Below is my attempt to solve a problem from a text book on ODEs. My answer for $y$ does not match the book's answer. I would like to know where I went wrong or if the book is wrong.
Thanks,
Bob
Solve the following system of differential equations.
\begin{eqnarray*}
x' + y' - x - 3y &=& e^t \\
x' + y' + x &=& e^{3t} \\
\end{eqnarray*}
Answer:
To solve this system of differential equations we will use the operator method.
\begin{eqnarray*}
(D-1)x + (D-3)y &=& e^t \\
(D+1)x + Dy &=& e^{3t} \\
(D+1)(D-1)x + (D+1)(D-3)y &=& (D+1)e^t = e^t + e^t = 2e^t \\
(D^2-1)x + (D^2-2D+3)y &=& 2e^t \\
(D-1)(D+1)x + D(D-1)y &=& (D-1)e^{3t} = 3e^{3t} - e^{3t} = 2e^{3t} \\
(D^2-1)x + D(D-1)y &=& 2e^{3t} \\
%
(D^2-2D+3)y - D(D-1)y &=& 2e^t - 2e^{3t} \\
(D^2-2D+3)y - (D^2-D)y &=& 2e^t - 2e^{3t} \\
(-D+3)y &=& 2e^t - 2e^{3t} \\
-\frac{dy}{dt} + 3y &=& 2e^t - 2e^{3t} \\
\frac{dy}{dt} - 3y &=& 2e^{3t} - 2e^t \\
\end{eqnarray*}
Now we have a first order differential equation and we look for an integrating factor $I(x)$.
\begin{eqnarray*}
I(t) &=& e ^ { \int -3 \,\, dt } = e^ {-3t} \\
e^ {-3t} \frac{dy}{dt} - 3e^ {-3t}y &=& 2 - 2e^{-2t} \\
D\left( e^ {-3t}y \right) &=& 2 - 2e^{-2t} \\
e^ {-3t}y &=& 2t + e^{-2t} + c_0 \\
y &=& 2te^{3t} + e^t +c_0e^{3t} \\
\end{eqnarray*}
The books answer is:
\begin{eqnarray*}
x &=& ce^{-3t} + \frac{e^t}{4} \\
y &=& -\frac{2e^{-3t}}{3} + \frac{e^{3t}}{3} - \frac{e^{t}}{2} \\
\end{eqnarray*}
This system is a simple one:
$$x' + y' - x - 3y = e^t \\ x' + y' + x = e^{3t}$$
Let's subtract the equations:
$$2x+3y=e^{3t}-e^t$$
So we immediately find:
$$x=-\frac{3}{2}y+\frac{1}{2} (e^{3t}-e^t)$$
$$x'=-\frac{3}{2}y'+\frac{1}{2} (3e^{3t}-e^t)$$
Substituting into the second equation:
$$-\frac{1}{2}y'+\frac{1}{2} (3e^{3t}-e^t)-\frac{3}{2}y+\frac{1}{2} (e^{3t}-e^t)=e^{3t}$$
Simplifying:
$$y'+3y=2(e^{3t}-e^t)$$
$$x=-\frac{3}{2}C_1 e^{-3t}-\frac{1}{2}e^{3t}+\frac{3}{4}e^t+\frac{1}{2} (e^{3t}-e^t)$$
Up to a constant (and a typo) the book answer is right.
In your solution you write: $(D+1)(D-3)=(D^2-2D+3)$, but it should be $(D^2-2D-3)$, I think this was your mistake.