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Solve this system of differential equations: $$y_1'' + y_2' + 3y_1 = e^{-x}$$ $$y_2''-4y_1' + 3y_2 = \sin(2x)$$

My try: Using $y_1' = p$ and $y_2'= q$ , we get: $$\begin{pmatrix} p' \\q'\\y_1'\\y_2'\end{pmatrix} =\begin{pmatrix} 0 & -1 & -3 & 0 \\ 4 & 0 & 0 & -3 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} p \\q\\y_1\\y_2\end{pmatrix} + \begin{pmatrix} e^{-x} \\\sin(2x)\\0\\0\end{pmatrix}$$

I solved the homogeneous part and find : $$ y_h = c_1e^{-ix}\begin{pmatrix} 1/2 \\-i\\i/2\\1\end{pmatrix} + c_2e^{ix}\begin{pmatrix} 1/2 \\i\\-i/2\\1\end{pmatrix} + c_3e^{3ix}\begin{pmatrix} -3/2 \\3i\\i/2\\1\end{pmatrix} + c_4e^{-3ix}\begin{pmatrix} -3/2 \\-3i\\-i/2\\1\end{pmatrix}$$

I don't know how to solve non-homogeneous part of the equation.

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  • $\begingroup$ @Moo Can you show the first steps please? I'm really confused because it involves lots of calculations and also complex numbers. $\endgroup$
    – S.H.W
    Jun 13, 2019 at 15:56
  • $\begingroup$ @Moo Yes but it is possible that I've made a mistake in my calculations. $\endgroup$
    – S.H.W
    Jun 13, 2019 at 16:16
  • $\begingroup$ @Moo Are you sure that it is simpler? I think that approach needs the same amount of work. $\endgroup$
    – S.H.W
    Jun 13, 2019 at 16:25
  • $\begingroup$ @Moo No , it doesn't have any initial conditions. $\endgroup$
    – S.H.W
    Jun 13, 2019 at 20:16
  • $\begingroup$ Yes, thanks a lot. $\endgroup$
    – S.H.W
    Jun 15, 2019 at 8:49

1 Answer 1

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I am going to rewrite the system as

$$\tag 1 x'' + y' + 3 x = e^{-t} \\ y'' - 4 x' + 3 y = \sin 2t $$

Using the first equation, we have

$$\tag 2 y' = -x'' - 3 x + e^{-t} \\ y'' = -x''' - 3 x' - e^{-t} \\ y''' = - x'''' - 3 x'' + e^{-t}$$

Taking the derivative of the second equation

$$\tag 3 y''' - 4 x'' + 3 y' = 2 \cos 2 t$$

Substituting $(2)$ into $(3)$

$$(- x'''' - 3 x'' + e^{-t}) - 4 x'' + 3 (-x'' - 3 x + e^{-t}) = 2 \cos 2t $$

This simplifies to

$$-x'''' - 10 x'' - 9x = 2 \cos 2 t - 4 e^{-t} $$

Solving this using your favorite method

$$x(t)= c_1 \cos 3t +c_2 \sin 3t+c_3 \cos t + c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$

Now, using the first equation and $x(t)$, we need to solve

$$\tag 4 y' = -x'' - 3 x + e^{-t} = 6 c_1 \cos 3t+6 c_2 \sin 3t-2 c_3 \cos t -2 c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$

By integrating $(4)$, we get

$$y(t) = 2 c_1 \sin 3t-2 c_2 \cos 3t-2 c_3 \sin t+2 c_4 \cos t-\dfrac{e^{-t}}{5}+\dfrac{1}{15} \sin 2t$$

You can substitute $(1)$ and $(2)$ into the original system and verify they are solutions.

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