Systems of Linear Differential Equations - Is this Correct? I have to solve the following first-order linear system, $x(t)$ represents one population and the $y(t)$ represents another population that lives in the same ecosystem: (Note: $'$ denotes prime)
Note: $x(t)$ and $y(t)$ must be found using the Reduction method.
$x'=−5x−20y$ (Equation 1)
$y'=5x+7y$ (Equation 2)
I start off with finding the derivative of (Equation 1) which gives me:
$x''=−5x'−20y'$
and then substitute $y'$ using the equations above (Equation 2):
$x''=−5x'−20(5x+7y)$
foil expansion:
$x''=−5x'−100x−140y$ (Equation 3)
so I find by by using (Equation 1):
$$
\begin{align}
x' &=-5x-20y\\
x'+5x &= -20y\\
\frac{x'+5x}{-20} &= y\\
y&=- \frac 1 {20}(x'+5x)
\end{align}$$
After finding this y, I insert it into (Equation 3)
$x''=-5x'-100x-140(-\frac 1 {20}(x'+5x))$
And expand thus getting:
$$x''=-5x'-100x+7x'+35x$$
Collect 'like' terms and put on LHS.
$$x''-2x'+65x = 0$$
Then put into auxillary form: 
$$r^(2)−2r+65=0$$ 
Find roots:
$$r=1±8i$$
I can now say that $x(t)= C_1 e^t \sin8t + C_2 e^t \cos8t$ (Note: $\frac {C_1} {C_2}$ are constants)
Which I simplified to: 
$$x(t) = e^t(C_1 \sin8t + C_2 \cos8t)$$
Now I need $y(t)$. I will begin with finding the derivative of $x(t)$.
$$
x'(t) = e^t((C_1-8C_2)\sin8t + (8C_1+C_2)\cos8t)$$
$$x'(t) = e^t (C_1 \sin8t - 8C_2 \sin8t + 8C_1 \cos8t + C_2 \cos8t)$$
I input this into the original equation (Equation 1):
$x'=3x-5y$ now becomes 
$$e^t C_1 \sin8t - e^t 8C_2\sin8t + e^t8C_1 \cos8t + e^t C_2 \cos8t = -5 (e^t C_1 \sin8t + e^t C_2 \cos8t - 20(b(t))$$
I expand, collect 'like' terms, find LCD, and determine that:
$$b(t) = \dfrac{4C_2 e^t \sin8t - 4 C_1 e^t \cos8t - 3 C_1 e^t \sin8t - 3 C_2 e^t \cos8t}{10} $$
My question: Are my $x(t)$ and $y(t)$ correct using the methods shown above (which is represented using an example in my book).
 A: We have:
$$x(t) = e^{t}(c_1 \sin(8t) + c_2 \cos(8t))$$
Now, we want to solve:
$$y(t) = -\dfrac{1}{20}\left(x'(t) + 5 x(t)\right)$$
We get:


*

*$x'(t) = e^t (c_1 \sin (8 t)+ c_2 \cos (8 t))+e^t (8 c_1 \cos (8 t)-8 c_2 \sin (8 t))$

*$5 x(t) = 5 e^t (c_1 \sin (8 t) + c_2 \cos (8 t)) $

*$x'(t) + 5 x(t) = 6 e^t (c_1 \sin (8 t)+c_2 \cos (8 t))+e^t (8 c_1 \cos (8 t)-8 c_2 \sin (8 t))$

*$y(t) = -\dfrac{1}{20}\left(x'(t) + 5 x(t)\right) =$


$$-\dfrac{1}{20}(6 e^t (c_1 \sin (8 t)+c_2 \cos (8 t))+e^t (8 c_1 \cos (8 t)-8 c_2 \sin (8 t)))$$
Simplification yields:
$$y(t) = -\dfrac{1}{10} e^t ((3 c_1-4 c_2) \sin (8 t)+(4 c_1+3 c_2) \cos (8 t))$$
Now, you can verify that these solutions simultaneously solve your system by calculating $x'(t)$ and verifying that it equals $(−5x(t)−20y(t))$ and finding $y'(t)$ and verifying that it equals $(5x(t)+7y(t))$.
The more general approach to solve these problems is the matrix approach with eigenvalues and eigenvectors.
A: Related problem. Here is the technique I referred to in the link in my comment. the system 
$$ x'=-5x-20y \\ 
y'= 5 x + 7 y $$  
can be written as
$$\implies Dx+5x+20y=0 \\ 
\quad \quad Dy-7y -5x = 0, $$
where $D=\frac{d}{dt}$. Rearranging the above system yields
$$\implies (D+5)x+20y=0 \\ 
\quad \quad  -5x + (D-7)y = 0 \longrightarrow (*). $$
Multiplying the first equation in $(*)$ by $5$ and applying the operator $D+5$ to the second equation in $(*)$, then adding the two equations results in a second order ode in $y$. 
I think you can carry on the calculations.    
