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Homework Four: Problem 1
Problem Statment and Given:  p.19-1 For the Legendre equation 1 on p.14-2 with $$n=0$$, the homogeneous solution $$u_1(x)=1$$. Use reduction of order method 2 (undetermined factor) to find $$u_2(x)$$, the second homogeneous solution.

Solution:  In general for any given L2-ODE-VC Non Homogenous: $$y'' + a(x)y' + b(x)y = f(x)$$ Apply the following transformation where u1(x) is a homogenous solution: $$y(x) = U(x)u_1(x)$$ $$y(x) = U(x)u_1(x)$$ $$y' = U'u_1 + Uu_1'$$ $$y = Uu_1 + 2U'u_1' + Uu_1''$$ $$Uu_1 + U'(2u_1'+ u_1 a(x)) + U(u_1+ u_1' a(x) + u_1 b(x)) = f(x)$$
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U''u_1 + U'(2u_1'+ u_1 a(x)) = f(x)$$
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For this problem: $$u_1(x) = 1$$ $$u_1'(x) = 0$$ $$f(x) = 0$$ $$a(x) = -2x/(1-x^2)$$

We obtain the following ODE: $$U''-2x/(1-x^2) U' = 0$$

Multiply by $$(1-x^2)$$ and let $$U' = Z$$: $$(1-x^2) Z'-2xZ = 0$$ $$\frac{d((1-x^2) Z)}{dx} = 0$$ $$(1-x^2) Z= C_1$$ $$U'= \frac{C_1}{1-x^2}$$ $$U= \int \frac{C_1}{(1+x)(1-x)} dx$$ $$U= \frac{C_1}{2}[ \int ( \frac{1}{1+x} + \frac{1}{1-x} ) dx]$$

$$U= \frac{C_1}{2}[ ln( (1+x)(1-x) ) + k]$$ We obtain the following Solution:


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y(x) = U(x)= ln( (1-x^2)^\frac{C_1}{2} ) + C_2$$ Apply  $$y(x) = U(x)u_1(x)$$ You get $$y(x) = U(x)$$
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 * style="width:30%; padding:10px; border:2px solid #000000" | $$ \displaystyle
 * $$\displaystyle $$
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