User:Egm6321.f09.Team06.sada/HW4

Problem 1
P.19-1 For the Legendre equation 1 on p.14-2 with n = 0, the homogeneous solution u1(x) = 1. Use reduction of order method 2 (undetermined factor) to find u2(x), the second homogeneous solution.

Solution
Equation 1 on p.14-2 with n=0, reduces to

$$\displaystyle (1-x^2)y''-2x y' = 0 $$

The above equation is rewritten as,

$$\displaystyle y'' - \frac{2x}{1-x^2} y' = 0 $$

Where, $$\displaystyle a_1(x)=\frac{-2x}{1-x^2} $$

From Lecture 17-4, Equation 2:

$$\displaystyle u_2 (x) = u_1(x) \int_{}^x {\frac{1}} \exp \left[ { - \int_{}^t {a_1 (s)ds} } \right]dt $$

$$\int {} a_1(s)ds=\int {} \frac{-2s}{(1-s^2)}ds= \log(1-s^2) $$

Substituting $$\displaystyle u_1(t)=1$$ and $$\int {} a_1(s)ds $$ in Equ.2

$$\displaystyle u_2 (x) = (1) \int_{}^x {\frac{1}{(1)^2} \exp \left[ - \log(1-s^2) \right]}dt = \int_{}^x {\frac{1}{1-s^2}}dt $$

$$\displaystyle \int_{}^x {\frac{1}{1-s^2}}dt = \int_{}^x \left[\frac{1/2}{1-s} +\frac{1/2}{1+s} \right]ds = -\frac{1}{2}\log(1-x)+\frac{1}{2}\log(1+x) $$

Thus, the second homogeneous solution by reduction of order method 2 is given by

$$ \displaystyle u_2(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) $$