The solution of the differential equation $x^3 \frac{d y}{d x}+4 x^2 \tan y=e^x \sec y$ satisfying $y(1)=0$, is
Answer & explanation
Correct answer: option 2
We have,
$x^3 \frac{d y}{d x}+4 x^2 \tan y=e^x \sec y$
$\Rightarrow x^3 \cos y \frac{d y}{d x}+4 x^2 \sin y=e^x $
$\Rightarrow x^4 \cos y d y+4 x^3 \sin y d x=x e^x d x$
$\Rightarrow d\left(x^4 \sin y\right)=x e^x d x$
On integrating, we get
$x^4 \sin y=(x-1) e^x+C$ .......(i)
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
$x^4 \sin y=(x-1) e^x \Rightarrow \sin y=e^x(x-1) x^{-4}$