Solving differential equations through matrix

[devoured_elysium]

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

 

[HallsofIvy]

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.

 

[devoured_elysium]

it's the latter one, as you thought. thanks by the response!

 

[devoured_elysium]

Hi again

Sorry but I could not follow the step X=(x, y) and why then it becomes the next equation.

 

[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)

Do you see how the matrix multiplication on the right works out?

 

[D H]

\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}<br /> \mathbf X &amp;\equiv \bmatrix x\\y\endbmatrix \\[12pt]<br /> \mathbf A &amp;\equiv \bmatrix 0&amp;&amp;1\\-w^2&amp;&amp;0\endbmatrix<br /> \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" ).

In this case,

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

where \mathbf I is the identity matrix. Thus

\begin{array}{rl}<br /> (\mathbf A t)^{2n} &amp;= (-1)^n (wt)^{2n} \mathbf I \\<br /> (\mathbf A t)^{2n+1} &amp;= (-1)^n \frac 1 w (wt)^{2n+1} \mathbf A\\<br /> \end{array}

The matrix exponential is thus

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