Solving a System of two ODEs

Question

Below is my attempt to solve the problem. However, my answer is different from the book's answer. I would like to know where I went wrong or maybe the book's answer is wrong.

Bob

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 system of differential equations we will use the operator method. \begin{eqnarray*} (5D-3)x + (D+1)y &=& 0 \\ (4D-3)x + Dy &=& -3t \\ D(5D-3)x + D(D+1)y &=& 0 \\ (D+1)(4D-3)x + D(D+1)y &=& (D+1)(-3t) = -3t - 3 \\ D(5D-3)x - (D+1)(4D-3)x &=& 3t + 3 \\ (5D^2 - 3D)x - (4D^2 + D - 3)x &=& 3t + 3 \\ (D^2 - 4D + 3)x &=& 3t + 3 \\ \end{eqnarray*} Now to solve this equation we use the method of undermined coefficients. \begin{eqnarray*} x &=& x_c + x_p \\ m^2 - 4m + 3 &=& 0 \\ (m-3)(m-1) &=& 0 \\ m = 3 &\text{ or }& m = 1 \\ x_c &=& c_1e^{t} + c_2e^{3t} \\ x_p &=& At + B \\ x'_p &=& A \\ -4A + 3(At+B) &=& 3t + 3 \\ 3At + 3B - 4A &=& 3t + 3 \\ 3A &=& 3 \\ A &=& 1 \\ 3B - 4A &=& 3 \\ 3B - 4(1) &=& 3 \\ 3B 4 &=& 3 \\ B &=& \frac{7}{3} \\ x &=& c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} \\ \end{eqnarray*} Now we need to find $y$. \begin{eqnarray*} (4D-3)x + Dy &=& -3t \\ (4D-3)(c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} ) + Dy &=& -3t \\ 4c_1e^t + 12c_2e^{3t} + 4 - 3c_1e^{t} - 3c_2e^{3t} - 3t - 7 + Dy &=& -3t \\ c_1e^t + 12c_2e^{3t} + 4 - 3c_2e^{3t} - 3t - 7 + Dy &=& -3t \\ c_1e^t + 9C_2e^{3t} - 3 + Dy &=& 0 \\ \end{eqnarray*} \begin{eqnarray*} \frac{dy}{dt} &=& 3 - c_1e^t - 9c_2e^{3t} \\ y &=& 3t - c_1e^t - 3c_2e^{3t} + c_3\\ \end{eqnarray*} Now we need to eliminate the constant $c_3$. \begin{eqnarray*} x' &=& c_1e^{t} + 3c_2e^{3t} + 1 \\ 5(c_1e^{t} + 3c_2e^{3t} + 1) + y' - 3x + y &=& 0 \\ 5(c_1e^{t} + 3c_2e^{3t} + 1) + 3 - c_1e^t - 9c_2e^{3t} - 3x + y &=& 0 \\ 5c_1e^{t} + 15c_2e^{3t} + 5 + 3 - c_1e^t - 9c_2e^{3t} - 3( c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} ) + y &=& 0 \\ 5c_1e^{t} + 15c_2e^{3t} + 8 - c_1e^t - 9c_2e^{3t} - 3( c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} ) + y &=& 0 \\ 4c_1e^{t} + 6c_2e^{3t} + 8 - 3( c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} ) + y &=& 0 \\ c_1e^{t} + 3c_2e^{3t} + 8 - 3t - 7 + 3t - c_1e^t - 3c_2e^{3t} + c_3 &=& 0 \\ c_1e^{t} + 3c_2e^{3t} + 8 - 7 - c_1e^t - 3c_2e^{3t} + c_3 &=& 0 \\ 3c_2e^{3t} + 1 - 3c_2e^{3t} + c_3 &=& 0 \\ c_3 &=& -1 \\ \end{eqnarray*} Hence the answer is: \begin{eqnarray*} y &=& 3t - c_1e^t - 3c_2e^{3t} - 1 \\ \end{eqnarray*}
The books answer is: \begin{eqnarray*} x &=& c_1e^{t} + c_2e^{3t} + t + \frac{7}{3} \\ y &=& -c_1e^{t} - 3c_2e^{3t} + 3t + 1 \\ \end{eqnarray*}

Answer 1

Your answer is the right one.

Checked with Mathematica by direct substitution:

In[1]:= x[t_] := c1 Exp[t] + c2 Exp[3 t] + t + 7/3;
y[t_] := -c1 Exp[t] - 3 c2 Exp[3 t] + 3 t - 1;
Simplify[5 D[x[t], t] + D[y[t], t] - 3 x[t] + y[t]]
Simplify[4 D[x[t], t] + D[y[t], t] - 3 x[t] + 3 t]

Out[3]= 0

Out[4]= 0

Your solution looks right to me as well, but I didn't go into the details too much. At least I see no mistake in the part where you find $c_3$.

In the same way, we can see that the book's answer is wrong:

In[5]:= x[t_] := c1 Exp[t] + c2 Exp[3 t] + t + 7/3;
y[t_] := -c1 Exp[t] - 3 c2 Exp[3 t] + 3 t + 1;
Simplify[5 D[x[t], t] + D[y[t], t] - 3 x[t] + y[t]]
Simplify[4 D[x[t], t] + D[y[t], t] - 3 x[t] + 3 t]

Out[7]= 2

Out[8]= 0