ODE4.2

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$y''-3y'+2y=3e^{{x}}\,$

Set up and solve the characteristic equation for the homogeneous solution.

$m^{2}-3m+2=0\,$

$m={\frac {3\pm {\sqrt {9-8}}}{2}}={\frac {3}{2}}\pm {\frac {1}{2}}=2,1\,$

So the solution to the homogeneous equation is

$y_{h}=c_{1}e^{{2x}}+c_{2}e^{{x}}\,$

Guess a particular solution similar to the structure of the nonhomogeneous part of the DE. Multiply the exponential term by $x$ since it is included in the homogeneous solution.

$y_{p}=Axe^{{x}}\,$

Take first and second derivatives of $y_{p}$.

$y_{p}'=Axe^{{x}}+Ae^{{x}}=Ae^{{x}}(x+1)\,$

$y_{p}''=Ae^{{x}}+Ae^{{x}}(x+1)=Ae^{{x}}(x+2)\,$

Plug these derivatives into the original problem.

$Ae^{{x}}(x+2-3(x+1)+2x)=3e^{{x}}\,$

$-Ae^{{x}}=3e^{{x}}\,$ so that $A=-3\,$ and so the particular solution is

$y_{p}=-3xe^{{x}}\,$

Finally, the general solution is

$y=y_{h}+y_{p}=c_{1}e^{{2x}}+c_{2}e^{{x}}-3xe^{{x}}\,$