# Solving differential equations through matrix

by devoured_elysium
Tags: differential, equations, matrix, solving
 P: 15 Hello I'd like to know how to solve the following equation with matrix, if possible at all: d ( x^2 ) / dt^2 + w^2 x = 0 I know how to solve it without having to use a matrix, but I heard it is possible to do it with matrix. How about doing it? How is this method called? Thanks
 PF Patron Sci Advisor Thanks Emeritus P: 38,412 Is that d ( x^2 ) / dt^2 + w^2 x = 0 or d^2 x/dt^2+ w^2x= 0? I'm going to assume it is the latter. Define y= dx/dt so dy/dt= d^2x/dt^2 and the equation becomes dy/dt= -w^2x. You now have the two equations dx/dt= y and dy/dt= -w^2x. If you write $$X= \left(\begin{array}{c} x \\ y\end{array}\right)$$ Then the two equations become the single matrix equation $$\frac{dX}{dt}= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)X$$ To solve that, find the eigenvalues of the coefficient array (they are $\pm w i$). The general solution then can be written as exponentials of those eigenvalues times t or, since they are imaginary, sine and cosine.
 P: 15 it's the latter one, as you thought. thanks by the response!
P: 15

## Solving differential equations through matrix

Hi again

Sorry but I could not follow the step X=(x, y) and why then it becomes the next equation.
 PF Patron Sci Advisor Thanks Emeritus P: 38,412 $$\left(\begin{array}{c}\frac{dx}{dt} \\ \frac{dy}{dt}\end{array}\right)= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)$$ Do you see how the matrix multiplication on the right works out?
Mentor
P: 13,612
 Quote by HallsofIvy $$\left(\begin{array}{c}\frac{dx}{dt} \\ \frac{dy}{dt}\end{array}\right)= \left(\begin{array}{ccc}0 && 1 \\-w^2 && 0\end{array}\right)\left(\begin{array}{c}x \\ y\end{array}\right)$$
Taking this one step further, define $\mathbf X$ and $\mathbf A$ as
$$\begin{array}{rl} \mathbf X &\equiv \bmatrix x\\y\endbmatrix \\[12pt] \mathbf A &\equiv \bmatrix 0&&1\\-w^2&&0\endbmatrix \endarray$$

then

$$\frac {d\mathbf X}{dt}= \mathbf A\mathbf X$$

If $\mathbf X$ and $\mathbf A$ were scalars, the solution to the above would be the exponential

$$\mathbf X = e^{\mathbf A t}\mathbf X|_{t=0}$$

The series expansion of the exponential function works for matrices as well as scalars (for example, see http://mathworld.wolfram.com/MatrixExponential.html or http://www.sosmath.com/matrix/expo/expo.html).

In this case,

$$\mathbf A^2 = -w^2 \mathbf I$$

where $\mathbf I$ is the identity matrix. Thus

$$\begin{array}{rl} (\mathbf A t)^{2n} &= (-1)^n (wt)^{2n} \mathbf I \\ (\mathbf A t)^{2n+1} &= (-1)^n \frac 1 w (wt)^{2n+1} \mathbf A\\ \end{array}$$

The matrix exponential is thus

$$\begin{array}{rl} e^{\mathbf A t} &=\sum_{n=0}^{\infty} \frac {(\mathbf At)^n}{n!} \\[12pt] &= \sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac {(\mathbf At)^{2n+1}}{(2n+1)!} \\[12pt] &= \sum_{n=0}^{\infty} (-1)^n\frac {(wt)^{2n}}{(2n)!} \mathbf I + \frac 1 w\sum_{n=0}^{\infty}(-1)^n\frac{(wt)^{2n+1}}{(2n+1)!} \mathbf A \\[12pt] &= \cos(wt) \mathbf I +\frac 1 w\sin(wt) \mathbf A \end{array}$$

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