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2 methods to solve system of differential equations

  • Thread starter Sparky_
  • Start date
  • #1

Homework Statement

Greetings –

I am wanting to solve the following system of differential equations:


K*\frac{d^2x}{dt^2} = -K2*\frac{dz}{dt}

K*\frac{dz^2}{dt^2} = K2*\frac{dx}{dt}

Homework Equations

The Attempt at a Solution

Now – I am not just wanting to solve it (I know what the solution is already) I am wanting for grins to solve it as if I was back in school –

I would like to solve this system 2 ways:

Am I correct that using Laplace transforms are a straightforward way to solve this system?

I have scanned my old diff. e. book and could not find great examples of solving systems.

Can I use the method where I would set:
x = i1
x’ = i2
x” = i3

then i1’ = i2
i2’ = i3

Likewise for z
z = j1
z’ = j2
z’’ = j3

j1’ = j2
j2’ = j3

ending up with:

K(i2’) = -(k2)j2
K(j2’) = (k2)i2

What I don’t know how to do is solve this system using a typical tool - something like pulling out the characteristic equation or so forth?

Can you help – how can I solve this system (a second method other than Laplace)?


Answers and Replies

  • #2
Homework Helper
Gold Member
No - you just have to realise you have time derivatives of x and z in the equation and you don't have x and y. So just express in terms of new variables which are those derivatives. u = dx/dt, v = dz/dt say. Not fairly obvious?

Then how to solve that, I posted yesterday or so.
  • #3
  • #4
Homework Helper
Gold Member
  • #5
Science Advisor
Homework Helper
Pretty much like epenguin, I tend to stop reading when I see "Laplace Transform". A very over rated technique in my opinion.

The real "two methods", in my opinion are:
1) Differentiate the first equation again to get
[tex]K\frac{d^3x}{dt^3}= -K_2\frac{d^2z}{dt^2}[/tex]

Now, from the second equation, substitute
for the second derivative of z. That gives
[tex]K\frac{d^3x}{dt^3}= -\frac{K_2^2}{K}\frac{dx}{dt}[/tex]
If we let u= dx/dt, that becomes a second order equation in u:
[tex]K\frac{d^2u}{dt^2}= -\frac{K_2^2}{K}u[/tex]
[tex]\frac{d^2u}{dt^2}+ \frac{K_2^2}{K^2}u= 0[/tex]
which is easily solved. Integrate u with respect to t to find x.

Once you know x, you can put it
[tex]K_2\frac{dz}{dt}= K_2\frac{d^2x}{dt^2}[/tex]
and integrate to find z.

Notice the solving the second order equation for u will give 2 undetermined constants. Then integrating to find x will introduce a third. Finally, integrating to find z will introduce a fourth undetermined constant. That is exactly what we would expect solving two second order equations.

2) Let w= dz/dt and let y= dx/dt so that the two equations become
[tex]K\frac{dw}{dt}= K_2y[/tex]
[tex]K\frac{dy}{dt}= K_2w[/tex]
so that we have four first order equations:
[tex]\frac{dx}{dt}= y[/tex]
[tex]\frac{dy}{dt}= \frac{K_2}{K}w[/tex]
[tex]\frac{dz}{dt}= w[/tex]
[tex]\frac{dw}{dt}= \frac{K_2}{K}y[/tex]

and we can write that as the single matrix equation:
[tex]\frac{d\begin{pmatrix}x \\ y\\ z\\ w\end{pmatrix}}{dt}= \begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 0 & \frac{K_2}{K}\\ 0 & 0 & 0 & 1 \\ 0 &\frac{K_2}{K}& 0 & 0\end{pmatrix}\begin{pmatrix}x \\ y \\ z \\ w\end{pmatrix}[/tex]
which can be solved by finding the eigenvalues and eigenvectors of the matrix.
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