Undergrad Series Solution for 2nd-Order Homogeneous ODE

[rghurst]

TL;DR

Can someone please explain why the solution provided by the characteristic equation does not entirely match the series solution? Thanks.

 

[pasmith]

Why do you think it doesn't?

Your solution with a_0 = 1 and a_1 = 0 is a solution with y(0) = 1 and y'(0) = 0. e^x doesn't satisfy that, but e^x - xe^x = 1 + \sum_{n=1}^\infty \left(\frac1{n!} - \frac{1}{(n-1)!}\right)x^n = 1 + \sum_{n=1}^\infty \frac{1 -n}{n!}x^n does. Is that series familiar?

Alternatively, if you solve the recurrence relation for a_n you find that <br /> a_n = \frac{(a_1 - a_0)n + a_0}{n!}, and then it is easy to see that <br /> \sum_{n=0}^\infty a_nx^n = (a_1 - a_0)xe^x + a_0e^x.

 

[rghurst]

Thanks. This makes better sense to me. I am still admittedly having a difficult
time seeing why the series solution would not match exactly that provided by the characteristic polynomial.

 

[pasmith]

You are representing a vector, y, with respect to two different bases. The first basis - the obvious one obtained from the characteristic polynomial - is \{e^x, xe^x\} and the components are c_1 and c_2. The second basis, obtained from the series solution - is \{e^x, e^x - xe^x\} and the components are a_0 and a_1. These components are related by <br /> \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = <br /> \begin{pmatrix} 1 &amp; 0 \\ -1 &amp; 1 \end{pmatrix}<br /> \begin{pmatrix} a_0 \\ a_1\end{pmatrix}. You should not expect different solution methods to give you exactly the same basis. However, for given initial or boundary conditions you should expect them to give the same solution.

 

[rghurst]

You are representing a vector, y, with respect to two different bases. The first basis - the obvious one obtained from the characteristic polynomial - is \{e^x, xe^x\} and the components are c_1 and c_2. The second basis, obtained from the series solution - is \{e^x, e^x - xe^x\} and the components are a_0 and a_1. These components are related by <br /> \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} =<br /> \begin{pmatrix} 1 &amp; 0 \\ -1 &amp; 1 \end{pmatrix}<br /> \begin{pmatrix} a_0 \\ a_1\end{pmatrix}. You should not expect different solution methods to give you exactly the same basis. However, for given initial or boundary conditions you should expect them to give the same solution.

You are representing a vector, y, with respect to two different bases. The first basis - the obvious one obtained from the characteristic polynomial - is \{e^x, xe^x\} and the components are c_1 and c_2. The second basis, obtained from the series solution - is \{e^x, e^x - xe^x\} and the components are a_0 and a_1. These components are related by <br /> \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} =<br /> \begin{pmatrix} 1 &amp; 0 \\ -1 &amp; 1 \end{pmatrix}<br /> \begin{pmatrix} a_0 \\ a_1\end{pmatrix}. You should not expect different solution methods to give you exactly the same basis. However, for given initial or boundary conditions you should expect them to give the same solution.

This makes full sense to me now. Thanks.