# Solve differential equation xy'+2y= -x^3+x, \ y(1)=2

Solve differential equation $$xy'+2y= -x^3+x, \ y(1)=2$$

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Divide both sides by x
$$y'+ \frac{2}{x}y= -x^2+1$$
This is first order linear differential equation of form
$$p(x)= \frac{2}{x}, \ q(x)= -x^2+1$$
Integrating factor is
$$=e^{\int p(x)dx}$$
$$=e^{\int \frac{2}{x}dx}$$
$$=e^{2 logx}$$
$$=e^{log x^2}$$
$$=x^2$$
equation is
$$y \cdot (I.F.)= \int (I.F.) q(x)dx+c$$
$$yx^2= \int x^2(-x+1)dx+c$$
$$yx^2=\int-x^4+x^2dx+c$$
$$x^2y= -\frac{x^5}{5}+\frac{x^3}{3}+c$$
$$y= -\frac{x^3}{5}+\frac{x}{3}+cx^{-2}$$
Given initial condition
$$y(1)=2$$
$$2= -\frac{1}{5}+\frac{1}{3}+\frac{c}{1}$$
$$2= \frac{-3+5}{15}+c$$
$$2= \frac{2}{15}+c$$
$$2- \frac{2}{15}=c$$
$$\frac{30-2}{15}=c$$
$$\frac{28}{15}=c$$
$$y= -\frac{x^3}{5}+\frac{x}{3}+\frac{28}{15x^2}$$