Question

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

Answer 1

Put $y-x=u$

$\frac{dy}{dx}-1=\frac{du}{dx}$

$\Rightarrow \frac{du}{dx}+1=x\tan u+1$

$\Rightarrow \frac{du}{dx}=x\tan u$

$\Rightarrow \frac{du}{\tan u}=xdx$

$\Rightarrow \frac{\cos udu}{\sin u}=xdx$

$\Rightarrow \int \frac{\cos udu}{\sin u}=\int xdx$

Let $t=\sin u\Rightarrow dt=\cos udu$

$\Rightarrow \int \frac{dt}{t}=\int xdx$

$\Rightarrow \log t=\frac{x^2}{2}+c$ where $c$ is the constant of integration

$\Rightarrow \log \sin u=\frac{x^2}{2}+c$ where $t=\sin u$

$\Rightarrow \log \sin (y-x)=\frac{x^2}{2}+c$ where $u=y-x$