Solve the system $\vec{x^{'}}=\begin{pmatrix}2 & -5\\1 & -2 \end{pmatrix}\vec{x}+ \begin{pmatrix} -\cos t\\ \sin t \end{pmatrix}$

The Eigenvalues are $(2-\lambda)(-2-\lambda)+5=0 \implies \lambda=\pm i$

The Eigevalue we choose is $\lambda=i$ and get a fundamental matrix of $$ \phi(t)=\begin{pmatrix} 5\cos t & 5\sin t\\ 2\cos t + \sin t & 2\sin t - \cos t \end{pmatrix}$$ We find the inverse by \begin{gather*} \phi^{-1}(t)=\dfrac{1}{5\cos t(2\sin t-\cos t)- 5\sin t(2 \cos t+ \sin t)} \begin{pmatrix} 2\sin t - \cos t & -5\sin t\\ -2 \cos t - \sin t & -5\cos t \end{pmatrix}=\\[8pt]\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & 5\cos t \end{pmatrix} \end{gather*}

We multiply by $g(t)$ and have \begin{gather*}\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & 5\cos t \end{pmatrix}\begin{pmatrix} -\cos t\\ \sin t \end{pmatrix}=\begin{pmatrix} -\cos^2 t+2\sin t\cos t +5 \sin^2 t\\ -2\cos^2 t -2\cos t \sin t +5\cos t \sin t \end{pmatrix}=\\[8pt]\begin{pmatrix} 5 \sin^2 t-\cos^2 t +2 \sin t \cos t\\ -2\cos^2 t +3\sin t \cos t \end{pmatrix}\end{gather*}

Integrating all of this yields $$\begin{pmatrix} \frac{5}{2}t - \frac{5}{4}\sin 2t - \frac{1}{2}t -\frac{1}{4}\sin 2t + \sin^2 t\\[8pt] -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t\end{pmatrix}=\begin{pmatrix} 2t-3\sin 2t +\sin^2 t\\ -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t \end{pmatrix}$$

Multiply by $\phi(t)$ and we have \begin{gather*}\begin{pmatrix} 5\cos t & 5\sin t\\ 2\cos t + \sin t & 2\sin t - \cos t \end{pmatrix}\begin{pmatrix} 2t-3\sin 2t +\sin^2 t\\ -t-\frac{1}{2}\sin 2t+\frac{3}{2}\sin^2 t \end{pmatrix}=\end{gather*} $$\begin{pmatrix} 10t\cos t -15 \sin 2t \cos t +5 \sin^2 t \cos t-5t\sin t -\frac{5}{2}\sin 2t \sin t+\frac{15}{2}\sin^3 t \\ 4t \cos t +2t \sin t -6 \sin 2t \cos t -3 \sin t \sin 2t+2\cos t \sin^2 t + \sin^3 t -2t\sin t +2t\cos t -\sin 2t \sin t+\frac{1}{2}\sin 2t \cos t +3\sin^3 t-\frac{3}{2}\sin^2 t \cos t \end{pmatrix} \dfrac{1}{5}$$

The book gets $x_h+\begin{pmatrix}2\\1 \end{pmatrix} t \cos t -\begin{pmatrix}1\\ 0 \end{pmatrix} t \sin t$

So how did they simplify all that?

  • $\begingroup$ I just realized I forgot to multiply by the integral of the bottom row of the matrix. However it doesn't change the fact that it is still complicated as hell. They got $c_1\begin{pmatrix} 5\cos t\\2\cos t+\sin t \end{pmatrix} +c_2\begin{pmatrix} 5\sin t \\ -\cos t +2 \sin t \end{pmatrix}$ $\endgroup$ – adam Apr 25 '14 at 2:30
  • $\begingroup$ So I reworked it with the missing factor... it makes it look worse actually heh $\endgroup$ – adam Apr 25 '14 at 2:48
  • $\begingroup$ dont tell me I messed up at the beggining $\endgroup$ – adam Apr 25 '14 at 3:06
  • $\begingroup$ does it have to do with me taking the inverse and missing a sign? $\endgroup$ – adam Apr 25 '14 at 3:08

Note: the author used $\lambda = i$ and $v_1 = (5, 2-i)$ for the calculations.

You have a sign issue:

$$\phi^{-1}(t)=\dfrac{1}{5}\begin{pmatrix} \cos t-2\sin t & 5\sin t\\ 2 \cos t + \sin t & -5\cos t \end{pmatrix}$$

Look at the sign of entry $\phi^{-1}(t)_{22}$, which should be a negative.

Multiplying by $g(t)$, we get:

$$\left( \begin{array}{c} \sin ^2(t)-\frac{1}{5} \cos (t) (\cos (t)-2 \sin (t)) \\ -\cos (t) \sin (t)-\frac{1}{5} \cos (t) (2 \cos (t)+\sin (t)) \\ \end{array} \right)$$

Integrating, we get:

$$\left( \begin{array}{c} \frac{2 t}{5}-\frac{1}{10} \cos (2 t)-\frac{3}{10} \sin (2 t) \\ \frac{3 \cos ^2(t)}{5}-\frac{1}{10} \sin (2 t)-\frac{t}{5} \\ \end{array} \right)$$

Expanding the trig terms, we get:

$$\left( \begin{array}{c} -\frac{1}{10} \cos ^2(t)-\frac{3}{5} \sin (t) \cos (t)+\frac{\sin ^2(t)}{10}+\frac{2 t}{5} \\ \frac{3 \cos ^2(t)}{10}-\frac{1}{5} \sin (t) \cos (t)-\frac{t}{5}-\frac{3 \sin ^2(t)}{10}+\frac{3}{10} \\ \end{array} \right)$$

Multiplying by $\phi(t)$, we get:

$$\left( \begin{array}{c} \left(2 t-\frac{1}{2}\right) \cos (t)-t \sin (t) \\ \frac{1}{10} (2 (5 t-4) \cos (t)+\sin (t)) \\ \end{array} \right)$$

Of course the $\cos t$ and $\sin t$ terms get absorbed into the constants and you are left with the author's result.

  • $\begingroup$ WOW! really like man all that work... $\endgroup$ – adam Apr 25 '14 at 3:10
  • $\begingroup$ I am also covering trajectories with non-linear system and autonomous systems. The only think I'm having issues with is how to connect each trajectory to the different critical points. Do you know any way to check to see if your trajectory is right by having something graph it? $\endgroup$ – adam Apr 25 '14 at 3:15
  • $\begingroup$ No for non-linear autonomous systems. Also I am still getting a complicated answer despite the sign change it doesn't look like its changing much $\endgroup$ – adam Apr 25 '14 at 3:21
  • $\begingroup$ how are you going to draw a phase portrait in latex? $\endgroup$ – adam Apr 25 '14 at 3:30
  • $\begingroup$ when you integrated what idenity did you use??? because I had double angles everywhere $\endgroup$ – adam Apr 25 '14 at 4:01

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