx + C \]

\[ y = \frac{1}{\mu(x)}\left[\int \mu(x)q(x)\,dx + C\right] \]

The second line is the payoff: the left side has collapsed into one derivative, so integrating it just removes the d by dx.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 411 — equations 4.20 to 4.23

Read the four lines as a story. First multiply by mu. Second, recognise that the left side is now the derivative of mu times y; this is the collapse. Third, integrate both sides: on the left the integral simply undoes the derivative, and on the right you get the integral of mu times q plus a constant. Fourth, divide by mu to get y by itself.

The collapse is the only step that needs thought, and it is guaranteed by the way mu was chosen. In practice you should still write the product rule out once to see that it really happens; if it does not, your factor is wrong.

The last line is the book's general solution formula, equation 4.23. You can use it directly, but most people find it more reliable to run through the four lines every time, because each step can be checked.

21. Reading the solution formula

Notation

Step through the notes with the formula in view. The structure matters more than the symbols: one over mu, times the bracket of an integral plus a constant.

The annotation on the constant is the practical one. Because the constant is inside the bracket, it gets divided by mu along with everything else. That is why the answer always contains C divided by the integrating factor: C times e to the minus 3x when the factor is e to the 3x, C over x cubed when the factor is x cubed.

The last note gives you the shape of every answer in this section. The general solution is one particular solution plus C times one over mu. Once you know that, you can spot many errors instantly: if your answer has C added on its own, not divided by the factor, something went wrong.

22. Match each equation to its integrating factor

Matching

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 420 — Exercises 218, 219, 221 and 222

Before matching, rewrite each equation in standard form. Only then can you read off p, and only then is the factor e to the integral of p.

The first is y prime minus x times y equal to 3, so p is minus x and the factor is e to the minus x squared over 2. The second is already in standard form with p equal to e to the x, whose antiderivative is e to the x again, giving e to the e to the x: an odd-looking factor, but a perfectly good one.

The third needs the antiderivative of the hyperbolic tangent, which is the log of the hyperbolic cosine. With the minus sign, the exponential of minus the log of cosh x is one over cosh x. The fourth has y on both sides; collect them and p is 3t minus e to the t.

23. Trap: skipping the division by the leading coefficient

Trap

24. Put the five steps in order

Ranking

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 412 — Problem-Solving Strategy

Put the steps in order before checking, and for each one ask what it depends on. That dependency chain is the reason the order is fixed.

You cannot compute the factor without p, and p only exists in standard form, so standard form comes first. The factor must exist before you multiply by it. The constant C first appears in the integration, so you cannot use the initial condition until after step 4.

One detail on the last step: use the initial condition on the final formula for y, after dividing by the factor. It is possible to use it on the integrated line instead, but then you must remember to evaluate the factor at the starting point too, and that is where mistakes creep in.

25. Solving: the collapse and the family

Section

Part 3

26. Example 4.16: a power of x as the factor

Worked example

\[ xy' + 3y = 4x^2 - 3x, \quad x > 0 \]

\[ y' + \frac{3}{x}\,y = 4x - 3 \]

\[ \mu = e^{\int 3/x\,dx} = e^{3\ln x} = x^3 \]

\[ x^3y' + 3x^2y = \frac{d}{dx}\left[x^3y\right] = 4x^4 - 3x^3 \]

\[ x^3y = \frac{4x^5}{5} - \frac{3x^4}{4} + C \]

\[ y = \frac{4x^2}{5} - \frac{3x}{4} + Cx^{-3} \]

\[ xy' = \tfrac85x^2 - \tfrac34x - 3Cx^{-3}, \quad 3y = \tfrac{12}{5}x^2 - \tfrac94x + 3Cx^{-3} \]

\[ xy' + 3y = 4x^2 - 3x \;\checkmark \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 412 — Example 4.16

Follow the five steps. Dividing by x gives p equal to 3 over x, so the integral of p is 3 ln x, and the factor is e to the 3 ln x, which is x cubed. This is the logarithm shortcut from the notation slide.

Multiplying through by x cubed gives x cubed y prime plus 3x squared y on the left, and that is exactly the product rule for x cubed times y. Integrating both sides with the power rule and keeping the constant gives x cubed y equal to four fifths x to the fifth minus three quarters x to the fourth plus C.

Dividing by x cubed turns C into C over x cubed. The check substitutes the answer back into the original equation. The C terms cancel, minus 3C over x cubed against plus 3C over x cubed, and what is left is exactly 4x squared minus 3x. That cancellation is the proof that every value of C works.

27. The family of Example 4.16

Picture it

\[ y = \underbrace{\tfrac45x^2 - \tfrac34x}_{C = 0} + \frac{C}{x^3} \]

Here is the whole family drawn. The highlighted curve is the solution with C equal to 0, which is just the parabola four fifths x squared minus three quarters x.

Every other solution is that parabola plus C over x cubed. Near x equal to 0 that extra term is enormous, so the curves shoot up for positive C and plunge down for negative C. Away from 0 it becomes tiny, so all the curves merge into the parabola.

This is a preview of an idea you will use all through the lesson: the general solution is one particular solution plus a multiple of one over the integrating factor. The shape of one over the factor, here one over x cubed, decides where solutions differ and where they agree.

28. What happens at x = 0, and for negative x

Concept

The book assumed x positive. Two things go wrong at zero, and the factor changes form for negative x.

\[ p(x) = \frac{3}{x} \quad\text{is undefined at } x = 0 \]

\[ \mu = e^{\int 3/x\,dx} = e^{3\ln|x|} = |x|^3 \]

For negative x the absolute value makes the factor minus x cubed, a constant multiple of x cubed, so the same family of formulas results; but a solution may not cross zero, because every one with C not zero blows up there.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 413 — the analysis after Example 4.16

The book assumed x positive, and this slide explains why. The coefficient p is 3 over x, which does not exist at x equal to 0, so the standard form breaks down there. Any solution you build must live on an interval that does not include 0.

For negative x the integral of 3 over x is 3 times the log of the absolute value of x, so the factor is the absolute value of x, cubed. On the negative axis that is minus x cubed, just a constant multiple of x cubed, and you already know constant multiples of the factor change nothing. So the same family of formulas works on either side of 0.

What does not work is crossing 0. Every solution except the parabola has a C over x cubed term that blows up there, so solutions on the left and on the right are separate pieces.

29. Find the error: C added after dividing

Error analysis

Look at the two lines and try to spot the problem before revealing the notes. The integration itself is fine; the error is about where the constant was put.

The constant should have appeared when the right side was integrated, giving x cubed y equal to the polynomial plus C. Dividing by x cubed then gives C over x cubed. Here the constant was forgotten and then tacked on at the very end, after the division, so it was never divided.

The last note shows how to catch this yourself: substitute. With C equal to 1, the wrong answer gives x y prime plus 3y equal to the right side plus 3, which is not the equation. A constant added on its own is almost never correct in this method.

30. Checkpoint 4.16: look before you compute

Step zero

\[ (x - 2)y' + y = 3x^2 + 2x, \quad x > 2 \]

Write down your answer before revealing. The habit this slide builds is to look at the equation for a moment before launching the five steps.

Apply the product rule to x minus 2 times y. The derivative of x minus 2 is 1, so you get x minus 2 times y prime plus y. That is exactly the left side of this equation.

So the equation already says that the derivative of x minus 2 times y equals 3x squared plus 2x. The integrating factor, if you compute it, turns out to be x minus 2, which just undoes the division into standard form. Recognising this saves two steps, and it is a good reminder of what the integrating factor is for: it turns the left side into this kind of perfect derivative.

31. Checkpoint 4.16: the factor was already there

Worked example

\[ (x - 2)y' + y = 3x^2 + 2x, \quad x > 2 \]

\[ y' + \frac{1}{x-2}\,y = \frac{3x^2 + 2x}{x-2} \]

\[ \mu = e^{\ln(x-2)} = x - 2 \]

\[ \frac{d}{dx}\left[(x-2)y\right] = 3x^2 + 2x \]

\[ (x-2)\,y = x^3 + x^2 + C \]

\[ y = \frac{x^3 + x^2 + C}{x - 2} \]

\[ \frac{d}{dx}\left[x^3 + x^2 + C\right] = 3x^2 + 2x \;\checkmark \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 413 — Checkpoint 4.16

Running the full method confirms what the previous slide spotted. Standard form has p equal to one over x minus 2; its antiderivative is the log of x minus 2, which is fine because x exceeds 2; and the factor is x minus 2.

Multiplying back gives the original left side, which is the derivative of x minus 2 times y. Integrate, keep the constant, and divide by x minus 2 to get the general solution.

The picture shows something interesting. Most solutions blow up at x equal to 2 because of the denominator. But when C is minus 12, the numerator also vanishes at 2, and the fraction simplifies to x squared plus 3x plus 6, a parabola that passes right through. The check differentiates the integrated line: the derivative of x cubed plus x squared plus C is exactly 3x squared plus 2x.

32. The equation separation could not solve

Worked example

Back to the warm-up equation.

\[ y' = x + y \]

\[ y' - y = x, \quad p(x) = -1 \]

\[ \mu = e^{-x} \]

\[ \frac{d}{dx}\left[e^{-x}y\right] = xe^{-x} \]

\[ e^{-x}y = -xe^{-x} - e^{-x} + C \]

\[ y = -x - 1 + Ce^{x} \]

\[ y' = -1 + Ce^{x}, \qquad x + y = -1 + Ce^{x} \;\checkmark \]

This is the warm-up equation, and now it takes five short lines. Standard form is y prime minus y equal to x, so p is minus 1 and the factor is e to the minus x.

The integration by parts is the only real work: the integral of x e to the minus x is minus x e to the minus x minus e to the minus x. Dividing by e to the minus x means multiplying by e to the x, which turns the constant into C e to the x.

The slope field shows the family. The solution with C equal to 0 is the straight line y equal to minus x minus 1, and you can check it directly: its slope is minus 1, and x plus y along the line is also minus 1. Every other solution peels away from that line, because C e to the x grows to the right. That exponential is exactly what separation could never have found.

33. Initial values, transient and steady state

Section

Part 4

34. Where do the solutions of Example 4.17 go?

Prediction

\[ y' + 3y = 2x - 1 \]

Make a genuine prediction before revealing. You have enough to reason it out: the general solution is always one particular solution plus C divided by the integrating factor.

Here p is 3, so the factor is e to the 3x and one over the factor is e to the minus 3x. That shrinks toward zero very quickly as x increases, so whatever C is, its contribution disappears. What remains is the particular solution, and it is the same for every member of the family.

If you chose that it depends on y of 0, you are right about the start and wrong about the finish. The initial value decides C, and C only matters for a short while.

35. Example 4.17: the general solution

Worked example

\[ y' + 3y = 2x - 1, \quad y(0) = 3 \]

\[ p(x) = 3, \quad q(x) = 2x - 1 \]

\[ \mu = e^{\int 3\,dx} = e^{3x} \]

\[ \frac{d}{dx}\left[ye^{3x}\right] = (2x-1)e^{3x} \]

\[ u = 2x - 1, \quad du = 2\,dx, \quad v = \tfrac13 e^{3x} \]

\[ ye^{3x} = \frac{(2x-1)e^{3x}}{3} - \int \frac23 e^{3x}\,dx \]

\[ ye^{3x} = \frac{(2x-1)e^{3x}}{3} - \frac{2e^{3x}}{9} + C \]

\[ \tfrac23e^{3x} + (2x-1)e^{3x} - \tfrac23e^{3x} = (2x-1)e^{3x} \;\checkmark \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 413 — Example 4.17

This is the book's first initial-value problem, split over two slides so that each step has room. The equation is already in standard form, so p is 3, q is 2x minus 1, and the factor is e to the 3x.

After multiplying and collapsing, the right side is 2x minus 1 times e to the 3x, and that needs integration by parts. The usual choice is to differentiate the polynomial and integrate the exponential: u is 2x minus 1, v is one third e to the 3x. Then the leftover integral is two thirds of e to the 3x, which integrates to two ninths of e to the 3x.

Before dividing, check the antiderivative by differentiating it. The product rule on the first term gives two thirds e to the 3x plus 2x minus 1 times e to the 3x; the second term gives minus two thirds e to the 3x. The two thirds cancel and exactly the integrand remains.

36. Example 4.17: fitting y(0) = 3

Worked example

\[ ye^{3x} = \frac{(2x-1)e^{3x}}{3} - \frac{2e^{3x}}{9} + C \]

\[ y = \frac{2x-1}{3} - \frac29 + Ce^{-3x} \]

\[ y = \frac23x - \frac59 + Ce^{-3x} \]

\[ 3 = 0 - \frac59 + C \]

\[ C = 3 + \frac59 = \frac{32}{9} \]

\[ y = \frac23x - \frac59 + \frac{32}{9}e^{-3x} \]

\[ y' + 3y = \tfrac23 - \tfrac{32}{3}e^{-3x} + 2x - \tfrac53 + \tfrac{32}{3}e^{-3x} = 2x - 1 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 413-414 — Example 4.17

Divide the integrated line by e to the 3x. The e to the 3x cancels from the first two terms, and C becomes C times e to the minus 3x. Combining minus one third with minus two ninths gives minus five ninths.

Now use the initial condition. At x equal to 0 the line part is minus five ninths and the exponential is 1, so 3 equals minus five ninths plus C, and C is 32 over 9. Always substitute into the final formula, after dividing, to avoid forgetting the factor.

The verification substitutes the solution back into the equation. The derivative of 32 ninths e to the minus 3x is minus 32 thirds e to the minus 3x, and three times the solution contributes plus 32 thirds e to the minus 3x, so the exponentials cancel. The remaining terms give two thirds plus 2x minus five thirds, which is 2x minus 1, exactly the right side. And at 0 the solution equals 3, so both requirements are met.

37. Transient and steady state

Concept

\[ y = \underbrace{\tfrac23x - \tfrac59}_{\text{steady state}} + \underbrace{\tfrac{32}{9}e^{-3x}}_{\text{transient}} \]

The transient is the only part that remembers the initial value, and it decays. The steady state is the same for every solution: it is what the forcing term alone produces.

The slope field makes the prediction visible. Every short segment tilts toward the dashed line, and every solution curve, whatever its starting point, bends onto that line within about one unit of x.

That gives a natural split of the solution into two parts. The transient, 32 ninths e to the minus 3x, carries all the information about the starting value, and it decays. The steady state, two thirds x minus five ninths, is what the forcing term alone produces, and every solution shares it.

This split will appear in every application in this lesson. In the ball problem the steady state is the terminal velocity; in the circuit it is the ongoing oscillating current; in the tank it is the final amount of salt. In each case the physical system forgets where it started.

38. Slide C and watch the transient die

Tweak it

Start at the default, which is close to the solution of Example 4.17, and slide C slowly toward negative values. Watch the left end of the curve: only the starting value y of 0 moves, since it equals C minus five ninths.

Now watch the right end. Whatever C you choose, by x equal to 2 the curve is back on the line, because e to the minus 6 is about a quarter of a hundredth, so even C equal to 4 contributes only about a hundredth.

That is what it means for a transient to die. The initial condition matters, but only for a limited time, and how long depends on the exponent. A factor of e to the 3x kills transients three times faster than e to the x would.

39. Checkpoint 4.17: when the exponential grows

Worked example

\[ y' - 2y = 4x + 3, \quad y(0) = -2 \]

\[ \mu = e^{\int -2\,dx} = e^{-2x} \]

\[ \frac{d}{dx}\left[ye^{-2x}\right] = (4x+3)e^{-2x} \]

\[ ye^{-2x} = -\frac{(4x+3)e^{-2x}}{2} - e^{-2x} + C \]

\[ y = -2x - \frac52 + Ce^{2x} \]

\[ -2 = -\frac52 + C \;\Longrightarrow\; C = \frac12 \]

\[ y' - 2y = (-2 + e^{2x}) - (-4x - 5 + e^{2x}) = 4x + 3 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 414 — Checkpoint 4.17

The method runs exactly as before. Here p is minus 2, so the factor is e to the minus 2x, and the integration by parts gives minus one half of 4x plus 3 times e to the minus 2x, minus e to the minus 2x.

Dividing by e to the minus 2x means multiplying by e to the 2x, so the constant becomes C e to the 2x. The initial value gives minus 2 equal to minus five halves plus C, so C is one half.

Now look at the picture, because it is the opposite of Example 4.17. The exponential e to the 2x grows rather than decays, so there is no steady state: every solution with C not zero runs away from the line, and a tiny change in the starting value changes the long-run behaviour completely. Whether a transient dies depends on the sign of p, and here p is negative. The check substitutes back, and the exponentials cancel just as before.

40. Exercise 241: a line as the steady state

Worked example

\[ y' + y = x, \quad y(0) = 3 \]

\[ \mu = e^{x} \]

\[ \frac{d}{dx}\left[ye^{x}\right] = xe^{x} \]

\[ ye^{x} = (x-1)e^{x} + C \]

\[ y = x - 1 + Ce^{-x} \]

\[ 3 = -1 + C \;\Longrightarrow\; C = 4 \]

\[ y' + y = (1 - 4e^{-x}) + (x - 1 + 4e^{-x}) = x \;\checkmark \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 420 — Exercise 241

This exercise is a clean example of the split into two parts. The factor is e to the x, the integral of x e to the x is x minus 1 times e to the x, and dividing gives x minus 1 plus C e to the minus x.

At x equal to 0 the line part is minus 1, so C must be 4 to start at 3. The check substitutes back: the derivative is 1 minus 4 e to the minus x, adding the solution cancels the exponentials, and x is left.

The figure separates the pieces. The dashed line x minus 1 is the steady state: it is itself a solution, and it is the one the forcing term produces. The dashed exponential 4 e to the minus x is the transient, the only part that knows the starting value 3. The solution is their sum, and by x equal to 4 it is indistinguishable from the line.

41. Where does every solution of y′ + 2y = 6 end up?

Estimation

\[ y' + 2y = 6 \]

\[ y' = 0 \;\Longrightarrow\; 2y = 6 \;\Longrightarrow\; y = 3, \qquad y = 3 + Ce^{-2x} \]

Commit to an answer before revealing. The quick route does not require solving the equation at all.

If the solutions settle down to a constant value, then at that value the derivative is zero. Putting y prime equal to 0 in the equation leaves 2y equal to 6, so y is 3. That is the steady state, and you can read it off in one line whenever the forcing is constant.

The general solution confirms it: 3 plus C e to the minus 2x. Because p is positive, the transient decays, so every solution really does approach 3. If p had been negative, 3 would still be a solution, but the others would run away from it rather than toward it.

42. Complete the transient table

Comparison

Fill in each blank before checking. The table links three things you have seen separately: the coefficient p, the integrating factor, and whether the transient dies.

The rule behind the last column is simple. The transient is C divided by the factor. If the factor grows, dividing by it makes the transient shrink; if the factor shrinks, the transient grows. For a constant p, that is the same as asking whether p is positive.

The last row is a reminder that factors need not be exponentials. For x y prime plus 3y, the factor is x cubed, and the transient C over x cubed dies as x grows, just more slowly than an exponential would.

43. Exercise 261: break the formula

Counterexample

\[ y' - y = e^{kt}, \quad y(0) = 0 \quad\Longrightarrow\quad y = \frac{e^{kt} - e^{t}}{k - 1} \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 421 — Exercise 261

Try to find the bad value of k yourself before revealing. The formula has k minus 1 in the denominator, so it cannot be used at k equal to 1.

That does not mean there is no solution at k equal to 1. Solve that case directly: the factor is e to the minus t, and the forcing is e to the t, so their product is exactly 1. Integrating 1 gives t, and the solution is t e to the t. This is the only place a factor of t can appear, and it appears because the forcing matches the natural behaviour of the equation exactly.

The figure shows that nothing dramatic happens to the solutions themselves as k approaches 1. The curves for k equal to 0, one half and 0.9 rise steadily toward the t e to the t curve. Only the formula breaks down, which is a good reminder to check every formula for values that make a denominator zero.

44. Falling with air resistance

Section

Part 5

45. Terminal velocity straight from the equation

Intuition

The steady state of the ball equation needs no solving. At terminal velocity the velocity stops changing, so its derivative is zero.

\[ m\frac{dv}{dt} = -kv - mg \]

\[ 0 = -kv - mg \;\Longrightarrow\; v = -\frac{mg}{k} \]

Gravity and drag balance exactly. The minus sign says the ball is then moving downward, and a heavier ball or a smaller drag constant means a faster fall.

\[ \text{racquetball: } -\frac{0.0427(9.8)}{0.5} \approx -0.8369\text{ m/s} \]

You can find the most important number in the ball problem without solving anything. Terminal velocity is the velocity the ball settles to, and once it has settled, the velocity is not changing, so its derivative is zero.

Set the derivative to zero in the equation and you get 0 equal to minus k v minus m g, so v is minus m g over k. It is negative because the ball is falling, with upward counted positive. The formula also matches intuition: a heavier ball falls faster, and a larger drag constant makes it fall more slowly.

For the racquetball, the numbers give about minus 0.837 metres per second. That is very slow, because a racquetball is light and hollow, so the drag is large compared with its weight. The full solution on the next slides will approach exactly this value.

46. Every velocity heads for the same value

Picture it

Four balls, four starting velocities, one destination. The transient is C times e to the minus 11.7096 t, which halves every six hundredths of a second, so the steady state takes over almost at once.

The figure shows the racquetball equation started from four different velocities, including the upward 2 metres per second of the book's example and a ball thrown downward at 2 metres per second.

They all end up on the same dashed line. The transient here is C times e to the minus 11.7096 t, which halves about every six hundredths of a second, so within a third of a second the starting velocity has been forgotten.

Notice that the ball thrown downward faster than the terminal speed slows down. That surprises people, but it follows from the equation: when the ball is moving down faster than terminal velocity, drag is stronger than gravity, so the net force points up.

47. Example 4.18(a): the racquetball's velocity

Worked example

Mass 0.0427 kg, drag 0.5v, hit upward at 2 m/s.

\[ 0.0427\,\frac{dv}{dt} = -0.5v - 0.0427(9.8), \quad v(0) = 2 \]

\[ \frac{dv}{dt} = -11.7096v - 9.8 \]

\[ \frac{dv}{dt} + 11.7096v = -9.8, \quad p = 11.7096, \; q = -9.8 \]

\[ \frac{d}{dt}\left[ve^{11.7096t}\right] = -9.8e^{11.7096t} \]

\[ v = -\frac{9.8}{11.7096} + Ce^{-11.7096t} = -0.8369 + Ce^{-11.7096t} \]

\[ 2 = -0.8369 + C \;\Longrightarrow\; v(t) = 2.8369e^{-11.7096t} - 0.8369 \]

\[ v' = -33.219e^{-11.7096t}, \quad -11.7096v - 9.8 = -33.219e^{-11.7096t} + 9.800 - 9.8 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 414-415 — Example 4.18a

Start by substituting the numbers into Newton's law. Dividing by the mass gives the equation with coefficient 11.7096, which is 0.5 divided by 0.0427. Moving that term to the left puts the equation in standard form with a constant p.

A constant p means the factor is simply e to the p t. After multiplying and collapsing, the right side is minus 9.8 times the same exponential, which integrates in one line. Dividing by the factor gives a constant, minus 0.8369, plus C times e to the minus 11.7096 t. That constant is the terminal velocity you found on the previous slides.

The initial velocity 2 gives C equal to 2.8369. For the check, differentiate: the derivative is minus 33.219 times the exponential. The right side of the equation gives the same exponential term plus 11.7096 times 0.8369, which is 9.800, minus 9.8. Those cancel to rounding, so the solution satisfies the equation.

48. Example 4.18(b): when does it stop rising?

Worked example

The ball is highest when its velocity is zero: positive before, negative after.

\[ 2.8369e^{-11.7096t} - 0.8369 = 0 \]

\[ e^{-11.7096t} = \frac{0.8369}{2.8369} \approx 0.2950 \]

\[ -11.7096t = \ln 0.2950 \approx -1.2208 \]

\[ t \approx 0.104\text{ s} \]

\[ t_{\text{no drag}} = \frac{2}{9.8} \approx 0.204\text{ s} > 0.104\text{ s} \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 415 — Example 4.18b

At the top of its flight the ball is momentarily neither rising nor falling, so its velocity is zero. Setting the velocity formula equal to zero and isolating the exponential gives e to the minus 11.7096 t equal to about 0.295.

Taking the natural log of both sides brings the exponent down. The log of 0.295 is about minus 1.221, and dividing by minus 11.7096 gives t about 0.104 seconds.

The check compares with a world without air. With gravity alone the velocity drops by 9.8 metres per second every second, so a ball launched at 2 would take 2 over 9.8, about 0.204 seconds, to stop. Drag adds to gravity while the ball rises, so it should stop sooner, and indeed it stops in about half the time.

49. Example 4.18(c): how high does it go?

Worked example

Hit from a height of 1 m. Height is the antiderivative of velocity.

\[ h'(t) = 2.8369e^{-11.7096t} - 0.8369, \quad h(0) = 1 \]

\[ h(t) = -0.2423e^{-11.7096t} - 0.8369t + C \]

\[ 1 = -0.2423 + C \;\Longrightarrow\; C = 1.2423 \]

\[ h(0.104) = -0.2423(0.2959) - 0.0870 + 1.2423 \approx 1.084\text{ m} \]

\[ 1 + \frac{2^2}{2(9.8)} \approx 1.204\text{ m} > 1.084\text{ m} \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 415-416 — Example 4.18c

Height is the antiderivative of velocity, so this part is a plain integration with an initial condition, not a new differential equation. The exponential integrates to the exponential divided by minus 11.7096, and 2.8369 over 11.7096 is about 0.2423.

The starting height of 1 metre fixes the constant at 1.2423. Then put in the time from part (b). The printed text writes this evaluation as h of 0.2, which is a misprint: the value 1.0836 it reports is the height at 0.104 seconds, the top of the flight. At 0.2 seconds the ball would already be falling, at about 1.052 metres.

The two panels show why the peak is where it is: the velocity crosses zero at exactly the moment the height is highest. The check compares with no drag, where the ball would climb 2 squared over twice 9.8, about 0.204 metres, reaching 1.204. Drag should cost height, and the answer of about 1.084 metres is indeed lower.

50. Checkpoint 4.18: a penny from the Empire State Building

Worked example

Mass 2.5 g, drag 0.0025v, dropped from rest, upward positive.

\[ 0.0025\,\frac{dv}{dt} = -0.0025v - 0.0025(9.8), \quad v(0) = 0 \]

\[ \frac{dv}{dt} + v = -9.8 \]

\[ \frac{d}{dt}\left[ve^{t}\right] = -9.8e^{t} \]

\[ v = -9.8 + Ce^{-t} \]

\[ v(t) = -9.8 + 9.8e^{-t} \;\longrightarrow\; -9.8\text{ m/s} \]

\[ v' = -9.8e^{-t}, \qquad -v - 9.8 = 9.8 - 9.8e^{-t} - 9.8 = -9.8e^{-t} \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 416 — Checkpoint 4.18

The set-up is the same as the racquetball, with a mass of 2.5 grams, which is 0.0025 kilograms, and a drag constant of 0.0025. Those two numbers are equal, which makes p exactly 1 after dividing by the mass.

The factor is e to the t, the integration is one line, and the starting velocity of 0 gives C equal to 9.8. So the velocity is minus 9.8 plus 9.8 e to the minus t, and as t grows the exponential dies, leaving a terminal velocity of minus 9.8 metres per second.

The picture compares this with the penny in a vacuum, whose velocity would keep growing along the dashed line. With drag the penny reaches its terminal speed in a few seconds. Working out the height, it would take about 38.7 seconds to fall the 369 metres, instead of about 8.7 seconds, and it would land at about 9.8 metres per second instead of about 85. The check substitutes back and both sides equal minus 9.8 e to the minus t.

51. Trap: switching the drag sign when the ball falls

Trap

52. Slide the drag constant

Tweak it

Start at k equal to 1 and slide it down toward 0.2. The curve drops much further before it levels off, and it takes much longer to level off. Then slide k up to 3 and watch the curve flatten almost immediately at a small speed.

Two things change together. The terminal speed is 9.8 over k, so it is inversely proportional to the drag constant. The time scale is one over k, the time for the transient to shrink by a factor of e. Doubling k halves both the terminal speed and the time it takes to reach it.

This is why a feather and a stone behave so differently. For the same mass, a large drag constant means a gentle terminal speed reached almost at once, while a small one lets the object keep accelerating for a long time.

53. A skydiver falling 5000 m

Real world

\[ m\,v' = mg - kv, \quad v(0) = 0 \quad (\text{down positive}) \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 421 — Exercises 251 to 253

Here downward is taken as positive, which is often more natural for something that only falls. Then gravity is plus m g and drag is minus k v, and the solution from rest is m g over k times one minus e to the minus k t over m.

With a mass of 100 kilograms and k equal to 4, the terminal speed is 245 metres per second and the time scale is 25 seconds. Integrating the velocity gives the distance fallen, and solving for when it reaches 5000 metres needs a calculator: about 40.5 seconds.

Compare that with about 31.9 seconds with no drag at all. The linear drag model makes a real difference, but 245 metres per second is still far faster than a real skydiver falls. That is because air resistance on a large object grows more like the square of the speed, which is the model Exercise 254 asks you to try. That equation is not linear, so it needs separation instead.

54. Circuits and mixing

Section

Part 6

55. One loop, three voltage drops

Picture it

A generator drives current round a closed loop through a resistor, an inductor and a capacitor, each of which uses up part of the voltage.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 417 — Figure 4.25

Trace the loop in the figure with your finger. Current leaves the generator, passes through the resistor, the inductor and the capacitor, and returns. Each element takes a share of the voltage, labelled beside it.

Kirchhoff's loop rule says the shares add up to exactly what the generator supplies. The resistor's drop is proportional to the current; the inductor's drop is proportional to the rate of change of the current; the capacitor's drop is proportional to the charge stored on it.

The next slide turns that picture into an equation, and removing one element at a time turns it into the two first-order linear equations this section solves.

56. Kirchhoff's loop rule gives linear equations

Concept

\[ E_L + E_R + E_C = E \]

\[ Li' + Ri + \frac{1}{C}q = E \qquad (i = q') \]

Drop the capacitor and the equation is first order in the current; drop the inductor and it is first order in the charge.

\[ \text{LR: } Li' + Ri = E(t) \qquad \text{RC: } Rq' + \frac{1}{C}q = E(t) \]

Both are linear with constant p: R over L for the first, one over RC for the second. So the integrating factor is always a plain exponential.

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 417 — equations 4.24 to 4.26

Substituting the three drops into the loop rule gives equation 4.24: L times the derivative of the current, plus R times the current, plus the charge over C, equals the source voltage. Remember that current is the rate of change of charge, so i is q prime.

With no capacitor, the equation involves only the current and its derivative, and it is first-order linear in i. With no inductor, it involves only the charge and its derivative, and it is first-order linear in q.

In both cases the coefficient p is a constant, R over L or one over R times C, because resistance, inductance and capacitance do not change with time. So the integrating factor is always a plain exponential, and the transient always decays, because p is positive. The only effort goes into integrating the exponential times the source voltage.

57. Example 4.19: the LR circuit's general solution

Worked example

E = 50 sin 20t volts, R = 5 ohms, L = 0.4 henrys, no current at first.

\[ 0.4i' + 5i = 50\sin 20t, \quad i(0) = 0 \]

\[ i' + 12.5i = 125\sin 20t \]

\[ \frac{d}{dt}\left[ie^{12.5t}\right] = 125e^{12.5t}\sin 20t \]

\[ \int e^{at}\sin bt\,dt = \frac{e^{at}(a\sin bt - b\cos bt)}{a^2+b^2} \]

\[ ie^{12.5t} = \frac{250\sin 20t - 400\cos 20t}{89}e^{12.5t} + C \]

\[ i = \frac{250\sin 20t - 400\cos 20t}{89} + Ce^{-12.5t} \]

\[ \frac{d}{dt}\left[e^{at}(a\sin bt - b\cos bt)\right] = (a^2+b^2)e^{at}\sin bt \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 417-418 — Example 4.19

Substituting the values gives 0.4 times the current's derivative plus 5 times the current equal to 50 sine 20t. Dividing by 0.4 gives the standard form with p equal to 12.5, so the factor is e to the 12.5t.

The hard part is the integral of e to the 12.5t times sine 20t. It takes integration by parts twice, and the result is the formula on the slide, which is worth knowing: e to the a t times a sine b t minus b cosine b t, all over a squared plus b squared. With a equal to 12.5 and b equal to 20, the denominator is 556.25, and the coefficients simplify to 250 over 89 and 400 over 89.

Dividing by the factor gives the general solution. The check verifies the integration formula itself by differentiating it: the product rule produces four terms, the two cosine terms cancel, and the two sine terms combine to a squared plus b squared times e to the a t times sine b t.

58. Example 4.19: the initial current and the phase

Worked example

\[ i = \frac{250\sin 20t - 400\cos 20t}{89} + Ce^{-12.5t} \]

\[ 0 = -\frac{400}{89} + C \;\Longrightarrow\; C = \frac{400}{89} \]

\[ \sqrt{250^2 + 400^2} = 50\sqrt{89} \]

\[ \cos\varphi = \frac{8}{\sqrt{89}}, \quad \sin\varphi = \frac{5}{\sqrt{89}}, \quad \varphi \approx 0.559 \]

\[ i(t) = -\frac{50\sqrt{89}}{89}\cos(20t + \varphi) + \frac{400}{89}e^{-12.5t} \]

\[ i(0) = -\frac{50\sqrt{89}}{89}\cdot\frac{8}{\sqrt{89}} + \frac{400}{89} = 0, \quad \frac{50}{\sqrt{89}} \approx 5.30 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, pp. 418-419 — Example 4.19 and Figure 4.26

The initial current is zero. The printed text writes this step as v of 0 equal to 2, a slip carried over from the ball example; the condition used, and the answer, come from i of 0 equal to 0. At t equal to 0 the sine is zero and the cosine is one, so C is 400 over 89.

The rest rewrites the sine-and-cosine combination as a single cosine, which shows its size and timing at a glance. The size of a combination of sine and cosine is the square root of the sum of the squares of their coefficients, here 50 root 89. Dividing by 89 gives an amplitude of about 5.30 amperes. The phase angle phi, about 0.559 radians, is the acute angle with cosine 8 over root 89.

The figure shows the physics. The current starts at zero, as it must, but the steady oscillation does not start at zero, so the transient fills the gap and then decays. After about 0.3 seconds the transient is under three hundredths of an ampere, and the circuit simply oscillates in step with the generator. The check confirms the rewritten formula still starts at zero.

59. Checkpoint 4.19: an RC circuit

Worked example

E = 20 sin 5t volts, C = 0.02 farads, R = 8 ohms, initial charge 4 coulombs.

\[ 8q' + \frac{1}{0.02}q = 20\sin 5t, \quad q(0) = 4 \]

\[ q' + 6.25q = 2.5\sin 5t \]

\[ \frac{d}{dt}\left[qe^{6.25t}\right] = 2.5e^{6.25t}\sin 5t \]

\[ q = \frac{10\sin 5t - 8\cos 5t}{41} + Ce^{-6.25t} \]

\[ 4 = -\frac{8}{41} + C \;\Longrightarrow\; C = \frac{172}{41} \]

\[ \frac{50\cos 5t + 40\sin 5t}{41} + \frac{62.5\sin 5t - 50\cos 5t}{41} = 2.5\sin 5t \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 419 — Checkpoint 4.19

With no inductor, the loop rule gives R times the charge's derivative plus the charge over C equal to the source. With R equal to 8 and C equal to 0.02, dividing by 8 gives p equal to 6.25 and the forcing 2.5 sine 5t.

The same integration formula applies, with a equal to 6.25 and b equal to 5. The numbers simplify neatly: the coefficients become 10 over 41 and 8 over 41. The initial charge of 4 coulombs gives C equal to 4 plus 8 over 41, which is 172 over 41.

The check splits the solution in two. The transient alone satisfies the equation with zero on the right, because it is C over the factor. So the sinusoid alone must produce the forcing: its derivative plus 6.25 times itself adds up to 2.5 sine 5t, with the cosine terms cancelling. The figure shows the large initial charge draining away within a second, leaving a small oscillation of amplitude about 0.31 coulombs.

60. Mixing: rate in minus rate out

Concept

Let A be the kilograms of salt in the tank. Its rate of change is what flows in per minute minus what flows out per minute.

\[ \frac{dA}{dt} = \underbrace{(4)(0.5)}_{\text{in, kg/min}} - \underbrace{4\cdot\frac{A}{100}}_{\text{out, kg/min}} \]

The outflow carries the tank's current concentration, A over 100, so it depends on A. The equation is linear with constant p, just like the circuits.

Mixing problems are not in this section of the book, but they are one of the most common uses of linear equations, so they are worth adding. The principle is bookkeeping: the salt in the tank changes at the rate salt comes in minus the rate it goes out.

Salt comes in at 4 litres per minute times 0.5 kilograms per litre, which is 2 kilograms per minute, a constant. Salt goes out at 4 litres per minute times the tank's current concentration, which is A kilograms spread through 100 litres. So the outflow is 4A over 100, which is A over 25.

That gives A prime equal to 2 minus A over 25. Because the rate out depends on A, you cannot just integrate; but the equation is linear with constant p equal to one over 25, so the integrating factor handles it at once.

61. Salt in a tank of pure water

Worked example

The tank starts with pure water, so there is no salt at first.

\[ \frac{dA}{dt} + \frac{A}{25} = 2, \quad A(0) = 0 \]

\[ \mu = e^{t/25} \]

\[ \frac{d}{dt}\left[Ae^{t/25}\right] = 2e^{t/25} \]

\[ Ae^{t/25} = 50e^{t/25} + C \]

\[ A(t) = 50 + Ce^{-t/25} = 50\left(1 - e^{-t/25}\right) \]

\[ A' = 2e^{-t/25}, \qquad 2 - \frac{A}{25} = 2 - 2 + 2e^{-t/25} = 2e^{-t/25} \]

The equation is in standard form with p equal to one over 25, so the factor is e to the t over 25. Multiplying and collapsing gives the derivative of A times the factor equal to 2 times the factor, and integrating gives 50 times the factor plus C.

Dividing gives 50 plus C times e to the minus t over 25, and starting with no salt forces C to be minus 50. So the salt approaches 50 kilograms, which is exactly the incoming concentration times the tank's volume. That makes sense: eventually the tank is full of liquid just like the incoming brine.

The check substitutes back and both sides equal 2 e to the minus t over 25. The figure also shows how fast it happens. The gap to 50 kilograms halves every 25 ln 2 minutes, about 17.3, so after about an hour the tank holds over 45 kilograms of salt.

62. Putting it together

Section

Part 7

63. Pattern: solving a first-order linear equation

Pattern

1. Standard form first. Divide by the coefficient of y prime and move the y term left; p carries its sign.
2. Build the factor from p alone, with no constant.
3. Multiply and collapse: the left side must become one derivative. If it does not, the factor is wrong.
4. Integrate, keep C, then divide everything by the factor.
5. Fix C from the initial value, and check by substituting back.

This is the method to carry away, in the order you should use it. The flow diagram shows it as a pipeline, with the two constants of integration marked: the one you may drop and the one you must keep.

Most errors happen at the very first step, so spend your care there. Divide by the coefficient of y prime and move the y term to the left before you read off p, and keep its sign.

Step three has a built-in check: after multiplying, the left side must be the derivative of mu times y. Write the product rule out and compare. If it does not match, stop and fix the factor before integrating anything. And always finish by substituting your answer into the original equation.

64. Check: the integrating factor

Check

The coefficient of y is 2 over x, so its antiderivative is 2 ln x, and e to the 2 ln x is x squared. The logarithm shortcut turns an exponential factor into a power of x.

The tempting wrong answers each come from a specific slip: e to the 2x treats p as just 2, 2 ln x forgets the exponential, and x to the minus 2 has the sign flipped. You can always test a candidate factor by multiplying the left side by it and checking whether it becomes the derivative of the factor times y.

65. Check: which is not linear?

Check

For each option, look only at how y and y prime appear. In three of them each appears to the first power and is multiplied by a function of x alone, however complicated.

In y prime equals x times y squared, the unknown is squared, which breaks linearity. It is separable, so you could still solve it, just not by this lesson's method. Option C is both linear and separable, which is a reminder that the two tests are independent.

66. Check: an initial-value problem

Check

The general solution is 3 plus C e to the minus 2x, with 3 as the steady state and C e to the minus 2x as the transient. Putting in x equal to 0 gives 3 plus C equal to 1, so C is minus 2.

Option D is the instructive distractor: it starts at the right value, but it does not solve the differential equation. Meeting the initial condition is only half the job. The other half is substituting into the equation, and that check rules it out immediately.

67. Explain the two constants

Explain it to yourself

Write your two sentences before revealing. This question gets at why the method works, not just how to run it.

The key word for the first constant is multiplied. A constant multiple of the integrating factor multiplies every term of the equation by the same nonzero number, so the equation is unchanged. The key word for the second is added. It sits beside the antiderivative, gets divided by the factor, and becomes the C over mu term that distinguishes one solution from another. That term is the transient in every application you have seen, and it is the only place the initial condition enters.

68. Exit ticket

Exit ticket

\[ xy' = y - 3x^3, \quad y(1) = 0 \]

OpenStax Calculus Volume 2, §4.5 First-order Linear Equations §4.5, p. 420 — Exercise 243

This pulls the lesson together in one problem: standard form, the factor, the collapse, the constant, and a check.

Dividing by x gives p equal to minus one over x, so the factor is e to the minus ln x, which is one over x. The left side collapses to the derivative of y over x, and the right side becomes minus 3x. Integrating gives y over x equal to minus three halves x squared plus C, and multiplying by x gives the general solution.

The initial condition y of 1 equal to 0 gives C equal to three halves. The check at x equal to 2 is worth doing on paper: y is minus 9, y prime is minus 16.5, so x times y prime is minus 33, and y minus 3 x cubed is minus 9 minus 24, also minus 33.

69. Recap

Recap

\[ y = \frac{1}{\mu(x)}\left[\int \mu(x)q(x)\,dx + C\right], \qquad \mu = e^{\int p(x)\,dx} \]

That completes Chapter 4: separation for products, the integrating factor for linear equations. Chapter 5 turns to sequences and series.

Stewart, Calculus: Early Transcendentals 8e, §9.5 Linear Equations §9.5, pp. 620-626 — the same method in Stewart

The whole lesson rests on one idea. Multiply a linear equation by the right factor and its left side becomes the derivative of a single product, which one integration undoes. The factor is e to the integral of p, where p is read from the standard form.

Every answer has the same shape: one particular solution plus C divided by the factor. When one over the factor decays, that second part is a transient and the first part is the steady state that every solution approaches. You can often read the steady state straight from the equation by setting the derivative to zero.

The applications all fit that shape: terminal velocity for a falling object, the steady oscillating current in a circuit, the final amount of salt in a tank. Stewart's Section 9.5 covers the same method if you want more practice.