# Solve system of two differential equations

1. Dec 22, 2013

### skrat

1. The problem statement, all variables and given/known data
Solve the system
$\dot{x}=-2x+y+5sint$ and $\dot{y}=x-2y+3$ for $x(0)=1$ and $y(0)=1$

2. Relevant equations

3. The attempt at a solution

Well, I tried expressing x from the second equation, derived it and inserted it in the first equation but this process is long. Just to be clear, I will write a few steps..:
from second equation $x=\dot{y}+2y-3$ therefore $\dot{x}=\ddot{y}+2\dot{y}$

Now inserting these two in the first expression gives me $\ddot{y}+4\dot{y}+3y=6+5sint$

Homogeneous part is: $y_h=Ae^{-t}+Be^{-3t}$

particular 1 (for the constant polynom 6): Constant C=2

particular 2 (for sin function): $y_p=\frac{1}{2}sint-cost$

All together, the solution is than: $y=Ae^{-t}+Be^{-3t}+\frac{1}{2}sint-cost+2$

and x from $x=\dot{y}+2y-3= Ae^{-t}-Be^{-3t}+\frac{1}{2}cost+2sint-3$

because $x(0)=1$ and $y(0)=1$, than $A=-B=\frac{7}{4}$

HOWEVER, I do know that this can be done using matrices... It goes somehow like this...:

$\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix}=\begin{bmatrix} -2 & 1\\ 1& -2 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}+\begin{bmatrix} 5sint\\ 3 \end{bmatrix}$

Eigenvalues $\lambda _1=-1$ and $\lambda _2=-3$ give eigenvectors for matrix $A=\begin{bmatrix} -2 & 1\\ 1& -2 \end{bmatrix}$:

$v_1=(-1,1)$ and $v_2=(1,1)$

SO, now my question:

What will the solution look like? If the system would me homogeneous it would be something like $\vec{r}=Pe^{Dt}\vec{c}$ where P is matrix from eigenvectors and D matrix from eigenvalues. C is vector of constants.
But my system also has 5sint and +3 .... What do I do now?

Thanks for all the help!

2. Dec 22, 2013

### haruspex

Not sure if it's what you're thinking of, but here's one way.
In matrix form, you have $\dot x = A x + B(t)$ where A is constant. Writing D for the differential operator, (D-A)x = B, x = -(A-D)-1B = A-1(I-A-1D)-1B.
Expanding binomially, x = A-1Æ©A-nDnB. If you write that out I think you'll see how to sum the series.

3. Dec 22, 2013

### Ray Vickson

If you know how to compute the matrix exponential $e^{uA}$ for scalar $u$ you can solve the linear system
$$\dot{x} = A x + h(t)$$
as
$$x(t) = e^{At} x(0) + \int_0^t e^{(t-s)A} h(s) \, ds.$$

The easiest way to get the matrix exponential is to note that for an nxn matrix with distinct eigenvalues $r_1, r_2, \ldots, r_n$ we have
$$f(A) = \sum_{j=1}^n E_j f(r_j)$$
for any analytical function $f(x) = c_0 + c_1 x + c_2 x^2 + \cdots$ (with $f(A)$ defined as $c_0 I + c_1 A + c_2 A^2 \cdots$). Here, the matrices $E_1, E_2, \ldots, E_n$ are the same for any function f, so we can find them by looking at some simple, convenient functions. In particular,
$$f(x) = 1 = x^0 \Longrightarrow A^0 = I = E_1 r_1^0 + E_2 r_2^0 = E_1 + E_2 \\ f(x) = x \Longrightarrow A^1 = A = E_1 r_1 + E_2 r_2$$
for the 2x2 case with two distinct nonzero eigenvalues $r_1, r_2$. Thus, we get $E_1,E_2$ by solving
$$E_1 + E_2 = I\\ r_1 E_1 + r_2 E_2 = A$$
Now we can determine the matrix exponential from
$$f(x) = e^{ux} \Longrightarrow e^{uA} = E_1 e^{u r_1} + E_2 e^{u r_2}.$$

4. Dec 22, 2013

### HallsofIvy

Staff Emeritus
Here's how I would do it. The matrix $\begin{pmatrix}-2 & 1 \\ 1 & -2 \end{pmatrix}$
has eigenvalues -1 and -3 with corresponding eigenvectors $\begin{pmatrix}1 \\ 1\end{pmatrix}$ and $\begin{pmatrix}1 \\ -1\end{pmatrix}$. That tells us that
$$\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}\begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}$$

($\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}$ is the inverse matrix to $\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}$, the matrix having the eigenvectors as columns.)

Multiplying both sides of $\begin{pmatrix}x' \\ y'\end{pmatrix}= \begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}+ \begin{pmatrix}5 sin(t) \\ 3\end{pmatrix}$ by $\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}$ gives
$$\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}x' \\ y'\end{pmatrix}= \begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}-2 & 1 \\ 1 & -2\end{pmatrix}\left(\begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2\end{pmatrix}\right)\begin{pmatrix}x \\ y\end{pmatrix}+ \begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}5 sin(t) \\ 3\end{pmatrix}[tex] where I have used the fact that [tex]\begin{pmatrix}1/2 & 1/2 \\ 1/2 & -1/2 \end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}= \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}$$ to be able to write that product in the middle.

That reduces to
$$\begin{pmatrix}(1/2)x'+ (1/2)y' \\ (1/2)x'- y'\end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}\begin{pmatrix}(1/2)x+ (1/2)y \\ (1/2)x- (1/2)y\end{pmatrix}+ \begin{pmatrix}(5/2)sin(t)+ 3/2 \\ (5/2)sin(t)- 3/2\end{pmatrix}$$

Now, let U= (1/2)x+ (1/2)y and V= (1/2)x- (1/2)y and that becomes
$$\begin{pmatrix}U' \\ V' \end{pmatrix}= \begin{pmatrix}-1 & 0 \\ 0 & -3\end{pmatrix}\begin{pmatrix}U \\ V \end{pmatrix}+ \begin{pmatrix}((1/2) sin(t)+ 3/2 \\ (1/2) sin(t)- 3/2\end{pmatrix}$$
That is, we have uncoupled the two equations. We can now write them as $U'= -U+ (1/2)sin(t)+ 3/2$ and $V'= -3V+ (1/2)sin(t)- 3/2$.

Once you have solved those, of course
$$\begin{pmatrix}x \\ y \end{pmatrix}= \begin{pmatrix}1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix}U \\ V\end{pmatrix}$$

5. Dec 23, 2013

### skrat

Big thanks to everybody!

I was able to get the same result. :)