Question

Solve the differential equation, $\frac{dy}{dx}-\frac{2xy}{1+x^2}=x^2+1$.

Answer 1

Using the equation-

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

The solution is given by -

$y(I.F)=\int Q(I.F)dx$

where integration factor=$I.F=e^{\int Pdx}$

Here,

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

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

$\int Pdx=\int \frac{-2x}{1+x^2}dx=\int \frac{-dz}{z}dz$

where $z=x^2+1⟹dz=2xdx$

$=-\log z=\log \frac{1}{z}=\log \frac{1}{1+x^2}$

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

Required solution is-

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

$⟹y=x(x^2+1)+c$