Question

$\frac{d^{2}y}{d{x}^2}+\cos x\frac{dy}{dx}+4y\cos c^{2}x=0$ it being given that $z=\log \tan \frac{x}{2}.$

• A. $\frac{d^{2}y}{d{z}^2}+4y=0$
• B. $\frac{d^{2}y}{d{z}^2}-4y=0$
• C. $\frac{d^{2}y}{d{z}^2}+2y=0$
• D. $\frac{d^{2}y}{d{z}^2}-2y=0$

Answer 1

The correct option is A $\frac{d^{2}y}{d{z}^2}+4y=0$

Given $z=\log \tan \frac{x}{2}$
$\therefore \frac{dz}{dx}=co\sec x\cdots \left(1\right)$ $[\because \int co\sec xde=\log \tan \frac{x}{2}]$
$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=co\sec x\frac{dy}{dz}=\frac{1}{\sin x}\frac{dy}{dz}$
$\Rightarrow \sin x\frac{dy}{dx}=\frac{dy}{dz}$
Differentiating w.r.t $x$
$\sin x\frac{d^{2}y}{d{x}^2}+\cos x\frac{dy}{dx}=\frac{d^{2}y}{dz}.\frac{dz}{dx}$

$=\frac{1}{\sin x}\frac{d^{2}y}{d{z}^2}$

$-4y=\frac{d^{2}y}{d{z}^2}$

$\therefore \frac{d^{2}y}{d{z}^2}+4y=0$