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}}\,

Ordinary Differential Equations

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