Question

Simultaneous Differential Equation
x" +y" + 3x = 15(e^-t)
y" - 4x' +3y = 15sin2t

x(0) = 7
x'(0) = 0
y(0) = 0
y'(0) = 7

Answer 1

So what I did was find out that if you make a basis: {e^-t, sin(2t), cos(2t)} you have these two equations equal to the vectors: \[\left(\begin{matrix}15 \\ 0 \\ 0\end{matrix}\right) and\left(\begin{matrix}0\\ 15 \\ 0\end{matrix}\right)\] so these just represent the coefficients on e^-t, in the first vector and the coefficient on sin(2t)

Now find the derivative matrix for these things:

\[D=\left[\begin{matrix}-1 & 0 & 0 \\ 0 & 0 & -2 \\0 & 2 & 0 \end{matrix}\right]\]

So if you multiply this by a vector like \[\left(\begin{matrix}1 \\ 2 \\ 3\end{matrix}\right)\] this is like taking the derivative of:

\[1*e^{-t}+2*\sin(2t)+3*\cos(3t)\] so show that to make sure you follow what I'm talking about real quick. You should get the vector \[\left(\begin{matrix}-1 \\ -6 \\ 4\end{matrix}\right)\]

But that's just an example I made up to show you that this truly is the derivative matrix for our basis so you understand. Now we can plug this all in using x and y as vectors:

\[D^2 \bar x +D^2 \bar y + 3I \bar x=\left(\begin{matrix}15 \\ 0 \\ 0\end{matrix}\right)\]\[D^2 \bar y -4 D \bar x + 3I \bar y=\left(\begin{matrix}0 \\ 15 \\ 0\end{matrix}\right)\]

So now try to solve it. =)

Answer 2

There might be an easier way to solve this honestly... hahaha maybe undetermined coefficients is what you wanted to see, does that sound like stuff you've heard about? Sorry i just sort of jumped the gun and just did something.

Answer 3

\[\left[\begin{matrix}D^2+3I & D^2 \\ -4D & D^2+3I\end{matrix}\right]\left(\begin{matrix}\bar x \\ \bar y\end{matrix}\right)=\left(\begin{matrix}\bar u \\ \bar v\end{matrix}\right)\]

Here I just called those two vectors u and v for the ones representing 15e^-t and 15sin(2t).

Just invert this nice 2x2 matrix to get a simple formula for calculating x or y. Tah dah!