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I want to find $x,y: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$x'(t)= -3x(t)+4y(t)$

$y'(t) = -x(t) + y(t)$

with the initial conditions $x(0) = y(0) = 1$

$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ -1 & 1 \end{bmatrix} \times \begin{bmatrix} x \\ y \end{bmatrix} $

Let the $A= \begin{bmatrix} -3 & 4 \\ -1 & 1 \end{bmatrix}$

$det(A-\lambda I) = (1+\lambda)^2 = 0 \implies \lambda_1 = \lambda_2 = -1$

The eigenvectors are $V_{\lambda_1} = V_{\lambda_2} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

From there I am stuck and I don't know what to do

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  • $\begingroup$ The eigenvector is wrong: try to multiply $A$ with it. $\endgroup$ – N74 May 10 '16 at 8:28
  • $\begingroup$ How to know it's a Cauchy problem? $\endgroup$ – Pranita Gupta Jan 30 at 15:04
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Using Laplace Transform, we can say that:

  • $$x'(t)=-3x(t)+4y(t)\Longleftrightarrow$$ $$\mathcal{L}_t\left[x'(t)\right]_{(s)}=\mathcal{L}_t\left[-3x(t)+4y(t)\right]_{(s)}\Longleftrightarrow$$ $$sx(s)-x(0)=-3x(s)+4y(s)\Longleftrightarrow$$ $$sx(s)-1=-3x(s)+4y(s)\Longleftrightarrow$$ $$sx(s)+3x(s)=4y(s)+1\Longleftrightarrow$$ $$x(s)\left[s+3\right]=4y(s)+1\Longleftrightarrow$$ $$x(s)=\frac{4y(s)+1}{s+3}$$
  • $$y'(t)=-x(t)+y(t)\Longleftrightarrow$$ $$\mathcal{L}_t\left[y'(t)\right]_{(s)}=\mathcal{L}_t\left[-x(t)+y(t)\right]_{(s)}\Longleftrightarrow$$ $$sy(s)-y(0)=-x(s)+y(s)\Longleftrightarrow$$ $$sy(s)-1=-x(s)+y(s)\Longleftrightarrow$$ $$sy(s)-y(s)=1-x(s)\Longleftrightarrow$$ $$y(s)\left[s-1\right]=1-x(s)\Longleftrightarrow$$ $$y(s)=\frac{1-x(s)}{s-1}$$

So, we can see that:

  • $$x(s)=\frac{4\cdot\frac{1-x(s)}{s-1}+1}{s+3}\Longleftrightarrow x(s)=\frac{s+3}{(s+1)^2}$$
  • $$y(s)=\frac{1-\frac{4y(s)+1}{s+3}}{s-1}\Longleftrightarrow y(s)=\frac{s+2}{(s+1)^2}$$

With inverse Laplace Transform, we can find:

  • $$x(t)=(1+2t)e^{-t}$$
  • $$y(t)=(1+t)e^{-t}$$
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  • $\begingroup$ There is only one eigenvector namely $[2,1]^t$ so we can not use the method mentioned in question? $\endgroup$ – Pranita Gupta Jan 30 at 15:27

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