Exercise 9.5 · Q1: Linear: $\frac{dy}{dx} + 2y = \sin x$

Basics you need

Solution

1. Recognise the linear form: $P = 2$⁠, $Q = \sin x$

   $$\frac{dy}{dx} + (2)\,y = \sin x.$$
   Compare with $\frac{dy}{dx} + Py = Q$⁠. The coefficient of $\frac{dy}{dx}$ is already 1, so nothing needs dividing. $y$ and $\frac{dy}{dx}$ appear to the first power only, $P = 2$ is a constant and $Q = \sin x$ depends on $x$ only, so the equation is linear. (It is not separable: $\frac{dy}{dx} = \sin x - 2y$ is not a product of a function of $x$ and a function of $y$⁠.)
   Section 9.4.3, step (i). Lesson: linear equations

2. The integrating factor: $e^{2x}$

   $$\text{I.F.} = e^{\int P\,dx} = e^{\int 2\,dx} = e^{2x}.$$
   Any one antiderivative of $P$ will do. A constant added to $2x$ would only multiply the I.F. by a nonzero constant, which cancels from both sides later, so no constant is written here.
   Section 9.4.3, step (ii). Basics: exponentials

3. Multiply by $e^{2x}$⁠: the left side is one derivative

   $$\begin{aligned} e^{2x}\frac{dy}{dx} + 2e^{2x}y &= e^{2x}\sin x \\ \frac{d}{dx}\left(ye^{2x}\right) &= e^{2x}\sin x. \end{aligned}$$
   By the product rule, $\frac{d}{dx}\left(ye^{2x}\right) = e^{2x}\frac{dy}{dx} + y\cdot 2e^{2x}$⁠, exactly the left side. This is what the integrating factor is for.
   Product rule (Chapter 5, Section 5.3). Basics: (y · g)′ = g(y′ + Py)

4. Integrate both sides, with one constant

   $$ye^{2x} = \int e^{2x}\sin x\,dx + C.\quad\ldots(1)$$
   This is $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. The $C$ is the one arbitrary constant of the general solution, so the integral found in Step 5 gets no constant of its own.
   Section 9.4.3, step (iii). Basics: one constant for order 1

5. Find $I = \int e^{2x}\sin x\,dx$⁠: by parts twice

   Take $\sin x$ as the first function and $e^{2x}$ as the second, so $\int e^{2x}\,dx = \frac{e^{2x}}{2}$⁠:

   $$I = \sin x\cdot\frac{e^{2x}}{2} - \int\cos x\cdot\frac{e^{2x}}{2}\,dx.$$

   Again by parts, now with $\cos x$ first:

   $$\begin{aligned} &\int\cos x\cdot\frac{e^{2x}}{2}\,dx \\ &= \frac{1}{2}\bigg[\cos x\cdot\frac{e^{2x}}{2} \\ &\qquad + \int\sin x\cdot\frac{e^{2x}}{2}\,dx\bigg] \\ &= \frac{e^{2x}\cos x}{4} + \frac{I}{4}. \end{aligned}$$

   So

   $$\begin{aligned} I &= \frac{e^{2x}\sin x}{2} - \frac{e^{2x}\cos x}{4} - \frac{I}{4} \\ \frac{5}{4}I &= \frac{e^{2x}(2\sin x - \cos x)}{4} \\ I &= \frac{e^{2x}(2\sin x - \cos x)}{5}. \end{aligned}$$

   Check, by differentiating back with the product rule:

   $$\begin{aligned} &\frac{d}{dx}\left[\frac{e^{2x}(2\sin x - \cos x)}{5}\right] \\ &= \frac{e^{2x}}{5}\big[2(2\sin x - \cos x) \\ &\qquad + (2\cos x + \sin x)\big] \\ &= \frac{e^{2x}}{5}\cdot 5\sin x = e^{2x}\sin x. \end{aligned}$$
   In the second integration by parts, the derivative of $\cos x$ is $-\sin x$⁠, so the minus sign of the formula and this minus sign make $+\int\sin x\cdot\frac{e^{2x}}{2}\,dx$⁠, which is $+\frac{I}{2}$⁠; with the factor $\frac12$ in front this is $\frac{I}{4}$⁠. After two rounds the original integral comes back, so we solve for $I$ like an ordinary unknown: add $\frac{I}{4}$ to both sides and multiply by $\frac45$⁠. Keep the trigonometric function as the first function both times; switching halfway only undoes the first step.
   Section 7.6, integration by parts. Basics: every method ends with an integral

6. Solve for $y$

   Putting $I$ into (1):

   $$\begin{aligned} ye^{2x} &= \frac{e^{2x}(2\sin x - \cos x)}{5} + C \\ y &= \frac{1}{5}(2\sin x - \cos x) + Ce^{-2x}. \end{aligned}$$
   Multiply both sides by $e^{-2x}$ (never zero): $e^{2x}e^{-2x} = 1$⁠, and $C$ becomes $Ce^{-2x}$⁠. The solution holds for every real $x$⁠, since $P$ and $Q$ are defined everywhere.
   Section 9.4.3: the general solution. Lesson: Exercise 9.5 player

7. Check by substituting back

   From $y = \frac{1}{5}(2\sin x - \cos x) + Ce^{-2x}$⁠:

   $$\begin{aligned} \frac{dy}{dx} &= \frac{2\cos x + \sin x}{5} - 2Ce^{-2x}, \\ 2y &= \frac{4\sin x - 2\cos x}{5} + 2Ce^{-2x}, \end{aligned}$$

   and adding,

   $$\frac{dy}{dx} + 2y = \frac{5\sin x}{5} = \sin x,$$

   for every $C$⁠: the $Ce^{-2x}$ terms cancel ($-2Ce^{-2x} + 2Ce^{-2x} = 0$⁠).

   Substituting back proves the answer and shows the role of each part: $Ce^{-2x}$ alone solves $\frac{dy}{dx} + 2y = 0$⁠, and $\frac{1}{5}(2\sin x - \cos x)$ supplies the $\sin x$⁠.
   A solution makes L.H.S. = R.H.S. (Section 9.3). Lesson: general and particular solutions

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