Exercise 5.7 · Q13: $y=3\cos(\log x)+4\sin(\log x)$ satisfies $x^2y_2+xy_1+y=0$

Basics you need

Solution

1. Domain

   $\log x$ needs $x \gt 0$⁠; $\sin$ and $\cos$ accept every real number. So $y$ is defined, and differentiable, exactly for $x \gt 0$⁠. Everything below is for $x \gt 0$⁠, and in particular $x \neq 0$⁠.

   We shall divide by $x$ and multiply equations by $x$⁠; knowing $x \gt 0$ makes both safe.

2. First derivative $y_1$

   Term by term with Basic 1:

   $$\begin{aligned} y_1 &= 3\left(-\frac{\sin(\log x)}{x}\right) + 4\cdot\frac{\cos(\log x)}{x} \\ &= \frac{4\cos(\log x) - 3\sin(\log x)}{x}. \end{aligned}$$
   Both terms are "trig of $\log x$", so both get the chain-rule factor $\frac{1}{x}$⁠. We combine them over the common denominator $x$⁠.
   Basic 1; sum rule, Section 5.3, rule (1); constant multiples (Class XI, Limits and Derivatives). Lesson: chain rule

3. Clear the denominator

   Multiply both sides by $x$⁠:

   $$x\,y_1 = 4\cos(\log x) - 3\sin(\log x).$$
   Now the right side is again a combination of $\cos(\log x)$ and $\sin(\log x)$ with no fraction, so differentiating it is as easy as Step 2. The left side is a simple product.

4. Differentiate both sides with respect to $x$

   Left side, product rule with $u = x$⁠, $v = y_1$⁠: $\frac{d}{dx}(x\,y_1) = 1\cdot y_1 + x\,y_2$⁠.

   Right side, Basic 1 again:

   $$\begin{aligned} &\frac{d}{dx}\big[4\cos(\log x) - 3\sin(\log x)\big] \\ &= -\frac{4\sin(\log x)}{x} - \frac{3\cos(\log x)}{x} \\ &= -\frac{3\cos(\log x) + 4\sin(\log x)}{x} = -\frac{y}{x}. \end{aligned}$$

   So

   $$x\,y_2 + y_1 = -\frac{y}{x}.$$
   Two expressions that are equal for every $x \gt 0$ have equal derivatives. On the right, after collecting, the bracket $3\cos(\log x) + 4\sin(\log x)$ is exactly the original $y$⁠. That is the whole trick of this question.
   Product rule (Basic 2); the derivative of $y_1$ is $y_2$ (Basic 4).

5. Multiply by $x$ and rearrange

   $$\begin{aligned} x^2y_2 + x\,y_1 &= -y, \\ x^2y_2 + x\,y_1 + y &= 0. \end{aligned}$$

   This is the required relation, for every $x \gt 0$⁠.

   Multiplying $x\,y_2 + y_1 = -\frac{y}{x}$ by $x$ (which is not zero) gives an equivalent equation without fractions; moving $-y$ to the left gives the form asked for.

6. Independent check: compute $y_2$ and each term of the left side

   Write $y_1 = \frac{u}{x}$ with $u = 4\cos(\log x) - 3\sin(\log x)$⁠. From Step 4, $u' = -\frac{3\cos(\log x) + 4\sin(\log x)}{x}$⁠. By the quotient rule,

   $$\begin{aligned} y_2 &= \frac{u'\cdot x - u\cdot 1}{x^2} \\ &= \frac{-3\cos(\log x) - 4\sin(\log x) - u}{x^2} \\ &= -\frac{7\cos(\log x) + \sin(\log x)}{x^2}, \end{aligned}$$

   since on top the $\cos(\log x)$ terms give $-3 - 4 = -7$ and the $\sin(\log x)$ terms give $-4 + 3 = -1$⁠. Now each term of the left side:

   $$\begin{aligned} x^2y_2 &= -7\cos(\log x) - \sin(\log x), \\ x\,y_1 &= 4\cos(\log x) - 3\sin(\log x), \\ y &= 3\cos(\log x) + 4\sin(\log x). \end{aligned}$$

   Adding: the $\cos(\log x)$ coefficients give $-7 + 4 + 3 = 0$ and the $\sin(\log x)$ coefficients give $-1 - 3 + 4 = 0$⁠. So the left side is $0$⁠, the right side.

   This route computes the left side completely, term by term, without the "differentiate an equation" trick. Both routes agree.
   Quotient rule, Section 5.3, rule (3). Basics: derivative rules

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