Solve the differential equation $${y}''- 2{y}'- 3y= e^{3x}\left ( x^{2}+ 1 \right )$$
by jp.Wolfram|Alpha that is $y= c_{1}e^{-x}+ c_{2}e^{3x}+ \frac{1}{12}e^{3x}x^{3}- \frac{1}{16}e^{3x}x^{2}+ \frac{9}{32}e^{3x}x.$ Can you explain to me why ? I realise that the result can be seperated into 2 parts actually the 1st one $c_{1}e^{-x}+ c_{2}e^{3x}$ received by solving the polynomial $a^{2}- 2a- 3$, I need to the help.
Rewrite the equation as $$(e^{-3x} y''-3e^{-3x}y')+(e^{-3x}y'-3e^{-3x}y)=x^2+1$$ Now integrating $$ e^{-3x} y'+e^{-3x}y=\frac{x^3}{3}+x+C$$ $$\iff e^xy'+e^xy=e^{4x}(\frac{x^3}{3}+x+C)$$ Now again integrate both sides and use Integration by parts in RHS..
You may use the fact that $e^xy'+e^xy=(e^xy)'$