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The equation

$$(1-x^2)y'' - 2xy'+n(n+1)y = 0$$

is called Legendre's equation of order $n$.

I need to show that this equation of order $1$ has $y=x$ as one solution and then I need to use this to find the general solution for $x$ greater than $-1$ but less than $1$ (not inclusive)

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  • $\begingroup$ Did you try to take $n=1$ and plug in $y(x) = x$ and see if it satisfies the equation? $\endgroup$ – Winther Jul 4 '16 at 1:39
  • $\begingroup$ if i plug that in there is still a y'' $\endgroup$ – mp12345 Jul 4 '16 at 1:43
  • $\begingroup$ and a y' in the equation $\endgroup$ – mp12345 Jul 4 '16 at 1:43
  • $\begingroup$ If $y(x) = x$ what is $y'(x)$ and $y''(x)$ ? In other words: what is the derivative of $x$ and the second derivative of $x$? $\endgroup$ – Winther Jul 4 '16 at 1:43
  • $\begingroup$ y'(x)=1 y''(x)=0 $\endgroup$ – mp12345 Jul 4 '16 at 1:44
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Here is a solution using power series.

\begin{equation} (1-x^2)y'' - 2xy'+n(n+1)y = 0 \end{equation} Let $y=\sum_{k=0}^{\infty}c_kx^k$, $y^\prime=\sum_{k=1}^{\infty}kc_kx^{k-1}$, $y^{\prime\prime}=\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}$

Then

\begin{equation} (1-x^2)\sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}-2x\sum_{k=1}^{\infty}kc_kx^{k-1}+n(n+1)\sum_{k=0}^{\infty}c_kx^k=0 \end{equation}

\begin{equation} \sum_{k=2}^{\infty}k(k-1)c_kx^{k-2}-\sum_{k=2}^{\infty}k(k-1)c_kx^{k}-\sum_{k=1}^{\infty}2kc_kx^{k}+\sum_{k=0}^{\infty}n(n+1)c_kx^k=0 \end{equation}

\begin{equation} \sum_{k=0}^{\infty}(k+2)(k+1)c_{k+2}x^{k}-\sum_{k=0}^{\infty}k(k-1)c_kx^{k}-\sum_{k=0}^{\infty}2kc_kx^{k}+\sum_{k=0}^{\infty}n(n+1)c_kx^k=0 \end{equation} \begin{equation} \sum_{k=0}^{\infty}[(k+2)(k+1)c_{k+2}-k(k-1)c_k-2kc_k+n(n+1)c_k]x^k=0 \end{equation}

Therefore

\begin{equation} (k+2)(k+1)c_{k+2}+(n(n+1)-k(k+1))c_k=0 \text{ for }k\ge0 \end{equation}

So \begin{equation} c_{k+2}=\frac{k(k+1)-n(n+1)}{(k+2)(k+1)}c_k \end{equation}

Solving for the coefficients $c_k$ of the solution yields $c_2=-c_0, c_3=0$. Thus all odd powers of the series greater than $1$ will have coefficient zero and the general solution will be of the form \begin{equation} y=c_1x+c_0(1-x^2+\cdots) \end{equation}

Therefore one linearly independent solution is $y_1=x$.

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  • $\begingroup$ not sure if i can use power series or not but thanks for your time $\endgroup$ – mp12345 Jul 4 '16 at 2:21
  • $\begingroup$ I see that this is massive overkill. You only had to verify that $y=x$ is a solution and use reduction of order to find a second linearly independent solution. Nevermind. $\endgroup$ – John Wayland Bales Jul 4 '16 at 2:21
  • $\begingroup$ yes do you know how to show that second part? $\endgroup$ – mp12345 Jul 4 '16 at 2:27
  • $\begingroup$ Here is a good youtube explanation of the process of "reduction of order" you can use to find a second solution to a second order DE given a first solution. youtube.com/watch?v=oQSFW8BIrY0 $\endgroup$ – John Wayland Bales Jul 4 '16 at 4:20

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