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Prove the following.
If y = 3cos(log x) + 4sin(log x), show that x² (d²y)/(dx²) + x (dy)/(dx) + y = 0
If $y = 3\cos(\log x) + 4\sin(\log x)$, show that $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0$
Given: $y = 3\cos(\log x) + 4\sin(\log x)$
To show: $x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0$
Answer
Answer
To show $x^2 y_2 + x y_1 + y = 0$ for $y = 3\cos(\log x) + 4\sin(\log x)$.
- $y = 3\cos(\log x) + 4\sin(\log x)$, given.
- Differentiate with respect to $x$ (chain rule, $\frac{d}{dx}\log x = \frac1x$): $\frac{dy}{dx} = \frac{-3\sin(\log x) + 4\cos(\log x)}{x}$.
- Multiply by $x$: $x\frac{dy}{dx} = -3\sin(\log x) + 4\cos(\log x)$.
- Differentiate again (product rule on the left, chain rule on the right): $x\frac{d^2y}{dx^2} + \frac{dy}{dx} = \frac{-3\cos(\log x) - 4\sin(\log x)}{x}$.
- The right side is $-\frac{y}{x}$, so $x\frac{d^2y}{dx^2} + \frac{dy}{dx} = -\frac{y}{x}$.
- Multiply by $x$: $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} = -y$.
- Hence $x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + y = 0$.