\begin{equation*}
y^{\prime \prime} - 3y^\prime + 2y = t
\end{equation*}
The initial values are \(y(0) = \frac{3}{4}\) and \(y^\prime(0) = \frac{3}{2}\text{.}\) The linear differential operator is
\begin{equation*}
L = \frac{d^2}{dt^2} + 3 \frac{d}{dt} + 2 \text{.}
\end{equation*}
I first solve the homogeneous case using the characteristic equation, finding the roots and writing the general solution.
\begin{align*}
Ly \amp = 0 \implies r^2 - 3r + 2 = 0 \\
(r-2)(r-1) \amp = 0 \implies r = 1,2 \\
y_h \amp = A e^{2t} + B e^t
\end{align*}
The force term is \(t\text{,}\) which is a polynomial of degree one. The guess for undetermined coefficients is a polynomial of the same degree.
\begin{equation*}
y_p = Ct + D
\end{equation*}
I’ll need the derivatives of the guess to put it into the differential equation.
\begin{align*}
y_p^\prime \amp = C \\
y_p^{\prime \prime} \amp = 0
\end{align*}
I put all the pieces into the non-homogeneous DE.
\begin{equation*}
0 - 3C + 2(Ct + D) = t \implies (2C-1)t + (D-3C) = 0
\end{equation*}
This is an equation of polynomials. It is true only if botht the coefficients on the left are zero. The first coefficient lets me calculate \(C\text{.}\)
\begin{equation*}
2C = 1 \implies C = \frac{1}{2}
\end{equation*}
Knowing \(C\text{,}\) I can use the second coefficient to calculate \(D\text{.}\)
\begin{equation*}
-3C + 2D = 0 \implies -3 \frac{1}{2} + 2 D = 0 \implies D
= \frac{3}{4}
\end{equation*}
Now that I have the coefficients, I know the particular solution.
\begin{equation*}
y_p = \frac{t}{2} + \frac{3}{4}
\end{equation*}
Finally, the general solution is the particular solution plus the homogeneous solution.
\begin{equation*}
y = \frac{t}{2} + \frac{3}{4} + A e^{2t} + B e^t
\end{equation*}
Now I put in the initial values. First I use the initial value of the function.
\begin{equation*}
y(0) = A + B + \frac{3}{4} = \frac{3}{4} \implies A =
-B
\end{equation*}
Then I use the initial value for the derivative. First I need to calculate the derivative of the full solution.
\begin{equation*}
y^\prime = 2Ae^{2t} + Be^t + \frac{1}{2}
\end{equation*}
Then I input the initial condition.
\begin{equation*}
y^\prime(0) = \frac{3}{2}
\implies
2A + B + \frac{1}{2} = \frac{3}{2} \implies 2A + B - 1 = 0
\end{equation*}
Now I have a system of two equations from the two initial conditions. The solution to the system is \(A=1\) and \(B = -1\text{.}\) Finally, I can write the specific solution to the non-homogeneous initial value problem.
\begin{equation*}
y = \frac{t}{2} + \frac{3}{4} + e^{2t} - e^t
\end{equation*}
