Solving a System of Two Differential Equations and getting the wrong answer Problem:
Solve the following system of differential equations.
\begin{eqnarray*}
5x' + y' - 3x + y &=& 0 \\
4x' + y' - 3x &=& -3t \\
\end{eqnarray*}
Answer:
To solve this, we use the operator method.
\begin{eqnarray*}
(5D-3)x + (D+1)y &=& 0 \\
(D-3)x + Dy &=& -3t \\
D(5D-3)x + D(D+1)y &=& 0 \\
(D+1)(D-3)x + D(D+1)y &=& (D+1)(-3t) = -3t - 3 \\
D(5D-3)x - (D+1)(D-3)x &=& 3t + 3 \\
(5D^2 - 3D)x - (D^2 - 2D + 3)x &=& 3t + 3 \\
(4D^2 - D - 3)x &=& 3t + 3 \\
\end{eqnarray*}
Now to solve this equation, we need to find both the complementary solution and the particular solution. We call the complementary solution $x_c$ and the particular solution $x_p$. To find the complementary solution, we
set up the characteristic equation.
\begin{eqnarray*}
4m^2 - m - 3 &=& 0 \\
m &=& \frac{1 \pm \sqrt{1 - 4(4)(-3)} }{2(4)} = \frac{1 \pm \sqrt{1+ 48} }{8} \\
m &=& \frac{1 \pm 7 }{8} \\
m &=& 1 \,\, \text{or} \,\, m = -\frac{3}{4} \\
x_c &=& c_1 e^{x} + c_2 e^{-\frac{3}{4}} \\
\end{eqnarray*}
Now we need to find $x_p$.
\begin{eqnarray*}
x_p &=& At + B \\
x'_p &=& A \\
x''_p &=& 0 \\
4(0) - A - 3(At + B) &=& 3t + 3 \\
-3A &=& 3 \\
A &=& -1 \\
-A - 3B &=& 3 \\
-(-1) - 3B &=& 3 \\
-3B &=& 3 - 1 = 2\\
B &=& -\frac{2}{3} \\
x_p &=& -t -\frac{2}{3}  \\
\end{eqnarray*}
The books answer is:
\begin{eqnarray*}
x &=& c_1 e^{t} + c_2 e ^{3t} + t + \frac{7}{3} \\
y &=& - c_1 e^{t} - 3c_2e^{3t} + 3t + 1 \\
\end{eqnarray*}
Since my answer for $x$ is going to be different from the book, I conclude that I went wrong some where. Where did I go wrong?
Bob

from the second equation we obtain:
$$y'=-3t+3x-4x'$$ plugging this in the first equation:
$$x'-y=3t$$ differentiating this with respect to $t$:
$$x''-y'=3$$ and again we have
$$x''-(-3t+3x-4x')=3$$ this is easy to solve.
