Question

Find the solution of $\frac{dy}{dx}-x\tan (y-x)=1$

• A. $\sin (y-x)=k{e}^{\frac{1}{2}{x}^2}$
• B. $\cos (y-x)=k{e}^{\frac{1}{2}{x}^2}$
• C. $\sin (y-x)=k{e}^{\frac{1}{2}x}$
• D. $\cos (y-x)=k{e}^{\frac{1}{2}x}$

Answer 1

The correct option is A $\sin (y-x)=k{e}^{\frac{1}{2}{x}^2}$

Given $\frac{dy}{dx}-x\tan (y-x)=1$

Substitute $y-x=v$

Differentiating w.r.t. x, we get,

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

So the given differential eqn becomes

$1+\frac{dv}{dx}-x\tan v=1$

$\Rightarrow \frac{dv}{dx}=x\tan v$

$\Rightarrow \cot vdv=xdx$

Integrating, we get

$\log \sin x=\frac{x^2}{2}+\log k$

$\log \left(\frac{\sin x}{k}\right)=\frac{x^2}{2}$

$\Rightarrow \sin v=k{e}^{x^2/2}$

$\Rightarrow \sin (y-x)=k{e}^{x^2/2}$