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Solve the following system subject to the given initial data

$$x’(t)= x(t)-4y(t), x(0)=1 $$

$$y’(t)=x(t)+y(t), y(0)=1$$

So far i have gotten to the step where $$x''(t)=x'(t)-4y'(t)$$


I am not sure on how to continue after this

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Here's one way:

  1. Rewrite the second equation, solving it for $x$ in terms of $y$ and $y'$.

  2. Differentiate, giving you an equation for $x'$ in terms of $y'$ and $y''$.

  3. Substitute what you've got into the first equation: now you have an equation with just $y$ and its derivatives, and I assume you will know how to solve that.

There's another way, which involves writing the two equations as a single matrix-vector equation --- does that sound like something you are supposed to know about?

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Differentiating $y'$ we have

$$ y''=x'+y'=x-4y+y'=(y'-y)-4y+y'=2y'-5y $$

We obtain equation


The characteristic equation is $\lambda^2-2\lambda+5=0$

This equation has complex roots. If you do not know get real solutions you can see this question

Another way to solve a system is :

The solution is $e^{tA}x_0$ where

$x_0 = \begin{bmatrix} 1\\ 1 \\ \end{bmatrix}$


$A=\begin{bmatrix} 1 & -4 \\ 1 & 1 \end{bmatrix}$

If you don't know calculate $e^{tA}$, here is a simple way

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