Differential equations first order

Question

How to find the particular the solution for this differential equation $x y'(x)-y(x)(1+x^2)=\frac{x^2}{2}$ first i solved it as homogenous differential equation as follows $xy'(x)-y(x)(1+x^2)=0$ then i got $y(x)=ce^{\frac{x^2}{2}}x$

Answer 1

Let $$ y'(x)+P(x)y(x)=Q(x)$$ then $$y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)Q(x)dx+c\right)$$ where $$\mu(x)=\exp\left(\int P(x)dx\right)$$ we have $$ y'(x)-\frac{(1+x^2)}{x}y(x)=\frac{x}{2}$$ $$\mu(x)=\exp\left(-\int \frac{1+x^2}{x}dx\right)=\frac{1}{x}e^{\frac{-x^2}{2}}$$ $$y(x)=xe^{\frac{x^2}{2}}\left(\int \frac{1}{2}e^{\frac{-x^2}{2}}dx+c\right)$$