Question

The solution of differential equation ${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$ is

• A. $\tan y=\frac{1}{2}\left(x^2-1\right)+c{e}^{-x^2}$
• B. $\tan y=\frac{1}{2}\left(x^2-1\right)+c{e}^{x^2}$
• C. $\tan y=\frac{1}{2}\left(x^2+1\right)+c{e}^{-x^2}$
• D. $\tan y=\frac{1}{2}\left(x^2+1\right)+c{e}^{x^2}$

Answer 1

Answer: (A)

${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$
Let $\tan y=v$
$\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{dv}{dx}$
The given equation becomes
$\frac{dv}{dx}+2xv=x^3$
Now, I.F. $=e^{\int 2xdx}=e^{x^2}$
Hence, the solution of the differential equation is
$v\cdot e^{x^2}=\int x^3\cdot e^{x^2}dx+c$
Let $x^2=t\Rightarrow dt=2xdx$
$\Rightarrow v\cdot e^{x^2}=\frac{1}{2}\int t{e}^{t}dt+c=\frac{1}{2}{e}^t(t-1)+c$
$\Rightarrow v\cdot e^{x^2}=\frac{1}{2}{e}^{x^2}\left(x^2-1\right)+c$
$\therefore \tan y=\frac{1}{2}\left(x^2-1\right)+c{e}^{-x^2}$