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I need help with solving this system of differential equations: $$ \dot{x}= \begin{pmatrix} -1 & -1 \\ -1 & -1\end{pmatrix} x+ e^{-2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$

I think it's solvable with the method of variation of parameters. But apart from calculating the Eigenvalues and Eigenvectors I don't know how to proceed. I've searched for similar tasks, but could't find anthing that I can understand. I would be very greatful if you can tell me what to do :-) !!

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  • $\begingroup$ It's always better to post what you tried ..you get more answers. $\endgroup$
    – user577215664
    Commented Jun 20, 2018 at 14:01

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$$\dot{X}= \begin{pmatrix} -1 & -1 \\ -1 & -1\end{pmatrix} X+ e^{-2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

Rewrite both equations with $x(t),y(t),\, X=\pmatrix {x(t) \\ y(t)}$ $$ \begin{cases} x'=-x-y+e^{-2t} \\ y'=-x-y \end{cases} $$

Then you can deduce that $$y'-x'=-e^{-2t}$$ Integrate $$\implies y(t)=x(t)+\frac 12e^{-2t}+K$$ Plug that in this equation and solve for $x(t)$ the differential equation $$x'=-x-y+e^{-2t}$$ $$\implies x'=-2x+\frac 12e^{-2t}-K$$ $$......$$ Can you take it from there ?

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  • $\begingroup$ thank you alot !! ..now it makes more sense to me..what i get for $ x(t)= (\frac{1}{2} +k) e^{-2t} $ $\endgroup$
    – pastapesto23
    Commented Jun 20, 2018 at 14:39
  • $\begingroup$ yw @pastapesto23 ... $\endgroup$
    – user577215664
    Commented Jun 20, 2018 at 15:25

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