Question

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

• A. $\tan\,y=Ce^{-x^2}+(x^2-1)$
• B. $\tan\,y=Ce^{-x^2}+\frac{1}{2}(x^2-1)$
• C. $\tan\,y=Ce^{-x^2}-\frac{1}{2}(x^2-1)$
• 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 $y = z$
$\therefore sec^{2}y \frac{dy}{dx}= \frac{dz}{dx}$
$\therefore \frac{dz}{dx}+2xz = x^{3}$
$\therefore e^{\int pdx} = e^{\int2x\,dx} = e^{x^2}$
$\therefore $ Sol. is $z e^{x^2} = \int e^{x^2}. x^{3}\,dx$
$= \frac{1}{2}\int e^{x^2} . x^{2} . 2xdx = \frac{1}{2} \int e^{t}.tdt$
$= \frac{1}{2}\int e^{t}\left(t-1\right)+C$
$\therefore \left(tan\,y\right)e^{x^2} = \frac{1}{2}e^{x^2} \left(x^{2}-1\right)+C$
$\Rightarrow tan \,y = Ce^{-x^2}+\frac {1}{2}(x^2-1)$ $\Rightarrow tan\,y = Ce^{-x^2}+\frac{1}{2} \left(x^{2}-1\right)$