Exercise 9.3 · Q15: Curve through $(0, 0)$ with $y' = e^x\sin x$

Basics you need

Solution

1. Separate the variables

   $y' = \frac{dy}{dx}$⁠, so the equation is $\frac{dy}{dx} = e^x\sin x$⁠, of order 1 and degree 1. Multiply by $dx$⁠:

   $$dy = e^x\sin x\,dx.$$
   The right side depends on $x$ only: $g(x) = e^x\sin x$⁠, $h(y) = 1$⁠, so nothing in $y$ is divided by and no value of $y$ is lost. $e^x\sin x$ is defined for every $x$⁠, so the solution is valid on the whole real line.
   Section 9.4.1: separate, then integrate. Lesson: separate, then integrate

2. Integrate by parts, once

   Let $I = \int e^x\sin x\,dx$⁠. Take $\sin x$ as the first function and $e^x$ as the second:

   $$\begin{aligned} I &= \sin x\cdot e^x - \int\cos x\cdot e^x\,dx \\ &= e^x\sin x - \int e^x\cos x\,dx. \end{aligned}$$
   The rule: first function times the integral of the second, minus the integral of (derivative of the first times the integral of the second). The integral of $e^x$ is $e^x$ and the derivative of $\sin x$ is $\cos x$⁠. The new integral looks no simpler, but one more round brings $I$ back.
   Section 7.6, integration by parts. Basics: every method ends with an integral

3. Integrate by parts again, and solve for $I$

   In $\int e^x\cos x\,dx$ take $\cos x$ as the first function:

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

   So

   $$\begin{aligned} I &= e^x\sin x - e^x\cos x - I \\ 2I &= e^x(\sin x - \cos x) \\ I &= \frac{1}{2}e^x(\sin x - \cos x). \end{aligned}$$
   The derivative of $\cos x$ is $-\sin x$⁠, and minus a minus gives $+\int e^x\sin x\,dx$⁠, which is $I$ again. Moving $-I$ across gives $2I$⁠. Keeping the trigonometric function as the first function both times is what makes $I$ return with the opposite sign. The book works the same integral taking $e^x$ first; the result is the same.
   Section 7.6; the book computes $\int e^x\sin x\,dx$ in Chapter 7, Example 21. Basics: every method ends with an integral

4. The general solution, then the point

   $$y = \frac{1}{2}e^x(\sin x - \cos x) + C.$$

   The curve passes through $(0, 0)$⁠: put $x = 0$⁠, $y = 0$⁠:

   $$\begin{aligned} 0 &= \frac{1}{2}e^0(\sin 0 - \cos 0) + C \\ &= -\frac{1}{2} + C \;\Rightarrow\; C = \frac{1}{2}. \end{aligned}$$

   So $y = \frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2}$⁠. Multiplying by 2 and moving the 1 across:

   $$2y - 1 = e^x(\sin x - \cos x).$$
   The constants from both sides of Step 1, and from the integration by parts, all combine into one $C$⁠: one constant for a first order equation. $e^0 = 1$⁠, $\sin 0 = 0$⁠, $\cos 0 = 1$⁠. The last form clears the fractions and matches the textbook's answer.
   Section 9.3: a particular solution gives the constant a value. Basics: one constant, one condition

5. Check: substitute back, and at the point

   Differentiate $2y - 1 = e^x(\sin x - \cos x)$ by the product rule:

   $$\begin{aligned} 2y' &= e^x(\sin x - \cos x) \\ &\quad + e^x(\cos x + \sin x) \\ &= 2e^x\sin x, \end{aligned}$$

   so $y' = e^x\sin x$⁠. At $x = 0$⁠: $2y - 1 = 1\cdot(0 - 1) = -1$⁠, so $y = 0$⁠.

   Both the equation and the point check out, so this is the curve asked for (Section 9.3). The $\cos x$ terms cancel, which is the same cancellation that made $I$ come back in Step 3.
   Product rule (Chapter 5). Basics: y′ is the slope

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