non-homegeneous linear ODE

Definition

$y^{\prime }+p(x)y=r(x)$

Solution

Want to hold: $\mu (t)y^{\prime }+p(x)\mu (t)y=\frac{d}{dt}[\mu (t)y]\iff p(t)\mu (t)={\mu }^{\prime }(t)$

$p(t)\mu (t)={\mu }^{\prime }(t)\Rightarrow p(t)dt=\frac{d\mu }{\mu }\Rightarrow \int p(t)dt=\ln \mu (t)\Rightarrow \mu (t)=e^{\int p(t)dt+k}$

Let $k=0$ and fix $\mu$, then $\mu (t)$^[integrating factor] $=e^{\int p(t)dt}$

Now, the original ODE becomes $e^{\int p(t)dt}{y}^{\prime }+p(t)e^{\int p(t)dt}y=e^{\int p(t)dt}r(t)$ $\Rightarrow (e^{\int p(t)dt}y{)}^{\prime }=e^{\int p(t)dt}r(t)$

Let $h:=\int p(t)dt$ for simplicity

$\int (e^{h}y{)}^{\prime }dt=\int e^{h}r(t)dt$ $\Rightarrow (e^{h}y{)}^{\prime }=\int e^{h}r(t)dt+c$ $\Rightarrow y=e^{-h}(\int e^{h}r(t)dt+c)=c{e}^{-h}+e^{-h}\int r(t)e^{h}dt$

Examples

&\frac{dy}{dx} = \frac{x^{3}-2y}{x}\\ &\frac{dy}{dx} + \frac{2}{x}y = x^{2}\\ &h(x) := \int\frac{2}{x}dx = 2\ln|x| \Rightarrow \mu(x) := C\exp(\ln|x^{2}|)= Cx^2\\ &\frac{d}{dx}(Cx^{2}y) = Cx^{4}\\ &Cx^{2}y = C\int x^{4}dx = \frac{C}{5}x^{5} + C_{1}\\ &y = \frac{1}{5}x^{3} + \frac{C_{1}}{x^{2}} \end{aligned}$$