Existence of solution legendre equation

[saxen]

Hi all, I have my exam in differential equations in one week so I will probably post a lot of question. I hope you won't get tired of me!

Homework Statement

This is Legendres differential equation of order n. Determine an interval [0 t_0] such that the basic existence theorem guarantees a solution.

(1-t^2) $\frac{d^2y}{dt^2}$ - 2t$\frac{dy}{dt}$+n(n-1)y = 0
y(0)=0
y'(0)=0

Homework Equations

Picard-Lindelöfs theorem
M = max |f(x)|

T_0 = min {T,δ/M}

The Attempt at a Solution

I thought, Picards theorem is for first order equation so thought I should first rewrite the system as x'(t)=A(t)x(t)

Write

$x_{1}$ = y --- >$dx_{1}/dt$ = dy/dt

$x_{2}$ = dy/dt --- > $dx_{2}/dt$ = d^2y/dt^2

dx_1/dt = $\frac{1}{1+t^2}$

dx_2/dt = $\frac{-n(n+1)}{1+t^2}$ + $\frac{2t)}{1+t^2}$


Picards theorem require the function to be lipschitz, which it seems to be.

Picards theorem also states that you should pick the interval according to

M = max |f(x)|

T_0 = min {T,δ/M}

Here is where I am stuck, assuming I correct so far.

M = max |f(x)|

T_0 = min {T,δ/M}

I think M is equal to n(n+1) but I have NO idea on how to compute T_0.


As always, all help is greatly appreciated!

 

[saxen]

dx_1/dt =x_2

dx_2/dt = $\frac{-n(n+1)}{1+t^2}$x_1 + $\frac{2t)}{1+t^2}$ x_2

Some corrections to my current solution