Question

Solve:
$\frac{dy}{dx}=x\tan (y-x)+1$

Answer 1

$\frac{dy}{dx}=s\tan (y-x)+1----\left(1\right)$

Let $z=y-x\Rightarrow \frac{dz}{dx}=\frac{dy}{dx}-1$

Substiyuting in $\left(1\right)\Rightarrow \frac{dz}{dx}+1=x\tan z+1$

$\Rightarrow \frac{dz}{dx}=x\tan z$

$\frac{dz}{\tan z}=xdx$

Integrating both sides. $\int \frac{dz}{\tan z}=\int xdx$

$\Rightarrow \log |\sin z|=\frac{x^2}{2}+C$

$\log \sin (y-x)=\frac{x^2}{2}+C$

$y=x+{\sin }^{-1}{e}^{\frac{x^2}{2}}+C$