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Q.
The solution of differential equation $\sec ^{2} y \frac{d y}{d x}+2 x \tan y=x^{3}$ is
Differential Equations
Solution:
$\sec ^{2} y \frac{d y}{d x}+2 x \tan y=x^{3}$
Let $\tan y=v$
$\Rightarrow \sec ^{2} y \frac{d y}{d x}=\frac{d v}{d x}$
The given equation becomes
$\frac{d v}{d x}+2 x v=x^{3}$
Now, I.F. $=e^{\int 2 x d x}=e^{x^{2}}$
Hence, the solution of the differential equation is
$ v \cdot e^{x^{2}}=\int x^{3} \cdot e^{x^{2}} d x+c$
Let $ x^{2}=t \Rightarrow d t=2 x d x $
$\Rightarrow v \cdot e^{x^{2}}=\frac{1}{2} \int t e^{t} d t+c=\frac{1}{2} e^{t}(t-1)+c $
$\Rightarrow v \cdot e^{x^{2}}=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c $
$\therefore \tan y=\frac{1}{2}\left(x^{2}-1\right)+c e^{-x^{2}}$