Why we need linear independence of all solutions of a nth order ODE?
What happens if they are not linearly independent?
Why we need linear independence of all solutions of a nth order ODE?
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linear-algebra
ordinary-differential-equations
1 Answers
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It happens that some of the initial conditions of your Cauchy problem won't allow a solution written in terms of those functions. For instance, consider the Cauchy problem $$\begin{cases}y''-2y'+1=0\\ y(0)=0\\ y'(0)=2\end{cases}$$
Now, let's work as if we were bad students who did not study carefully linear ODE. The method I did not study well "says" (according to my very bad judgement) that:
I must find the roots of the polynomial $\lambda^2-2\lambda+1$
I must look for a solution in the form $c_1 e^{\lambda_1x}+c_2 e^{\lambda_2 x}$.
So, we get $\lambda_1=\lambda_2=1$ and hence proceed to look for a function $y(x)=(c_1+c_2)e^{x}$ such that $y(0)=0$ and $y'(0)=2$.
But $y(0)=0\implies c_1+c_2=0\implies c_1=-c_2$ and $y'(0)=2\implies c_1+c_2=2$, which leads us to a surprising non-existence result: so much for that big theorem with all those names and weird hypothesis about locally-uniform Lipschitz continuity! Here we are finding counterexamples in our Calculus I test!
Where does our procedure fail? We did not pick two linearly independent functions that satisfy $y''-2y'+1=0$. In fact, we should have looked for a solution in the form $c_1e^x+c_2xe^x$, as the correct version of the theorem on the solution of linear ODE tells us.