Question

Solve the differential equation: $\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y.$

• A. $2x=\sin y(1-2c{x}^2)$
• B. $2x=\sin y(1+2c{x}^2)$
• C. $2x+\sin y(1-2c{x}^2)=0$
• D. None of these.

Answer 1

The correct option is A $2x=\sin y(1-2c{x}^2)$

$\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$

$\Rightarrow \cot y\csc y\frac{dy}{dx}+\frac{1}{x}\csc y=\frac{1}{x^2}$
Put $-\csc y=v\Rightarrow \cot y\csc ydy=dx$
$\therefore \frac{dv}{dx}-\frac{1}{x}.v=\frac{1}{x^2}$ ...(1)
Here $P=-\frac{1}{x}\Rightarrow \int Pdx=-\int \frac{1}{x}dx=-\log x=\log \frac{1}{x}$
$\therefore I.F.=e^{\log \frac{1}{x}}=\frac{1}{x}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{x}\frac{dv}{dx}-\frac{1}{x^2}.v=\frac{1}{x^3}$
Integrating both sides
$\frac{v}{x}=\int \frac{1}{x^3}dx+c\Rightarrow -\frac{1}{2x^2}+c$

$\Rightarrow 2x=\sin y(1-2c{x}^2)$