Question

Solve the differential equation $x\frac{dy}{dx}-y=2x^2\csc 2x$.

• A. $y=cx+x\ln tanx$
• B. $y=cx+x\ln cotx$
• C. $y=cx-y\ln tanx$
• D. $y=cx-y\ln cotx$

Answer 1

The correct option is A $y=cx+x\ln tanx$

$x\frac{dy}{dx}-y=2x^2\csc 2x\Rightarrow \frac{dy}{dx}-\frac{y}{x}=2x\csc 2x$ ...(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
$\frac{1}{x}\frac{dy}{dx}-\frac{y}{x^2}=2\csc 2x$
Integrating both sides w.r.t $x$ we get
$\frac{v}{x}=2\int \csc 2xdx=\log \tan x+c\Rightarrow y=cx+x\log \tan x$