Question:

Solve for the following simultaneous differential equations:

\[  (1): \frac{dx}{dt}+ay = e^{bt}   \]

\[   (2): \frac{dy}{dt}-ax = 0     \]

where \(a\) and \(b\) are real numbers.

Answer:

Take derivative of equation (1).

\[ x” + ay’ = be^{bt} \]

Use (2) to replace \(y’\).

\[ x” + a^2x = be^{bt} \]

The general solution of this is given when right hand side is equal to zero as follows:

\[ x” + a^2x = 0 \]

Thus, this solution is:

\[ x(t) = C_1\cos at + C_2\sin at \]

To solve for the particular solution, due to the expression of the right hand side, we can assume that it is \(x=Ae^{bt}\). Then, substitute it into the above equation: \(x” + a^2x = 0\)

\[ (Ae^{bt})” + a^2(Ae^{bt})’ = Ab^2e^{bt} + Aa^2e^{bt} = (a^2 + b^2)Ae^{bt} \]

This must be equal to \(be^{bt}\), so

\[ A = \frac{b}{a^2 + b^2} \]

Since \(x=Ae^{bt}\), the particular solution becomes \(\frac{b}{a^2 + b^2}e^{bt}\). Therefore, the total solution is:

\[ x(t) = C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt}\]

Plug this solution into (1) to obtain the solution of \(y(t)\).

\[ (C_1\cos at + C_2\sin at + \frac{b}{a^2 + b^2}e^{bt})’+ay=e^{bt}  \]

\[ -C_1a\sin at + C_2a\cos at + \frac{b^2}{a^2 + b^2}e^{bt} + ay = e^{bt} \]

\[  ay = C_1a\sin at – C_2a\cos at – \frac{b^2}{a^2 + b^2}e^{bt} + e^{bt}  \]

Calculate the last two terms, and divide both sides by a. The constant a will be included into \(C_1\) and \(C_2\).

The total solution of \(y\) becomes

\[  y(t) = C_1\sin at – C_2\cos at + \frac{a}{a^2 + b^2}e^{bt} \]