Question

If the transformation $z=\log \tan \left(\frac{x}{2}\right)$ reduces the differential equation $\frac{d^{2}y}{d{x}^2}+\cos x\frac{dy}{dx}+4y{\text{cosec}}^{2}x=0$ into $\frac{d^{2}y}{d{x}^2}+Ay=0$ then the value of $A$ is

Answer 1

We have
$\frac{dz}{dx}=\frac{1}{2}\frac{{\sec }^2\left(x/2\right)}{\tan \left(x/2\right)}=\frac{1}{\sin x}=cosecx$
$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=cosecx\frac{dy}{dz}$
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dz}\left(\frac{dy}{dx}\right)\frac{dz}{dx}$
$=\frac{d}{dz}\left(cosecx\frac{dy}{dz}\right)cosecx$
$=(cosecx\frac{d^{2}y}{d{x}^2}-cosecx\cot x\frac{dx}{dz}\frac{dy}{dz})cosecx$
$=cose{c}^{2}x\frac{d^{2}y}{d{z}^2}-cosecx\cot x\frac{dy}{dz}$
Putting these values in the given differential equation, we have
$\frac{d^{2}y}{d{x}^2}+\cot x\frac{dy}{dx}+4ycose{c}^{2}x=0$
$\iff$ $cose{c}^{2}x\frac{d^{2}y}{d{x}^2}-cosecx\cot x\frac{dy}{dz}+\cot xcosecx\frac{dy}{dz}+4ycose{c}^{2}x=0$
$\iff$ $cose{c}^{2}x(\frac{d^{2}y}{d{z}^2}+4y)=0$ $\iff$ $\frac{d^{2}y}{d{z}^2}+4y=0$
$\Rightarrow A=4$