# Thread: Reducing Systems of higher order equations

1. ## Reducing Systems of higher order equations

I missed the lecture and was wondering if somebody could clue me in on the method. I as far as I can tell (and to prove that I am trying over here ) We introduce
x1=x(t) x2=x'(t) x3=x''(t)
and because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t). But I dont understand where I go from here.

Silent

I missed the lecture and was wondering if somebody could clue me in on the method. I as far as I can tell (and to prove that I am trying over here ) We introduce
x1=x(t) x2=x'(t) x3=x''(t)
and because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t). But I dont understand where I go from here.

Silent
First you are going to have to write out the question clearly. You say first
" We introduce x1=x(t) x2=x'(t) x3=x''(t) "
which has x as dependent variable and t as independent variable, and then
"because my 2 equations contain dy/dx and d2y/dx2 terms I assume I have to introduce y1=y(t) y2=y'(t) y3=y''(t)" where your equation seems to have y as dependent variable and x as independent variable but then you have switched to "t" as independent variable in the second line.

What is the specific problem you are trying to solve?

3. sorry, that was quite clumsy you are quite right though. I should have said d2x/dt2 and d2y/dt2.

The problem is:

5x''(t)+ 3(t)^0.5y'(t)-7x-8y=sinht
tan(t) y''(t)+7x'(t)+9ln(t)x=cosht

But this is just an example I picked off a work sheet, im more directly after the method.

Silent

4. You have:

$\displaystyle 5x''+3t^{1/2}y'-7x+8y=\sinh(t)$

$\displaystyle \tan(t)y''+7x'+9\ln(t)x=\cosh(t)$

I can then let $\displaystyle x'=u$ and $\displaystyle y'=v$ and arrive at the following first-order system of four equations:

$\displaystyle x'=u$

$\displaystyle 5u'+3t^{1/2}v-7x+8y=\sinh(t)$

$\displaystyle y'=v$

$\displaystyle \tan(t)v'+7x'+9\ln(t)x=\cosh(t)$

5. I was expecting to arrive at a matrix M(t) and an inhomogenous term H(t)
can this be drawn from your equations?

6. Yes: $\displaystyle \textbf{X}^{'}=\textbf{M}(t)\textbf{X}+\textbf{H}( t)$

Line them all up in the form of x, y, u, v ok. So the first row in the matrix corresponding to the first equation would be (0,0,1,0) right? The second row would be (0,0,0,1). You can do the rest.

8. $\displaystyle \left[\begin{array}{c} x \\ y \\ u \\ v\end{array}\right]^{'}$ $\displaystyle =\left[\begin{array}{cccc}0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 7/5 & -8/5 & 0 & -3/5 t^{1/2} \\-\frac{9\ln(t)}{\tan(t)} & 0 & -\frac{7}{\tan(t)} & 0\end{array}\right]$ $\displaystyle \left[\begin{array}{c} x\\ y \\ u \\ v\end{array}\right]$ $\displaystyle =\left[\begin{array}{c}0 \\ 0 \\ \frac{\sinh(t)}{5} \\ \frac{\cosh(t)}{\tan(t)}\end{array}\right]$