# Systems of differential equations...

• May 9th 2009, 09:43 PM
Raidan
Systems of differential equations...
Determine the General solution to the differential Equation system X'=AX
A=
0 -4
4 0
• May 10th 2009, 05:32 AM
Jester
Quote:

Originally Posted by Raidan
Determine the General solution to the differential Equation system X'=AX
A=
0 -4
4 0

For this problem $\displaystyle x' = -4y, y' = 4x$, it's actually easier to eliminate a variable, say y, giving

$\displaystyle x'' + 16 x = 0$

giving the solution $\displaystyle x = c_1 \sin 4 x + c_2 \cos 4x$ then use $\displaystyle y = - \frac{x'}{4}$ to find $\displaystyle y$.
• May 10th 2009, 06:19 AM
HallsofIvy
For a simple problem like this, danny arrigo's suggestion is best, but you can do as a matrix problem.

Find the eigenvalues of $\displaystyle A= \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}$. It should be easy to see that they are 4i and -4i. Since those are distinct, the matrix can be diagonalize (over the complex numbers).

With eigenvalue 4i, we have $\displaystyle \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 4ix & 4iy\end{bmatrix}$ which gives the two equations -4y= 4ix and 4x= 4iy. Those both reduce to y=ix so one eigenvector is <1, i>.

With eigenvalue -4i, we have $\displaystyle \begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} -4ix & -4iy\end{bmatrix}$ which gives the two equations -4y= -4ix and 4x= -4iy. Those both reduce to y=-ix so another, independent, eigenvector is <1, i>.

If we use those as columns in matrix P, then $\displaystyle P= \begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix}$ and $\displaystyle P^{-1}= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}$

Then it is easy to see that $\displaystyle P^{-1}AP= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}\begin{bmatrix}0 & -4 \\ 4 & 0\end{bmatrix}\begin{bmatrix}1 & 1 \\ i & -i\end{bmatrix}= \begin{bmatrix}4i & 0 \\0 & -4i\end{bmatrix}$.

If we multiply, on the left, by the constant matrix $\displaystyle P^{-1}$, we have $\displaystyle P^{-1}X'= (P^{-1}X)'= P^{-1}AX$.

Let $\displaystyle Y= P^{-1}X$ so that $\displaystyle X= PY$ and substitue PY for X in the equation: $\displaystyle Y'= P^{-1}A(PY)= (P^{-1}AP) Y$ so our differential equation becomes $\displaystyle Y'= \begin{bmatrix}2i & 0 \\ 0 & -2i\end{bmatrix}Y$.

The general solution to that, just like any y'= ay, is $\displaystyle Y= Ce^{\begin{bmatrix}2it & 0 \\0 & -2it\end{bmatrix}}$ and because that is a diagonal matrix it is easy to show that we get $\displaystyle Y= C\begin{bmatrix}e^{2it} & 0 \\ 0 & e^{-2it}\end{bmatrix}$

$\displaystyle = \begin{bmatrix}C_1 & C_2\end{bmatrix}\begin{bmatrix}e^{2it} & 0 \\ 0 & e^{-2it}\end{bmatrix}$

$\displaystyle = \begin{bmatrix}C_1e^{2it}\\ C_2e^{-2it}\end{bmatrix}$
Now we use $\displaystyle X= P^{-1}Y$ to write $\displaystyle X(t)= \begin{bmatrix}\frac{1}{2} & -\frac{1}{2}i \\ \frac{1}{2} & \frac{1}{2}i\end{bmatrix}\begin{bmatrix}C_1e^{2it} \\ C_2e^{-2it}\end{bmatrix}$$\displaystyle = \begin{bmatrix}\frac{C_1}{2}e^{2it}- \frac{C_2}{2}e^{-2it} \\ \frac{C_1}{2}e^{2it}+ \frac{C2}{2}e^{-2it}\end{bmatrix}$.

And, then use the fact that $\displaystyle e^{2it}= cos(2t)+ i sin(2t)$, $\displaystyle e^{-2it}= cos(t2)- i sin(2t)$ to change into sine and cosine, getting danny aririgo's answer.

Whew! Well, I said that, for a simple problem like this danny arrigo's solution was best!
• May 10th 2009, 06:36 AM
Raidan
Ahh.. thanx guys..I got it now..(Nod)