Question

Solve the differential equation : $\frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2+2$

Answer 1

Given differential equation is
$\frac{dy}{dx}-\frac{2x}{1+x^2}y=x^2+2$
Comparing it with the linear differential equation of the form
$\frac{dy}{dx}+Py=Q,$ we have
$P=-\frac{2x}{1+x^2}\text{and}Q=x^2+2$
Integrating factor $\text{I.F.}=e^{\int Pdx}$
$=\exp \left(\int \frac{-2x}{1+x^2}dx\right)$
$=e^{-\log (1+x^2)}[\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\log \left(f\right(x\left)\right)]\Rightarrow \text{I.F.}=e^{\log \frac{1}{1+x^2}}=\frac{1}{1+x^2}[\because e^{\log x}=x]$

Hence, the solution of the given differential equation is
$y\times \text{I.F.}=\int (Q\times \text{I.F.})dx\Rightarrow y\times \frac{1}{1+x^2}=\int \frac{x^2+2}{1+x^2}dx\Rightarrow \frac{y}{1+x^2}=\int (\frac{1+x^2}{1+x^2}+\frac{1}{1+x^2})dx\Rightarrow \frac{y}{1+x^2}=\int (1+\frac{1}{1+x^2})dx\Rightarrow \frac{y}{1+x^2}=x+ta{n}^{-1}x+C\Rightarrow y=(1+x^2)(x+ta{n}^{-1}x+C)$
which is the required general solution of the given differential equation.