# Solution for system of diff equations

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...

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Matterwave
Gold Member
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...

HallsofIvy
Homework Helper
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.

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

AlephZero
Homework Helper
Matterwave has seen the problem with the OP's idea.

You can convert the two equations into second order equations