# Same roots

1. Oct 18, 2005

### asdf1

why is it that when you have the same roots to an O.D.E., you usually add an x or x^2 to get a basis?

2. Oct 18, 2005

### mathmike

because when you multipy anything by a variable it will not be a scaler multiple of the original, and you need two unique so;utions (or more) for an ode

3. Oct 18, 2005

### cronxeh

Because if Wronskian=0 then you have linearly dependent solutions, and you need linearly independent solutions (i.e. W not equal 0).

4. Oct 19, 2005

### asdf1

but why pick x or x^2 to multiply? why not something like xy or x^y or e^x or something?

5. Oct 19, 2005

### mathmike

you could but why make it more complicated than you have to

6. Oct 19, 2005

### Tide

Let D be a differential operator such that a is a double root:

$$(D-a)^2 y = 0[/itex] Consider the slightly modified DE: [tex](D-a + \epsilon)(D-a-\epsilon)$$

Solve this second equation subject to the initial conditions $y(0) = y_0$ and $\dot y(0) = \dot y_0$ then pass to the limit of $\epsilon$ going to zero. You'll find the answer to your question in the result! :)

7. Oct 20, 2005

### asdf1

could you explain that part a little clearer?

8. Oct 20, 2005

### Tide

This ODE

$$\left( \frac {d^2}{dt^2} - 2 a \frac {d}{dt} + a^2 - \epsilon ^2 \right) y= 0$$

has repeated roots when $\epsilon \rightarrow 0$. Solve the equation as it is subject to specific initial conditions. When you're all done, pass to the limit $\epsilon \rightarrow 0$ and you'll see why the basis functions are the way they are when you have repeated roots! (Sorry - I don't have the time to type in the algebra but it is straightforward.)

9. Oct 21, 2005

### asdf1

thanks!!! :)

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