Example 19: $y = c_1e^{ax}\cos bx + c_2e^{ax}\sin bx$ solves $y'' - 2ay' + (a^2+b^2)y = 0$

Basics you need

Solution

1. Write the function as one product

   $$y = e^{ax}\left[c_1\cos bx + c_2\sin bx\right].\quad\ldots(1)$$
   Taking out the common factor $e^{ax}$ leaves one product of two factors, so each derivative is one use of the product rule.
   Factorising; the product rule needs a product. Basics: derivatives

2. The first derivative

   $$\begin{aligned} \frac{dy}{dx} &= e^{ax}\left[-bc_1\sin bx + bc_2\cos bx\right] \\ &\quad + \left[c_1\cos bx + c_2\sin bx\right]e^{ax}\cdot a \\ &= e^{ax}\big[(bc_2 + ac_1)\cos bx \\ &\qquad + (ac_2 - bc_1)\sin bx\big].\quad\ldots(2) \end{aligned}$$
   Product rule: (first)(derivative of second) + (second)(derivative of first). Then collect the $\cos bx$ terms ($bc_2$ and $ac_1$⁠) and the $\sin bx$ terms ($-bc_1$ and $ac_2$⁠).
   Product and chain rules (Chapter 5, Sections 5.3 and 5.3.1). Basics: derivatives

3. The second derivative

   Differentiate (2) the same way:

   $$\begin{aligned} &\frac{d^2y}{dx^2} \\ &= e^{ax}\big[(bc_2 + ac_1)(-b\sin bx) \\ &\qquad + (ac_2 - bc_1)(b\cos bx)\big] \\ &\quad + \big[(bc_2 + ac_1)\cos bx \\ &\qquad + (ac_2 - bc_1)\sin bx\big]e^{ax}\cdot a \\ &= e^{ax}\big[(a^2c_2 - 2abc_1 - b^2c_2)\sin bx \\ &\qquad + (a^2c_1 + 2abc_2 - b^2c_1)\cos bx\big]. \end{aligned}$$
   Collect again. The $\sin bx$ terms: $-b(bc_2 + ac_1) + a(ac_2 - bc_1)$ $= a^2c_2 - 2abc_1 - b^2c_2$⁠. The $\cos bx$ terms: $b(ac_2 - bc_1) + a(bc_2 + ac_1)$ $= a^2c_1 + 2abc_2 - b^2c_1$⁠.
   Second order derivative (Chapter 5, Section 5.7). Basics: y″ is how fast the slope changes

4. Substitute and collect $\sin bx$ and $\cos bx$

   In $\frac{d^2y}{dx^2} - 2a\frac{dy}{dx} + (a^2 + b^2)y$⁠, every term has the factor $e^{ax}$⁠. The coefficient of $e^{ax}\sin bx$ is

   $$\begin{aligned} &(a^2c_2 - 2abc_1 - b^2c_2) - 2a(ac_2 - bc_1) \\ &\quad + (a^2 + b^2)c_2 \\ &= a^2c_2 - 2abc_1 - b^2c_2 - 2a^2c_2 \\ &\quad + 2abc_1 + a^2c_2 + b^2c_2 = 0, \end{aligned}$$

   and the coefficient of $e^{ax}\cos bx$ is

   $$\begin{aligned} &(a^2c_1 + 2abc_2 - b^2c_1) - 2a(bc_2 + ac_1) \\ &\quad + (a^2 + b^2)c_1 \\ &= a^2c_1 + 2abc_2 - b^2c_1 - 2abc_2 \\ &\quad - 2a^2c_1 + a^2c_1 + b^2c_1 = 0. \end{aligned}$$
   The three pieces come from $\frac{d^2y}{dx^2}$⁠, from $-2a$ times (2), and from $(a^2 + b^2)$ times (1). In each line every term cancels with another: $a^2c_2 - 2a^2c_2 + a^2c_2 = 0$⁠, $-2abc_1 + 2abc_1 = 0$⁠, $-b^2c_2 + b^2c_2 = 0$⁠, and the same pattern in the second. (In the book's line for the L.H.S., the opening bracket before $a^2c_2$ is missing; the working is unaffected.)
   Substituting into the equation (Section 9.3). Lesson: is it a solution? Examples 2 and 3

5. Conclude

   $$\begin{aligned} \text{L.H.S.} &= e^{ax}\left[0\times\sin bx + 0\times\cos bx\right] \\ &= 0 = \text{R.H.S.} \end{aligned}$$

   Hence the given function is a solution of the given differential equation, for every $c_1$⁠, $c_2$ and every real $x$⁠.

   Both coefficients are 0 identically, whatever $a$⁠, $b$⁠, $c_1$⁠, $c_2$ and $x$ are, so the left side is 0 for all $x$⁠.
   Section 9.3: a function that makes L.H.S. = R.H.S. is a solution. Lesson: general and particular solutions

6. A shorter route: $y = e^{ax}u$ with $u'' = -b^2u$

   Let $u = c_1\cos bx + c_2\sin bx$⁠. Then $u'' = -b^2u$⁠, and by the product rule

   $$\begin{aligned} y' &= e^{ax}(au + u'), \\ y'' &= e^{ax}(a^2u + 2au' + u''). \end{aligned}$$

   So

   $$\begin{aligned} &y'' - 2ay' + (a^2 + b^2)y \\ &= e^{ax}\left[a^2u + 2au' + u''\right. \\ &\qquad\left. - 2a^2u - 2au' + (a^2 + b^2)u\right] \\ &= e^{ax}(u'' + b^2u) = 0. \end{aligned}$$
   This keeps $u$ packed together instead of expanding it, so the cancellation is visible at once: only $u'' + b^2u$ survives, and that is 0 because $u$ is a combination of $\cos bx$ and $\sin bx$⁠. It is the same verification, just organised better; the exam answer above uses it.
   Product rule; $(\cos bx)'' = -b^2\cos bx$⁠, $(\sin bx)'' = -b^2\sin bx$⁠. Lesson: Examples 19 to 22

7. Check with numbers

   Take $a = 1$⁠, $b = 2$⁠, $c_1 = c_2 = 1$⁠: $y = e^x(\cos 2x + \sin 2x)$⁠, and the equation is $y'' - 2y' + 5y = 0$⁠. At $x = 0$⁠, (1), (2) and Step 3 give

   $$\begin{aligned} y &= c_1 = 1, \\ y' &= bc_2 + ac_1 = 3, \\ y'' &= a^2c_1 + 2abc_2 - b^2c_1 = 1, \end{aligned}$$

   and $1 - 2\cdot3 + 5\cdot1 = 0$⁠. The numbers in the box below check $y'(0) = 3$ and $y''(0) = 1$ directly from values of $y$⁠.

   A numerical check catches sign slips in the long algebra: a wrong sign in (2) or in Step 3 would change $y'(0)$ or $y''(0)$⁠, and the difference quotients below would disagree.
   Derivative as a limit of difference quotients (Chapter 5, Section 5.3). Basics: derivatives

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