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I'm working on the system:

$dx/dt=x-2y$, $x(0)=-1$

$dy/dt=5x-y$, $y(0)=6$

This seems like it should be easy to solve and yet I'm struggling. I am trying to use the Laplace transform and I have done this to every term but I get stuck when you I need to put everything in terms of $X(s)$ and $Y(s)$.

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  • $\begingroup$ Why have you chosen to use the Laplace transform for this question? $\endgroup$
    – Theo C.
    Nov 16, 2017 at 21:27

1 Answer 1

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We are given

$$\tag 1 x'=x-2y \\ y' =5x-y\\ x(0)=-1, ~~y(0)=6$$

We will use the Laplace Transform and Paul's Online Math Notes as a guide.

Taking the Laplace transform of $(1)$ yields:

$$\begin{align} s X(s) - x(0) &= X(s) - 2 Y(s) \\ s Y(s) - y(0) &= 5X(s) - Y(s) \end{align}$$

This reduces to the system:

$$\begin{bmatrix} s-1 & 2 \\ -5 & s+1 \end{bmatrix} \begin{bmatrix} X(s) \\ Y(s)\end{bmatrix} = \begin{bmatrix} -1 \\ 6\end{bmatrix}$$

Next, solve for $X(s)$ and $Y(s)$ and then find the inverse Laplace Transform.

Can you proceed?

You should arrive at

$$x(t) = - \cos (3 t) - \dfrac{13}{3} \sin (3 t) \\ y(t) = 6 \cos (3 t) -\dfrac{11}{3} \sin (3 t)$$

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