Existence of solution legendre equation

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SUMMARY

The discussion focuses on solving Legendre's differential equation of order n, specifically the equation (1-t^2) d²y/dt² - 2t dy/dt + n(n-1)y = 0 with initial conditions y(0)=0 and y'(0)=0. The participant aims to determine an interval [0, t₀] that guarantees a solution based on Picard-Lindelöf's theorem. The participant correctly identifies the need to rewrite the system in first-order form and recognizes that the function is Lipschitz. However, they express uncertainty in computing T₀, which is derived from M = max |f(x)| and T₀ = min {T, δ/M}.

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saxen
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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!
 
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saxen said:
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
 

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