Question

By using transformation $z=\log \tan \left(\frac{x}{z}\right)$ the differential equation $y_2+\cot x\times y_1+4y{\csc }^{2}x=0$ reduces to

• A. $y_2-4y=0$
• B. $y_1-4y=0$
• C. $y_2+4y=0$
• D. None of these

Answer 1

The correct option is C $y_2+4y=0$

We have $\frac{dz}{dx}=\frac{1}{2}\frac{{\sec }^2\left(\frac{x}{z}\right)}{\tan \left(\frac{x}{z}\right)}=\frac{1}{\sin x}=\csc x$

$\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=\csc x\frac{dy}{dz}$

$\frac{d^{2}y}{{dx}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dz}\left(\frac{dy}{dx}\right)\frac{dz}{dx}=\frac{d}{dx}\left(\csc x\frac{dy}{dx}\right)\csc x$

$=\left(\csc \frac{d^{2}y}{{dx}^2}-\csc x\cot x\frac{dx}{dz}\frac{dy}{dz}\right)\csc x={\cos }^{2}x\frac{d^{2}y}{{dz}^2}-\csc x\cot x\frac{dy}{dz}$

$\frac{d^{2}y}{{dx}^2}+\cot x\frac{dy}{dx}+4y{\csc }^{2}x=0$

$\Rightarrow {\csc }^{2}x\frac{d^{2}y}{{dz}^2}-\csc x\cot x\frac{dy}{dz}+\cot x\csc x\frac{dy}{dz}+4y{\csc }^{2}x=0$

$\Rightarrow {\csc }^{2}x(\frac{d^{2}y}{{dz}^2}+4y)m=0\iff \frac{d^{2}y}{{dz}^2}+4y=0$