Find two linearly independent solutions to the pair of coupled ODEs

$\frac{dx}{dt} = 2x + 3y$,

$\frac{dy}{ dt} =-3x+2y$.

I figured the eigenvalues/eigenvectors of the corresponding matrix to be $$e^{(2+3i)t} \quad\text{with}\quad \left[\begin{array}{r} i \\ 1 \end{array}\right] \quad\text{and} \quad e^{(2-3i)t} \quad\text{with}\quad\left[\begin{array}{r} -i \\ 1 \end{array}\right].$$ But apparently the solutions are $$\left[\begin{array}{r} x \\ y \end{array}\right] =e^{2t} \left[\begin{array}{r} \cos3t \\ -\sin3t \end{array}\right] ,\left[\begin{array}{r} x \\ y \end{array}\right] =e^{2t} \left[\begin{array}{r} \sin3t \\ \cos3t \end{array}\right].$$

Even after transforming the exponential eigenvalues to trigonometric functions I still don't get the desired results. I checked the eigenvectors/values with a computer and they seem to be correct. Please help.

  • 1
    $\begingroup$ You need to find real solutions. And it's standard trick: $\cos z = (e^{iz}+e^{-iz})/2$ $\endgroup$ – Michael Galuza Aug 16 '15 at 9:10
  • $\begingroup$ Why just real ? $\endgroup$ – user117498 Aug 16 '15 at 9:15
  • $\begingroup$ Form of solution say us about it;) Anyway, this solutions are the same, because $c_1e^{a+bt}+c_2e^{a-bt}=e^{at}[(c_1+c_2)\cos bt + i(c_1-c_2)\sin bt]$ $\endgroup$ – Michael Galuza Aug 16 '15 at 9:18
  • $\begingroup$ Sorry, i don't get how that brings me to the desired result $\endgroup$ – user117498 Aug 16 '15 at 9:25

For $\lambda = 2 + 3i$, you should get an eigenvector of: $$ \vec v = \begin{bmatrix} -i \\ 1 \end{bmatrix} $$ The two linearly independent (real) solutions are the real and imaginary parts of $e^{\lambda t}\vec v$, which can be found by applying Euler's formula: \begin{align*} e^{(2 + 3i)t}\begin{bmatrix} -i \\ 1 \end{bmatrix} &= e^{2t}(\cos 3t + i\sin 3t)\begin{bmatrix} -i \\ 1 \end{bmatrix} \\ &= e^{2t}\begin{bmatrix} -i\cos 3t + \sin 3t \\ \cos 3t + i\sin 3t \end{bmatrix} \\ &= \underbrace{e^{2t}\begin{bmatrix} \sin 3t \\ \cos 3t \end{bmatrix}}_{\text{real part}} + i \cdot \underbrace{e^{2t}\begin{bmatrix} -\cos 3t \\ \sin 3t \end{bmatrix}}_{\text{imaginary part}} \end{align*}


Notice, we have $$\frac{dx}{dt}=2x+3y\tag 1$$ $$\frac{dy}{dt}=-3x+2y\tag 2$$

  1. Diving (1) by (2), we get $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}=\frac{2x+3y}{-3x+2y}$$ $$\frac{dx}{dy}=\frac{2\frac{x}{y}+3}{-3\frac{x}{y}+2}$$ Now, let $x=uy\implies \frac{dx}{dy}=u+y\frac{du}{dy}$, setting the values we get $$u+y\frac{du}{dy}=\frac{2u+3}{-3u+2}$$ $$y\frac{du}{dy}=\frac{2u+3}{-3u+2}-u$$ $$y\frac{du}{dy}=\frac{3(1+u^2)}{2-3u}$$ $$\frac{2-3u}{3(1+u^2)}du=\frac{dy}{y}$$ $$\frac{2}{3}\int \frac{du}{1+u^2}-\frac{1}{2}\int \frac{2udu}{1+u^2}=\int \frac{dy}{y}$$ $$\frac{2}{3}\tan^{-1}(u)-\frac{1}{2}\ln(1+u^2)=\ln y+C_1$$

  2. Diving (2) by (1), we get $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3x+2y}{ 2x+3y}$$ $$\frac{dy}{dx}=\frac{-3+2\frac{y}{x}}{2+3\frac{y}{x}}$$

    Now, let $y=ux\implies \frac{dy}{dx}=u+x\frac{du}{dx}$, setting the values we get $$u+x\frac{du}{dx}=\frac{2+3u}{-3+2u}$$ $$x\frac{du}{dx}=\frac{2+3u}{-3+2u}-u$$ $$x\frac{du}{dx}=\frac{2+6u-2u^2}{-3+2u}$$ $$\frac{-3+2u}{2+6u-u^2}du=\frac{dx}{x}$$ I hope you can solve further.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.