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rghurst
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- Can someone please explain why the solution provided by the characteristic equation does not entirely match the series solution? Thanks.
pasmith said:You are representing a vector, [itex]y[/itex], with respect to two different bases. The first basis - the obvious one obtained from the characteristic polynomial - is [itex]\{e^x, xe^x\}[/itex] and the components are [itex]c_1[/itex] and [itex]c_2[/itex]. The second basis, obtained from the series solution - is [itex]\{e^x, e^x - xe^x\}[/itex] and the components are [itex]a_0[/itex] and [itex]a_1[/itex]. These components are related by [tex]
\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}
\begin{pmatrix} a_0 \\ a_1\end{pmatrix}.[/tex] 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.pasmith said:You are representing a vector, [itex]y[/itex], with respect to two different bases. The first basis - the obvious one obtained from the characteristic polynomial - is [itex]\{e^x, xe^x\}[/itex] and the components are [itex]c_1[/itex] and [itex]c_2[/itex]. The second basis, obtained from the series solution - is [itex]\{e^x, e^x - xe^x\}[/itex] and the components are [itex]a_0[/itex] and [itex]a_1[/itex]. These components are related by [tex]
\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}
\begin{pmatrix} a_0 \\ a_1\end{pmatrix}.[/tex] 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.