Solution for system of diff equations

1. May 6, 2014

Jhenrique

Given system like $\begin{bmatrix} x'(t)\\ y'(t)\\ \end{bmatrix} = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} \begin{bmatrix} x(t)\\ y(t)\\ \end{bmatrix}$ can be rewritten as $(-1)u''(t) + (bd)u'(t) + (ad+c)u(t) = 0$ with $\begin{bmatrix} x(t)\\ y(t)\\ \end{bmatrix} = \begin{bmatrix} u(t)\\ u'(t)\\ \end{bmatrix}$

In other words, if a diff equation of 2nd order can be disassembled in a sytem of 1nd order, thus a system of 1nd order can be assembled in a diff equation of 2nd order, correct? This way, I don't need to find the eigenvalues and eigenvectors...

2. May 6, 2014

Matterwave

How do you guarantee that $y(t)=u'(t)=x'(t)$ will always work? I don't see why this would be so... they look independent to me in the first equation...

3. May 7, 2014

HallsofIvy

Yes, that is true. Dropping the matrix notation, if x'= ax+ by and y'= cx+ dy. then x''= ax'+ by'= ax'+ b(cx+ dy)= ax'+ bcx+ bdy. From the first equation, by= x'- ax so that
x''= ax'+ bcx+ d(x'- ax)= (a+ d)x'+ (bc- ad)x. We can always convert a system of n first order differential equations to a single nth order differential equation.

4. May 7, 2014

Jhenrique

And this method isn't more simple than compute the eigenvalues and eigenvectors of a matrix?

5. May 7, 2014

AlephZero

Matterwave has seen the problem with the OP's idea.

You can convert the two equations into second order equations

But it doesn't follow that y = x'. You can choose two independent boundary conditions for each equation.

If it did follow that y = x', it would also follow that x = y', and that clearly means "something is wrong here".

In fact the OP's idea is used in reverse: to solve a single differential equation of order n, it is often easiest to convert it into a system of n first-order equations.