Respuesta :

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The homogeneous part of the ODE has characteristic equation

[tex]r^2+4=0[/tex]

with roots at [tex]r=\pm2i[/tex], which means the characteristic solution is

[tex]y_c=C_1\cos2t+C_2\sin2t[/tex]

For the nonhomogeneous part, we can use as a trial solution

[tex]y_p=a_2t^2+a_1t+a_0+be^t[/tex]
[tex]{y_p}''=2a_2+be^t[/tex]

Substituting into the ODE gives

[tex](2a_2+be^t)+4(a_2t^2+a_1t+a_0+be^t)=t^2+3e^t[/tex]
[tex]\implies\begin{cases}4a_2=1\\a_1=0\\2a_2+4a_0=0\\5b=3\end{cases}[/tex]
[tex]\implies a_2=\dfrac14,a_1=0,a_0=-\dfrac18,b=\dfrac35[/tex]

and so the particular solution is

[tex]y_p=\dfrac14t^2-\dfrac18+\dfrac35e^t[/tex]

which makes the general solution

[tex]y=y_c+y_p[/tex]
[tex]y=C_1\cos2t+C_2\sin2t+\dfrac14t^2-\dfrac18+\dfrac35e^t[/tex]

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