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

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

The correct answer is option (2) : $sin \, y =e^x (x-1) x^{-4}$

$x^3\frac{dy}{dx}+4x^2  tan\, y = e^xsec\, y $

$⇒x^3cos\, y \frac{dy}{dx}+4x^2 sin y = e^x$

$⇒x^4 \, cos \, y \, dy + 4x^3 \, sin y \, dx = xe^x\, dx $

$⇒d(x^4 \, sin y ) = xe^x + dx$

On integrating, we get

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

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

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