Tedious differential equation, any tips? Let's say i have an equation like this one:
$\nu''+3\nu'=e^x(x^3-1)$
I have tried to search a particular solution using one of tose:
$(ax^3+bx^2+cx+d)\lambda e^x$
Calculus are tedious, there must be another way! 
 A: The solution to linear differential equation with constant coefficients can be done without any tricks following the standard procedure. And guessing the particular solution is not the right approach. The particular solution needs to be obtained directly. Just integrate your equation to get $$\nu'+3\nu=(x^3-3x^2+6x-7)e^x+c$$ and multiplying by $e^{3x}$ we get $$(\nu e^{3x})'=(x^3-3x^2+6x-7)e^{4x}+ce^{3x}$$ and integrating this one more time you get your $\nu$. Integration of a product of polynomial and exponential is particularly easy and can be done almost mechanically. Thus $$\nu e^{3x}=ce^{3x}+\frac{x^3-3x^2+6x-7}{4}e^{4x}-\frac{3x^2-6x+6}{16}e^{4x}+\frac{6x-6}{64}e^{4x}-\frac{6}{256}e^{4x}+d$$ or $$\nu=c+de^{-3x}+e^{x}\left(\frac{1}{4}x^3-\frac{15}{16}x^2+\frac{63}{32}x-\frac{427} {128}\right)$$
A: This is basically a first order ODE in disguise. Let $y (x)=\nu'(x) $ and then apply an integrating factor. You still get a tedious integration by parts but it is very direct. Then just integrate your result once more to recover $\nu (x) $
