# How can i solve this System of differential equations?

My Problem is this given system of differential equations: $$y_{1}^{\prime}=y_{1}-y_{2}$$ $$y_{2}^{\prime}=5y_{1}+3y_{2}$$ I am looking for the solution.

My Approach was: this seems to be a system of first-order differential equations. they are ordinary differential equations.

I built the corresponding matrix:

$$\underbrace{\pmatrix{ y_1^{\prime} \\ y_2^{\prime}}}_{\large{ {\vec y^{\prime}}}} = \underbrace{\pmatrix{1 & -1 \\ 5 & 3}}_{\large{\mathbf A}}\underbrace{\pmatrix{y_1\\y_2}}_{\large{\vec y}}$$ Now i need to find the eigenvalues of this matrix in order to determine the eigenvectors And here i am stuck. I failed in finding the eigenvalues. Every eigenvalue i find seems to be no number. so cannot calculate with it. But if the eigenvalues are anything other than numbers, (for example a complex number) how can i find the solution for the system of differential equations in this case?

• The eigenvalues are complex numbers. – Thomas Andrews Jul 1 '13 at 20:14
• Don't be so deterred by the fact that the eigenvalues are complex numbers :) – Ataraxia Jul 1 '13 at 20:16
• if the eigenvalues are anything other than numbers, (for example a complex number). Complex numbers are numbers. – user5402 Jul 1 '13 at 20:47

## 2 Answers

The equation $$\frac{\mathrm{d}}{\mathrm{d}t}\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}1&-1\\5&3\end{bmatrix}\begin{bmatrix}y_1\\y_2\end{bmatrix}\tag{1}$$ is correct. Note that $$\begin{bmatrix}1&-1\\5&3\end{bmatrix}^2 =4\begin{bmatrix}1&-1\\5&3\end{bmatrix} -8\begin{bmatrix}1&0\\0&1\end{bmatrix}\tag{2}$$ Solving the recurrence yields \begin{align} \begin{bmatrix}1&-1\\5&3\end{bmatrix}^k &=\frac14\begin{bmatrix}2+i&i\\-5i&2-i\end{bmatrix}(2+2i)^k\\ &+\frac14\begin{bmatrix}2-i&-i\\5i&2+i\end{bmatrix}(2-2i)^k\tag{3} \end{align} This gives us \begin{align} \exp\left(t\begin{bmatrix}1&-1\\5&3\end{bmatrix}\right) &=\frac14\begin{bmatrix}2+i&i\\-5i&2-i\end{bmatrix}e^{t(2+2i)}\\ &+\frac14\begin{bmatrix}2-i&-i\\5i&2+i\end{bmatrix}e^{t(2-2i)}\\ &=2\,\mathrm{Re}\left(\frac14\begin{bmatrix}2+i&i\\-5i&2-i\end{bmatrix}e^{(2+2i)t}\right)\\ &=\frac{e^{2t}}{2}\left(\begin{bmatrix}2&0\\0&2\end{bmatrix}\cos(2t) +\begin{bmatrix}-1&-1\\5&1\end{bmatrix}\sin(2t)\right)\tag{4} \end{align} Therefore, $$\begin{bmatrix}y_1(t)\\y_2(t)\end{bmatrix}=\frac{e^{2t}}{2}\left(\begin{bmatrix}2&0\\0&2\end{bmatrix}\cos(2t) +\begin{bmatrix}-1&-1\\5&1\end{bmatrix}\sin(2t)\right)\begin{bmatrix}y_1(0)\\y_2(0)\end{bmatrix}\tag{5}$$

• Why did you take $\Re$? – Kaster Jul 2 '13 at 1:40
• @Kaster: Because the series for the exponential is broken into two conjugate sums by $(3)$ and $z+\bar{z}=2\,\mathrm{Re}(z)$. I added an extra line to $(4)$ to hopefully make this clearer. – robjohn Jul 2 '13 at 5:40

From the first equation, we get $y_2=y_1-y'_1$. By substituting this into second equation, we get $y''_1-4y'_1+8y_1=0$, and after solving characteristic equation ($a^2-4a+8=0\Rightarrow a=2\pm 2i$), we get $$y_1=C_1e^{2x}\cos{2x}+C_2e^{2x}\sin{2x},$$ $$y_2=y_1-y'_1=-(2C_2+C_1)e^{2x}\cos{2x}+(2C_1-C_2)e^{2x}\sin{2x}.$$