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Find the general solution of the nonhomogeneous ODE
\begin{equation}
y^{\prime\prime}-y^\prime-6y=\sin{t}.
\end{equation}
where the complimentary solution is
\begin{equation}
y_c(t)=C_1e^{-2t}+C_2e^{3t}.
\end{equation}
Since the nonhomogeneous term is not a solution to the homogeneous equation, the particular solution takes the form
\begin{equation}
y_p(t)=A\cos{t}+B\sin{t}.
\end{equation}

This is substituted into the ODE and the coefficients are solved for:
\begin{equation}
\begin{split}
(-A\cos{t}-B\sin{t})-(-A\sin{t}+B\cos{t})-6(A\cos{t}+B\sin{t})=\sin{t}
\end{split}
\end{equation}
This is rearranged to give
\begin{equation}
(-A-B-6A)\cos{t}+(-B+A-6B)\sin{t}=\sin{t}.
\end{equation}
For this to hold for all $t$ one must have
\begin{align}
-7A-B=0&\\
A-7B=1&
\end{align}
The solution to this system of equations is $A=\frac{1}{50}$ and  $B=-\frac{7}{50}$. Thus the general solution is
\begin{equation}
y(t)=y_c(t)+y_p(t)=C_1e^{-2t}+C_2e^{3t}+\frac{1}{50}\cos{t}-\frac{7}{50}\sin{t}.
\end{equation}

Sun, 10 Nov 2024 22:18 GMT