Simultaneous ordinary differential equation 1

Question:
Solve for the following simultaneous differential equations:
\[ \left\{ \begin{array}{ll}
           \frac{dx}{dt}+ay = e^{bt}  & (1)\\
           \frac{dy}{dt}-ax = 0  & (2)
          \end{array}
   \right.   \]
where $a$ and $b$ are real numbers.

Answer:
Take differentiation of equation (1).
\[ x'' + ay' = be^{bt} \]
Use (2).
\[ x'' + a^2x = be^{bt} \]
The general solution of this is given when right hand side is equal to zero.
\[ x'' + a^2x = 0 \]
Thus,
\[ x(t) = C_1\cos at + C_2\sin at \]
To solve for the particular solution, due to the expression of right hand side, we can assume that it is $x=Ae^{bt}$. Then, substitute it into the above equation:
\[ (Ae^{bt})'' + a^2(Ae^{bt})' = Ab^2e^{bt} + Aa^2e^{bt} = (a^2 + b^2)Ae^{bt} \]
This must be equal to $be^{bt}$, so
\[ A = \frac{b}{a^2 + b^2} \]
The particular solution becomes $\frac{b}{a^2 + b^2}e^{bt}$. Therefore,
\[ x(t) = C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt}\]
Plug this solution into (1) to obtain the solution of $y(t)$.
\[ (C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt})'+ay=e^{bt}  \\
   -C_1a\sin at + C_2a\cos at + \frac{b^2}{a^2 + b^2}e^{bt} + ay = e^{bt} \\
  ay = C_1a\sin at - C_2a\cos at - \frac{b^2}{a^2 + b^2}e^{bt} + e^{bt}
\]
The total solution of $y$ becomes
\[  y(t) = C_1\sin at - C_2\cos at + \frac{a}{a^2 + b^2}e^{bt} \]

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