Example 14: Linear: $\frac{dy}{dx} - y = \cos x$

Basics you need

Solution

1. Recognise the linear form: $P = -1$⁠, $Q = \cos x$

   $$\frac{dy}{dx} + (-1)\,y = \cos x.$$
   Compare with $\frac{dy}{dx} + Py = Q$⁠. Here $y$ and $\frac{dy}{dx}$ appear to the first power only, the coefficient of $y$ is the constant $-1$⁠, and the right side $\cos x$ depends on $x$ only. So the equation is linear with $P = -1$⁠, $Q = \cos x$⁠. (It is not separable: $\cos x + y$ does not split as a product of a function of $x$ and a function of $y$⁠.)
   Section 9.4.3: first order linear equation, step (i). Lesson: linear equations

2. The integrating factor

   $$\text{I.F.} = e^{\int P\,dx} = e^{\int -1\,dx} = e^{-x}.$$
   Any one antiderivative of $P$ will do: a constant added to $-x$ would only multiply the I.F. by a nonzero constant, which cancels later. So no constant of integration is written here; the solution's one constant comes in Step 4.
   Section 9.4.3, step (ii). Basics: (y · g)′ = g(y′ + Py)

3. Multiply by the I.F.: the left side is one derivative

   $$\begin{aligned} e^{-x}\frac{dy}{dx} - e^{-x}y &= e^{-x}\cos x \\ \frac{d}{dx}\left(ye^{-x}\right) &= e^{-x}\cos x. \end{aligned}$$
   By the product rule, $\frac{d}{dx}\left(ye^{-x}\right) = e^{-x}\frac{dy}{dx} + y\cdot(-e^{-x})$⁠, which is exactly the left side. This is the whole point of the integrating factor.
   Product rule (Chapter 5, Section 5.3); Section 9.4.3. Basics: (y · g)′ = g(y′ + Py)

4. Integrate both sides, with one constant

   $$ye^{-x} = \int e^{-x}\cos x\,dx + C.\quad\ldots(1)$$
   A function whose derivative is $e^{-x}\cos x$ is an antiderivative of it, plus an arbitrary constant. This $C$ is the one arbitrary constant of the general solution, so when we find the integral in Step 5 we add no further constant.
   Section 9.4.3, step (iii): $y\,(\text{I.F.}) = \int(Q\times\text{I.F.})\,dx + C$⁠. Basics: one constant for order 1

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

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

   $$\begin{aligned} I &= \cos x\,(-e^{-x}) \\ &\quad - \int(-\sin x)(-e^{-x})\,dx \\ &= -e^{-x}\cos x - \int e^{-x}\sin x\,dx. \end{aligned}$$

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

   $$\begin{aligned} &\int e^{-x}\sin x\,dx \\ &= \sin x\,(-e^{-x}) \\ &\quad - \int\cos x\,(-e^{-x})\,dx \\ &= -e^{-x}\sin x + I. \end{aligned}$$

   So

   $$\begin{aligned} I &= -e^{-x}\cos x + e^{-x}\sin x - I \\ 2I &= (\sin x - \cos x)e^{-x} \\ I &= \frac{(\sin x - \cos x)e^{-x}}{2}. \end{aligned}$$
   Each integration by parts turns cosine into sine or back, and the exponential stays an exponential. After two rounds the original integral $I$ reappears with a minus sign, so we can solve for $I$ like an ordinary unknown. The first function must be the trigonometric one both times (or the exponential both times); switching halfway just 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):

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

7. Check by substituting back

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

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

   for every $C$⁠: the $Ce^x$ terms cancel.

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

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