Question

The solution of $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$ is

• A. $ 3x(1+{{y}^{2}})=4{{y}^{3}}+c $
• B. $ 3y(1+{{x}^{2}})=4{{x}^{3}}+c $
• C. $ 3x(1-{{y}^{2}})=4{{y}^{3}}+c $
• D. $ 3y(1+{{y}^{2}})=4{{x}^{3}}+c $

Answer 1

Answer: (B)

Given differential equation can be rewritten as
$\frac{d y}{d x}+\frac{2 x}{1+x^{2}} y=\frac{4 x^{2}}{1+x^{2}}$
It is a linear differential equation of the form
$\frac{d y}{d x}+P y=Q$
$\therefore =e^{\int \frac{2 x}{1+x^{2}} d x}=\left(1+x^{2}\right)$
$\therefore $ Solution is
$y .\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \cdot \frac{4 x^{2}}{1+x^{2}} d x$
$\Rightarrow y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}+c_{1}$
$\Rightarrow 3 y\left(1+x^{2}\right)=4 x^{3}+c$