Question

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

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

Answer 1

Answer: (B)

$\frac{dy}{dx}+\frac{1}{x}\tan y=\frac{1}{x^2}\tan y\sin y$
$\Rightarrow$ $\cot y\cos ecy\frac{dy}{dx}+\frac{1}{x}\cos ecy=\frac{1}{x^2}$ ...(i)
Let cosec $y=-v$
$\Rightarrow$ $-\cos ecy.\cot y\frac{dy}{dx}=\frac{-dv}{dx}$
Then $\frac{dv}{dx}-\frac{1}{x}v=\frac{1}{x^2}$
Here, $P=-\frac{1}{x},Q=\frac{1}{x^2}$
$IF=e^{\int Pdx}$
$=e^{\int \left(-\frac{1}{x}\right)dx}=e^{-\log x}$
$=x^{-1}=1/x$
Hence, solution of the given differential equation is
$v.\left(\frac{1}{x}\right)=\int \frac{1}{x^2}.\left(\frac{1}{x}\right)dx-c$
$=\int \frac{1}{x^3}dx-c$
$\Rightarrow$ $\frac{v}{x}=\left(-\frac{1}{2x^2}\right)-c$
$\Rightarrow$ $\frac{v}{x}=-\frac{1}{2x^2}-c$
$\Rightarrow$ $2xv=-(1+2c{x}^2)$
$\Rightarrow$ $-2x\cos ecy=-(1+2c{x}^2)$
$\Rightarrow$ $2x=\sin y(1+2c{x}^2)$