The solution of the differential equation $x^3\frac{dy}{dx}+4x^2\tan y=e^{x}secy$ satisfying $y\left(1\right)=0,$ is

We have,

$\begin{array}{l} x^3\frac{dy}{dx}+4x^2\tan y=e^x\sec y\\ \Rightarrow x^3\cos y\frac{dy}{dx}+4x^2\sin y=e^x\\ \Rightarrow x^4\cos ydy+4x^3\sin ydx=x{e}^{x}dx\\ \Rightarrow d\left(x^4\sin y\right)=x{e}^{x}dx\end{array}$

On integrating, we get 

$x^4\sin y=(x-1)e^x+C$

It is given that $y=0$ where $x=1.$

Putting $x=1$ and $y=0$ in (i), we get $C=0$

Putting $C=0$ in (i) , we get

$x^4\sin y=(x-1)e^x\Rightarrow \sin y=e^x(x-1)x^{-4}$