Question

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

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

Answer 1

Answer: (B)

Given differential equation can be rewritten as
$\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{4x^2}{1+x^2}$
It is a linear differential equation of the form
$\frac{dy}{dx}+Py=Q$
$\therefore =e^{\int \frac{2x}{1+x^2}dx}=\left(1+x^2\right)$
$\therefore$ Solution is
$y.\left(1+x^2\right)=\int \left(1+x^2\right)\cdot \frac{4x^2}{1+x^2}dx$
$\Rightarrow y\left(1+x^2\right)=\frac{4x^3}{3}+c_1$
$\Rightarrow 3y\left(1+x^2\right)=4x^3+c$