Toronto Math Forum

a,First solve the homogenous part:
\begin{aligned}
r^{2}-8r+25 &=0\\
r^{2}-8r+16&=-9\\
(r-4)^{2} &=9i^2\\
r_1 &=3i+4 \\r_2 &=-3i+4
\end{aligned}
So the solution to homogenous part is:
\begin{aligned}
y_c(x) =c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)
\end{aligned}
Next we solve
\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=18e^{4x}\\
\end{aligned}
Let
\begin{aligned}
y_p(x)&=Ae^{4x}\\
y_p^{\prime}(x) &= 4Ae^{4x}\\
y_p^{\prime\prime}(x) &= 16Ae^{4x}\\
                                 
\end{aligned}
Thus we can have
\begin{aligned}
16Ae^{4x}-32Ae^{4x}+25Ae^{4x} &=18e^{4x}\\
A &= 2\\
\end{aligned}
Therefore, we can have:
\begin{aligned}
y_p(x) &=2e^{4x}\\
\end{aligned}

\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=104Cos(3x)\\
y_p(x) &= Acos(3x)+Bsin(3x)\\
y_p^{\prime}(x) &= -3Asin(3x)+3Bcos(3x)\\
y_p^{\prime\prime}(x) &= -9Acos(3x)-9Bsin(3x)\\
\end{aligned}
Thus we can have
\begin{aligned}
-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25Acos(3x)+25Bsin(3x) &=104cos(3x)\\
(16A-24B)cos(3x)+(16B+24A)sin(3x) &= 104cos(3x)\\
A &= 2\\
B &=-3\\
y_p(x) &= 2cos(3x)-3sin(3x)
\end{aligned}
From the above, we get
\begin{aligned}
y &= y_c(x)+y_p(x)=c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}

b,
Since
\begin{aligned}
y(0) &= 0\\
y^{\prime}(0) &= 0\\
\end{aligned}
we can get
\begin{aligned}
c_1 &= -4\\
c_2 &= 1/3
\end{aligned}
So the solution is
\begin{aligned}
y &= -4e^{4x}cos(3x)+1/3e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}