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$Legendre$'s equation is the following: $(1-x^2)y''-2xy'+n(n+1)y=0$. For our purposes, we want to find the general solution when $n = 1$ for $x \in (-1,1)$. The answer should contain one natural logarithm, one square root, and no absolute values. I should use the reduction of order substitution $y = ux^m$ for a strategically chosen $m$.

My attempt: $$(1-x^2)y''-2xy'+2y=0$$ rewrite $$y''-\frac{2xy'}{1-x^2}+\frac{2y}{1-x^2}=0$$ Since I see that $x^2$ and $2x$ appear, I think it is correct to let $m = 2 $

let $y=ux^2 \rightarrow y' = 2ux + u'x^2 \rightarrow y'' = 2u+4u'x+u''x^2$

substitute and find LCD $$\frac{2u(1-x^2)}{1-x^2}+\frac{4u'x(1-x^2)}{1-x^2}+\frac{u''x^2(1-x^2)}{1-x^2}-\frac{2ux^2+2u'x^3}{1-x^2}=0$$ combine like terms $$\frac{2u+4u'x + u''x^2-u''x^4-4ux^2-6u'x^3}{1-x^2}=0$$

So I don't really see a path from here. So I am wondering if I am missing something here, or perhaps my choice of $m$ is incorrect. I am not sure what $m$ would work well though, I feel like I am lacking some sort of intuition here. Any help is appreciated, thank you.

Lutterbach
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2 Answers2

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Hint: In order to use reduction of order we need to know one solution of the ODE. You are told that a solution is $y=x^m$ for some $m$. Step 1 is to find the $m$ such that the ODE is satisfied. To get you started, for $m\neq 0,1$ we have \begin{align*} (1-x^2)m(m-1)x^{m-2}-2mx^m + 2x^m = 0. \end{align*} Do any $m\neq0,1$ satisfy this?

Timothy Hedgeworth
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  • Won't this depend on x? – Lutterbach Feb 12 '19 at 02:24
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    It needs to equal $0$ for all values of $x$. So the answer is no $m\neq 0,1$ satisfies this! If $m=1$ then $y'' = 0$ and the above ODE is satisfed! Thus the correct choice is $m=1$, you should be able to take it from here! – Timothy Hedgeworth Feb 12 '19 at 03:09
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Better try $y=\sum_{k=1}^{\infty} a_k x^k$ and then derive two solutions:

(1) $y = a_1 x \text{ if the series terminates}$

(2) $y = a_0(\frac{x}{2} ln \frac{1+x}{1-x}-1) \text{ if the series does not terminate}$

See About the Legendre differential equation

Maestro13
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