Question

Solution of ${\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^{3}is$

• A. $\tan y=C{e}^{-x^2}+(x^2-1)$
• B. $\tan y=C{e}^{-x^2}+\frac{1}{2}(x^2-1)$
• C. $\tan y=C{e}^{-x^2}-\frac{1}{2}(x^2-1)$
• D. none of these.

Answer 1

Answer: (B)

This is of the form
$f\left(y\right)\frac{dy}{dx}+f\left(y\right).P\left(x\right)=Q\left(x\right)$
Put tan $y=z$
$\therefore se{c}^{2}y\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore \frac{dz}{dx}+2xz=x^3$
$\therefore e^{\int pdx}=e^{\int 2xdx}=e^{x^2}$
$\therefore$ Sol. is $z{e}^{x^2}=\int e^{x^2}.x^{3}dx$
$=\frac{1}{2}\int e^{x^2}.x^2.2xdx=\frac{1}{2}\int e^t.tdt$
$=\frac{1}{2}\int e^t\left(t-1\right)+C$
$\therefore \left(tany\right)e^{x^2}=\frac{1}{2}{e}^{x^2}\left(x^2-1\right)+C$
$\Rightarrow tany=C{e}^{-x^2}+\frac{1}{2}(x^2-1)$ $\Rightarrow tany=C{e}^{-x^2}+\frac{1}{2}\left(x^2-1\right)$