Question

$\frac{dy}{dx}+(2x{\tan }^{-1}y-x^3)(1+y^2)=0.$

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

Answer 1

The correct option is B $2{\tan }^{-1}y=x^2-1+2c{e}^{-x^2}$

$\frac{dy}{dx}+(2x{\tan }^{-1}y-x^3)(1+y^2)=0$

$\Rightarrow \frac{1}{1+y^2}\frac{dy}{dx}+2x{\tan }^{-1}y=x^3$

Put ${\tan }^{-1}y=v\Rightarrow \frac{1}{1+y^2}\frac{dy}{dx}=\frac{dv}{dx}$

$\therefore \frac{dv}{dx}+2xv=x^3$ ...(1)

Here $P=2x$

$\Rightarrow \int PdP=\int 2xdx=x^2$

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

Multiplying (1) by $I.F.$ we get

$e^{x^2}.\frac{dv}{dx}+e^{x^2}.2xv=e^{x^2}.x^3$

Integrating both sides, we get

$v{e}^{x^2}=\frac{1}{2}{e}^{x^2}(x^2-1)+c$

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