i need someone to give me exact graph of the general solution for

$$x'=\begin{bmatrix} 1 & 1\\ 4& 1 \end{bmatrix}x$$

i solved it manually , the general solution is like this

$$x(t)=c_1\begin{pmatrix} 1\\ 2 \end{pmatrix} e^{3t} +c_1\begin{pmatrix} 1\\ -2 \end{pmatrix} e^{-t} $$

and the graph will be something like this http://prntscr.com/52xmbe

I'll appreciate any help , thanks for advance

  • $\begingroup$ The eigenvectors are perpendicular in the picture, that's not right, just saying. $\endgroup$ – Git Gud Nov 4 '14 at 13:59
  • $\begingroup$ lol its like not good pic -,- but its my try , i need exact graph to work with $\endgroup$ – Bswan Nov 4 '14 at 14:05
  • $\begingroup$ @Amzoti thanks for the link , but i have matrix ao idk how would that work $\endgroup$ – Bswan Nov 4 '14 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.