Question

Solution of $\frac{dy}{dx}-x\tan (y-x)=1$ is:

• A. $\tan (y-x)=c{e}^{\frac{x^2}{2}}$
• B. $sin(y-x)=e^{\frac{x^2}{2}}+c$
• C. $sec2(y-x)=c{e}^{\frac{x^2}{2}}$
• D. $\cos (y-x)e^{\frac{x^2}{2}}=c$

Answer 1

Answer: (B)

$sin(y-x)=e^{\frac{x^2}{2}}+c$

Put $y-x=v$
$y=v+x$
$\Rightarrow \frac{dy}{dx}=1+\frac{dv}{dx}$
$\Rightarrow 1+\frac{dv}{dx}-x\tan v=x$
$\Rightarrow \int \cot vdv=\int xdx$
$\Rightarrow \log \sin v=\frac{x^2}{2}$
$\Rightarrow \sin v=e^{\frac{x^2}{2}}+c$
$\Rightarrow \sin (y-x)=e^{\frac{x^2}{2}}+c$