Question

The solution of $(1+x^2)\frac{dy}{dx}+2xy-4x^2=0$ 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. $3y(1+y^2)=4y^3+c$

Answer 1

Answer: (B)

$3y(1+x^2)=4x^3+c$

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

$\Rightarrow \frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{4x^2}{1+x^2}$

$P=\frac{2x}{1+x^2}$

$Q=\frac{4x^2}{1+x^2}$

$I.F=e^{\int pdx=e^{\int \frac{2x}{1+x^2}dx}}$

$=e^{\log \left(1+x^2\right)}$

$=1+x^2$

$y\times I.F=\int Q\times I.Fdx+c$

$\Rightarrow y\left(1+x^2\right)=\int \frac{4x^2}{1+x^2}\times \left(1+x^2\right)dx+c$

$\Rightarrow y\left(1+x^2\right)=4\frac{x^3}{3}+c$

$\Rightarrow 3y\left(1+x^2\right)=4x^3+c$