Finding a Basis for O.D.Es with Same Roots

[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?

 

[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

 

[cronxeh]

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

 

[asdf1]

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

 

[mathmike]

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

 

[Tide]

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

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

 

[asdf1]

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! :)

could you explain that part a little clearer?

 

[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.)

 

[asdf1]

thanks! :)