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I have the differential system

$x'=-2x+5y$

$y'=-2x+4y$

I calculated the eigenvectors but they turned out to be imaginary. I want the general solution to this system in terms of sine and cosine but I have no idea how to convert to this from complex numbers.

Anything helps. Thanks.

John Smith
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  • You should include your work so that we can see specifically what step you're stuck on. In general, Euler's formula and the properties of the exponential function give $$e^{a+bi}=e^a\left(\cos(b)+i\sin(b)\right)$$ – Dave Dec 10 '17 at 18:46
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    Hint: If the eigenvalues are $a\pm ib$, then the coefficient matrix is similar to $\tiny{\begin{bmatrix}a&-b\b&a\end{bmatrix}}$. – amd Dec 10 '17 at 18:56

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You don’t really need the eigenvectors to generate a solution to this system. The characteristic polynomial of the coefficient matrix $A$ is $\lambda^2-2\lambda+2 = (\lambda -1)^2+1$, therefore $(A-I)^2+I=0$ by Cayley-Hamilton. Set $B=A-I$. Since the identity matrix commutes with everything $e^{tA}=e^{t(I+B)}=e^te^{tB}$. Expand the power series of $e^{tB}$ and use the fact that $B^2=-I$: $$\begin{align} e^{tB} &= I+tB+{t^2\over2!}B^2+{t^3\over3!}B^3+{t^4\over4!}B^4+\cdots \\ &= I + tB-{t^2\over2!}I-{t^3\over3!}B+{t^4\over4!}I+\cdots\end{align}$$ Gathering up terms involving $I$ and $B$, we find that the coefficient of $I$ is the power series for $\cos t$ and the coefficient of $B$ is the power series for $\sin t$. Putting all this together, $$e^{tA}=e^t\cos(t)\,I+e^t\sin(t)\,B = e^t\begin{bmatrix}\cos t-3\sin t & 5\sin t \\ -2\sin t & \cos t + 3\sin t\end{bmatrix}.$$

More generally, you can use a similar calculation to find that if the eigenvalues of a $2\times2$ matrix $A$ are $\alpha\pm i\beta$, then $$e^{tA} = e^{\alpha t}\left(\cos{\beta t} \, I+{\sin{\beta t}\over\beta}B\right),$$ where $B=A-\alpha I$.

That this looks a lot like $e^{\alpha+i\beta}$ is no coincidence. Under matrix multiplication, matrices of the form $\tiny{\begin{bmatrix}\alpha&-\beta\\\beta&\alpha\end{bmatrix}}$ (conformal matrices) are isomorphic to the complex numbers, and a $2\times2$ real matrix with complex eigenvalues is similar to such a matrix.

amd
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For any linear homogeneous ODE system ($\hat{A}\in\mathbb{R}^{n\times n}$) of dimension $n$ on domain $S\subseteq\mathbb{R}$: $$ \frac{\text{d}Y(t)}{\text{d}t} = \hat{A}Y(t)\tag{1} $$ If some solution curve $Y_C(t)$ is the general complex solution of $(1)$, then so is the linear combination of its real and complex parts: $$ Y_R(t) = a_1\Re\{Y_C(t)\} + a_2\Im\{Y_C(t)\};\quad a_1,a_2\in\mathbb{R} $$ Which is a real-valued solution curve $Y_R:S\rightarrow \mathbb{R}^n$.

This can be shown by first proving that the real and complex parts of the eigenvalues are linearly independent (See proof) and then exploiting the linearity of the system.

G B
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