Question

The solution of the differential equation $\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{\tan y\sin y}{x^2}$ is

• A. $\frac{1}{y}cosecx=\frac{1}{2x^2}+k$
• B. $\frac{1}{x}cosecy=\frac{1}{2x^2}+k$
• C. $\frac{1}{x}\cos y=\frac{1}{2x^2}+k$
• D. $\frac{1}{x}\cot y=\frac{1}{2x^2}+k$

Answer 1

The correct option is D $\frac{1}{x}\cot y=\frac{1}{2x^2}+k$

We have
$\cot y.cosecy\frac{dy}{dx}+cosecy.\frac{1}{x}=\frac{1}{x^2}$

$|\begin{array}{c}\text{Put cosec y = z}\\ -\text{cosec y.cot y}\frac{dy}{dx}=\frac{dz}{dx}\end{array}$

$\frac{-dz}{dx}+z.\frac{1}{x}=\frac{1}{x^2}$

$\Rightarrow \frac{dz}{dx}-\frac{1}{x}.z=\frac{-1}{x^2}$

$\therefore \text{I.F.}=e^{\int -\frac{1}{x}dx}=\frac{1}{x}$

& $\therefore$ Solution is $z.\frac{1}{x}=\int -\frac{1}{x^3}dx+k$

or $\frac{1}{x}cosecy=\frac{1}{2x^2}+k$

Where k is constant of integration.