# Verify the real solution of a linear system of differential equation

I'm trying to solve $Y' = AY$ where $A= \left[ \begin{array}{ c c } -2 & 6 \\ -3 & 4 \end{array} \right]$

I have found the eigenvalue $1 \pm 3i$ with eigenvector for $1+3i:$

$v = \left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right]$

Which seems to be correct by testing $Av = (1+3i)v$ but when I try to write it as a real solution I don't seem to get the right answer.

$$\left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right](\cos 3t + i\sin 3t) = \left[ \begin{array}{ c c } \cos 3t + \sin 3t - i\cos 3t + i\sin 3t \\ \cos 3t + i\sin 3t \end{array} \right]$$

$$v_1 = \left[ \begin{array}{ c c } \cos 3t + \sin 3t \\ \cos 3t \end{array} \right],\ v_2 = \left[ \begin{array}{ c c } \cos 3t + \sin 3t \\ \sin 3t \end{array} \right]$$

If I now verify by $v_1' = Av_1$

$$\left[ \begin{array}{ c c } 3(\cos 3t - \sin 3t) \\ -3\sin 3t \end{array} \right] \neq \left[ \begin{array}{ c c } 4\cos 3t - 2\sin 3t \\ \cos 3t - 3\sin 3t \end{array} \right]$$

You are still missing an exponential term, we have:

$e^{\lambda t}v_1 = e^{(1+3i)t}\begin{bmatrix}1-i\\1\end{bmatrix} = e^te^{3it}\begin{bmatrix}1-i\\1\end{bmatrix} = e^t(\cos 3t + i \sin 3t)\begin{bmatrix}1-i\\1\end{bmatrix} = \begin{bmatrix}e^t(\sin 3t + \cos 3t+i (\sin 3 t-\cos 3 t))\\ e^t(\cos 3t + i \sin 3t) \end{bmatrix}$

So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):

$$Y(t) = c_1 e^t\begin{bmatrix}\sin 3t + \cos 3t\\ \cos 3t \end{bmatrix}+ c_2e^t\begin{bmatrix}\sin 3t - \cos 3t\\ \sin 3t \end{bmatrix}$$

• @BabakS.: It takes me 5x longer to post answers than it does to find them! :-) Thank you my friend! – Amzoti Jun 8 '13 at 7:19
• @Amzoti: Isee. You know, sometimes, I learn some noble points of your solutions here. Believe me or not. – mrs Jun 8 '13 at 7:21
• @BabakS.: You are too kind my friend! Have a great day! – Amzoti Jun 8 '13 at 13:16

The last term in the first component in the vector, you have no $i$. Here is the correction

$$\left[ \begin{array}{ c c } 1-i \\ 1 \end{array} \right](\cos 3t + i \sin3t) = \left[ \begin{array}{ c c } \cos 3t + \sin 3t - i\cos3t + i\sin 3t \\ \cos 3t + i \sin 3t \end{array} \right] .$$

$$\lambda_1=1+3\,i,\, v_1= (1-i,1)$$

$$\lambda_1=1-3\,i,\, v_1= (1+i,1)$$

Note that, to find the general solution you need to consider the two eigenvectors. It seems you are considering only one. The general solution has the form

$$Y(t) = c_1v_1 e^{\lambda_1t} + c_2 v_2 e^{\lambda_2 t}.$$

Have you given initial conditions? If yes, then you need to use them to find the constants $c_1$ and $c_2$.

• I'm not sure if i follow $(1-i)(cos 3t + i sin 3t) = cos 3t + i sin 3t - i cos 3t -i^2 sin 3t = cos 3t + sin 3t + i(sin 3t - cos 3t)$ – shardy Jun 8 '13 at 0:12
• @shardy: $(1-i)( a+ib )=( a+ib -i(a+ib) ) = a+ib-ia-i^2b=a+ib-ia+b.$ – Mhenni Benghorbal Jun 8 '13 at 0:17
• @shardy: You are right. – Mhenni Benghorbal Jun 8 '13 at 0:22
• @shardy: Let me check if you got the right eigenvalues. – Mhenni Benghorbal Jun 8 '13 at 0:23