<s>[INST] solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}{dx}=\frac{x^4}{12}+\frac{x^3}{3}+x^2+c_1x+c_2$$
This is what I got to but I am unsure beyond here 
 [/INST] $$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
Substitute $z=y''$
$$z'+z=x^2+2x+2$$
$$(z-x^2-2)'+(z-x^2-2)=0$$
$$v'=-v$$
Where $v=z-x^2-2$
$$\implies \ln|v|=-x+K \implies v=K_1e^{-x} $$
$$\implies z=x^2+2+K_1e^{-x}$$
Integrate twice to get y
$$\boxed{y=K_1e^{-x}+K_2x+K_3+x^2+\frac {x^4}{12}}$$
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