Question

If $y=3\cos \left(\log x\right)+4\sin \left(\log x\right),$ show that $x^2{y}_2+x{y}_1+y=0$

Answer 1

Finding $\frac{dy}{dx}$
Let $y=3\cos \left(\log x\right)+4\sin \left(\log x\right)$
Here, $\frac{dy}{dx}=y_1$ and $\frac{d^{2}y}{d{x}^2}=y_2$
Differentiating both sides w.r.t. $x$ we get
$\frac{dy}{dx}=-3\sin \left(\log x\right)\times \frac{1}{x}+4\cos \left(\log x\right)\times \frac{1}{x}$
$\{\text{Using chain rule}\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}\}$
$\Rightarrow x\frac{dy}{dx}=-3\sin \left(\log x\right)+4\cos \left(\log x\right)\cdots \left(i\right)$

Finding $\frac{d^{2}y}{d{x}^2}$
Again, differentiating both sides of (i) w.r.t. $x$, we get
${\left(x\frac{dy}{dx}\right)}^{\prime }=(-3\sin (\log x){)}^{\prime }+(4\cos \left(\log x\right){)}^{\prime }$
$\Rightarrow {\left(x\frac{dy}{dx}\right)}^{\prime }=-3\cos \left(\log x\right)\times (\log x{)}^{\prime }+4(-\sin \left(\log x\right))\times (\log x{)}^{\prime }$
$\Rightarrow {\left(x\frac{dy}{dx}\right)}^{\prime }=-3\cos \left(\log x\right)\times \frac{1}{x}-4\sin \left(\log x\right)\times \frac{1}{x}$
$\Rightarrow x^{\prime }\frac{dy}{dx}+{\left(\frac{dy}{dx}\right)}^{\prime }=-3\cos \left(\log x\right)\times \frac{1}{x}-4\sin \left(\log x\right)\times \frac{1}{x}$
$\Rightarrow \frac{dy}{dx}+x\frac{d^{2}y}{d{x}^2}=\frac{-1}{x}\left(3\cos \right(\log x)+4\sin (\log x\left)\right)$
Where $y=3\cos \left(\log x\right)+4\sin \left(\log x\right)$
$\Rightarrow \frac{dy}{dx}+x\frac{d^{2}y}{d{x}^2}=\frac{-1}{x}\times y$
$\Rightarrow x\frac{dy}{dx}+x^2\frac{d^{2}y}{d{x}^2}=-y$
$\Rightarrow x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}+y=0$
$\Rightarrow x^2{y}_2+x{y}_1+y=0$
Hence proved.