Question

A solution of differential equation (sec2 y)dy/dx+ 2x tan y = x3 is

• A. $2\tan y = c.e^{- x^{2}} + x^{2} - 1$
• B. $\tan y = ce^{- x^{2}} + x^{2} - 1$
• C. $\tan y = ce^{x^{2}} + x^{2} - 1$
• D. None of these

Answer 1

Answer: (A)

$2\tan y = c.e^{- x^{2}} + x^{2} - 1$

Answer 2

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

Put tan y = z $\Rightarrow$ sec2 $y\frac{dy}{dx} = \frac{dz}{dx}$

Hence, given equation becomes, $\frac{dz}{dx} + 2x.z = x^{3}$

I.F. =$e^{\int_{}^{}{2xdx}} = e^{x^{2}}$

$\therefore$ Solution is $z.e^{x^{2}} = \int_{}^{}{x^{3}e^{x^{2}}}dx = \frac{1}{2}\int_{}^{}{te^{t}}dt$

Let $t = x^{2} \Rightarrow dt = 2xdx$

$\Rightarrow$ $\tan ye^{x^{2}} = \frac{1}{2}\lbrack t.e^{t} - e^{t}) + k$

$\Rightarrow$ $\tan y = ke^{- x^{2}} + \frac{e^{- x^{2}}.e^{x^{2}}}{2}(x^{2} - 1)$

$\Rightarrow$2 $\tan y = ce^{- x^{2}} + x^{2} - 1.$