# System of differential equations - a big problem solving it

1. Nov 24, 2004

### ILens

Hi!

I have a serious problem solving the following system of differential equations:

$$m_1\ddot{x_1}=-k_1x_1-k(x_1-x_2)$$

$$m_2\ddot{x_2}=-k_2x_2-k(x_2-x_1)$$

Does anybody have an idea how to solve it?

Thanks.

2. Nov 24, 2004

### dextercioby

It can be easily decoupled obtaing a 4-th order ODE in one of the coordinates.Which can be solved.

3. Nov 24, 2004

### ILens

Thank you, but you told me something that I already know. I have problems deriving these equations. Would you mind being more explicit in your answer.

4. Nov 24, 2004

### dextercioby

From the first equation express x2 as a function of x1 and its second derivatives and puti it in the second equation.You'll get the 4-th order LODE.Make simplifications assuming the masses are the same.in that case,formulas will get smaller in size.

5. Nov 24, 2004

### ILens

Thanks for your answer.

You have just confirmed that the solution I have derived is correct :rofl:

6. Nov 24, 2004

### Dr Transport

The equations can be solved by two methods, set both $$x_{1}$$ and $$x_{2}$$ equal to complex exponentials with different constant coefficients then plug away, you'll get an algebraic set of equations to solve, the other way is to take sums and differences of them and use another coordinate system like $$y_{i} = x_{1} \pm x_{2}$$ and work thru using the method in the first part of this reply.

There is no need to get into a 4th order ODE.

7. Nov 24, 2004

### Tide

Laplace Transforms, anyone? :-)

8. Nov 24, 2004

### Dr Transport

Let $$x_{i} = a_{i}e^{i\omega t}$$, substitute and get the following equations

$$-a_{1}m_{1}\omega^{2} = -a_{1}k_{1} - k(a_{1} - a_{2})$$

$$-a_{2}m_{2}\omega^{2} = -a_{2}k_{2} - k(a_{2} - a_{1})$$

Solve this set of equations for $$\omega$$ the use the boundary and initial conditions to obtain the $$a_{1}$$ and $$a_{2}$$. This is a complicated set of equations because of the 3 distinct spring constants. at first glance it was exactly solvable almost trivial, but 3 constants makes it an order of magnitude more difficult.

Last edited: Nov 24, 2004
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