Question

Solution of the differential equation $(x^2+1)y^{\prime }+2xy=4x^2$ is

• A. $y(1+x^2)=\frac{4x^3}{3}+C$
• B. $y(1-x^2)=x^3+C$
• C. $y(1-x^2)=\frac{x^3}{2}+C$
• D. None of these

Answer 1

The correct option is A $y(1+x^2)=\frac{4x^3}{3}+C$

We have,

$(x^2+1)y^{\prime }+2xy=4x^2$

$\Rightarrow (x^2+1)\frac{dy}{dx}+2xy=4x^2$

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

On comparing that,

$\frac{dy}{dx}+Py=Q$

Now,

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

I.F.$=e^{\int pdx}$

$=e^{\int \frac{2x}{1+x^2}dx}$

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

$=(x^2+1)$

Now,

$y\times I.F.=\int Q.IFdx+C$

$y\times (x^2+1)=\int \frac{4x^2}{(x^2+1)}\times (x^2+1)dx+C$

$y(x^2+1)=\int 4x^{2}dx+C$

$y(x^2+1)=4\int x^{2}dx+C$

$y(x^2+1)=4\frac{x^3}{3}+C$

$y(x^2+1)=\frac{4x^3}{3}+C$

Hence, this is the answer.