Solving differential equations through matrixby devoured_elysium Tags: differential, equations, matrix, solving 

#1
May907, 11:43 AM

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 



#2
May907, 01:39 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

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 [tex]X= \left(\begin{array}{c} x \\ y\end{array}\right)[/tex] Then the two equations become the single matrix equation [tex]\frac{dX}{dt}= \left(\begin{array}{ccc}0 && 1 \\w^2 && 0\end{array}\right)X[/tex] To solve that, find the eigenvalues of the coefficient array (they are [itex]\pm w i[/itex]). The general solution then can be written as exponentials of those eigenvalues times t or, since they are imaginary, sine and cosine. 



#3
May907, 03:13 PM

P: 15

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




#4
May1007, 06:32 AM

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. 



#5
May1007, 07:36 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

[tex]\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)[/tex]
Do you see how the matrix multiplication on the right works out? 



#6
May1007, 08:32 AM

Mentor
P: 14,442

[tex]\begin{array}{rl} \mathbf X &\equiv \bmatrix x\\y\endbmatrix \\[12pt] \mathbf A &\equiv \bmatrix 0&&1\\w^2&&0\endbmatrix \endarray[/tex] then [tex]\frac {d\mathbf X}{dt}= \mathbf A\mathbf X[/tex] If [itex]\mathbf X[/itex] and [itex]\mathbf A[/itex] were scalars, the solution to the above would be the exponential [tex]\mathbf X = e^{\mathbf A t}\mathbf X_{t=0}[/tex] 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, [tex]\mathbf A^2 = w^2 \mathbf I[/tex] where [itex]\mathbf I[/itex] is the identity matrix. Thus [tex]\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}[/tex] The matrix exponential is thus [tex]\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}[/tex] 


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