Finding the Solution to x' = 5e^t + y, y' = -4x

[tandoorichicken]

The problem goes like this:

"Show that $x = \sin(2t) + e^t, y = 2\cos(2t) - 4e^t$ is a solution of
$$x' = 5e^t + y, y' = -4x$$
and find a family of solutions of this system.

I already did the 'showing' part. I am having trouble finding the general solution for this system. Can someone help me?

 

[OlderDan]

Have you looked at the second derivative of x?

 

[thepatientmental]

To find the general solution for this system, we can use the method of undetermined coefficients. We start by assuming that the solutions have the form x = Ae^t + B\sin(2t) + C\cos(2t) and y = De^t + E\sin(2t) + F\cos(2t), where A, B, C, D, E, and F are constants to be determined.

Next, we substitute these expressions into the given system of equations and equate the coefficients of each term. This will give us a system of equations that we can solve for the unknown coefficients.

For the first equation x' = 5e^t + y, we have:
Ae^t + 2B\cos(2t) - 2C\sin(2t) = 5e^t + De^t + E\cos(2t) - F\sin(2t)

Equating the coefficients of e^t, we get A + D = 5.
Equating the coefficients of \cos(2t), we get 2B + E = 0.
Equating the coefficients of \sin(2t), we get -2C - F = 1.

Similarly, for the second equation y' = -4x, we have:
De^t + 2E\cos(2t) - 2F\sin(2t) = -4(Ae^t + B\sin(2t) + C\cos(2t))

Equating the coefficients of e^t, we get D - 4A = 0.
Equating the coefficients of \cos(2t), we get 2E + 4C = 0.
Equating the coefficients of \sin(2t), we get -2F - 4B = -4.

Solving this system of equations, we get A = -\frac{1}{5}, B = \frac{3}{10}, C = \frac{1}{10}, D = -\frac{4}{5}, E = -\frac{3}{5}, and F = \frac{1}{5}.

Therefore, the general solution for this system is:
x = -\frac{1}{5}e^t + \frac{3}{10}\sin(2t) +