Proving Linear Independence of 2nd DE Solutions

sigh1342
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Homework Statement


suppose v(t) , u(t)are two linearly independent solution of the 2nd DE.
(6t^2-t-1)y''+t^2e^ty'-(3t^3-t-1)y=t^2e^t-3t^3+1
satisfying the condition v(0)=u(0)=1, prove thatu'(0) ≠ v'(0)


Homework Equations





The Attempt at a Solution


I've tried to use Wronskian, but it seems to fail.
 
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Welcome to PF;
If u and v are solutions to the non-homogeneous DE, then u-v is...
 
You need to use a theorem on when solutions to an ODE are unique.
 
thank you Simon, yep , u-v should be one of the general solution of the corresponding homogeneous equation. But how can I find the other one ,?

deluks ,what do you mean ? (or what theorem?

In this question,let the particular solution be x(t), then
u-x, v-x is the general solution of the corresponding homogeneous equation . Are they linearly independent ? (for general case, Are they always linearly independent ?)
 
I was thinking: if ##y_1## and ##y_2## are solutions to the corresponding homogeneous equations, then surely ##u-v=c_1y_1+c_2y_2## ... what happens if you differentiate both sides? What has to happen to make u'=v'?

This approach may not be rigorous enough for your course...

You really need to show that, according to some method for demonstrating that two solutions are independent, that they cannot have the same derivative. i.e. If u and v are independent, then they satisfy some criteria ... if their derivatives are the same they cannot satisfy those criteria.

You know what it takes to make two solutions independent right?
You must know more than one way of testing independence?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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