Question

(a) From (33) and (34) of Section 6.4 we know that when n=0, Legendre's differential equation

$(1-x^2)y"-2xy\text{'}=0$ has the polynomial solution $y=P_{-}\left(0\right)\left(x\right)=1$ Use (5) of Section 4.2 to show that a second Legendre function satisfying the DE for -1<x<1 is

$y=\frac{1}{2}In\left(\frac{1+x}{1-x}\right)$

(b) We also know from (33) and (34) of Section 6.4 that when n=1, Legendre's differential equation $(1-x^2)y"-2xy+2y=0$possesses the polynomial solution $y=P_1\left(x\right)=x$. Use (5) of Section 4.2 to show that a second Legendre function satisfying the DE for -1<x<1 is

$y=\frac{x}{2}|n\left(\frac{1+x}{1-x}\right)-1,$

.

(c) Use a graphing utility to graph the logarithmic Legendre functions given in parts (a) and (b).

Short answer

Graph.

Step-by-step solution

Definition of Power series solutions

If$X=X_0$is an ordinary point of the differential equation , we can always find two linearly independent solutions in the form of a power series centered at$X_n$that is,

$y=\sum _{n=0}^x{C}_n{(x-x_0)}^n$

A power series solution converges at least on some interval defined by$\left|x-x_0\right|<R$, where R is the distance from$X_0$to the closest singular point.

Using Legendre’s equation to solve the problem

a) Putting the given Legendre's equation in standard form

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

Identifying $p\left(x\right),f\left(x\right)$and compute integrating factor

role="math" localid="1664272258871" $P\left(x\right)=-\frac{2\times }{1-x^2}$

$f\left(x\right)=0$

$\begin{array}{l}1-x^2=t\\ -2xdx==dt\end{array}$

$\int \frac{-2x}{1-x^2}dx=\int \frac{dt}{t}$

role="math" localid="1664272515115" $=\mathrm{In}\left|t\right|$

$=\mathrm{In}|1-x^2|(*)$

Find the second Legendre’s function

Finding the second Legendre's function

$y_2\left(x\right)=P_0\left(X\right)\int \frac{e^{\int P\left(x\right)dx}}{P_0^2\left(x\right)}dx$

localid="1664282453997" $=\int e^{-\int P\left(x\right)dx}dx$

localid="1664279264139" $\stackrel{(*)}{=}\int e^{-In\left|1-x^2\right|}dx$

localid="1664282466790" $=\int \frac{1}{1-x^2}\mathrm{dx}$

localid="1664282475590" $=\frac{1}{2}\int \frac{1}{1-x}dx+\frac{1}{2}\int \frac{1}{1+x}dx$

localid="1664282483690" $=-\frac{1}{2}In(1-x\}+\frac{1}{2}In\left(1+x\right)$

localid="1664282636980" $\frac{1}{2}In\left(\frac{1+x}{1-x}\right)$

A graph of the above equation will be shown bellow

Substitute the Legendre’s function

b) Putting the given Legendre's equation in standard form

$y"-\frac{2x}{1-x^2}y\text{'}+\frac{2}{1-x^2}=0$

Identifyingand compute integrating factor

$P\left(x\right)=-\frac{2x}{1-x^2}$

$f\left(x\right)=0$

By (*), we have

$\int P\left(x\right)\mathrm{dx}=|\mathrm{n}|1-x^2|$

Finding the second Legendare's function

$y_2\left(x\right)=P_1\left(x\right)\int \frac{e^{-\int P\left(x\right)dx}}{i_1^{-2}\left(x\right)}dx$

$=x\int \frac{e^{-\int P\left(x\right)dx}}{x^2}dx$

$\stackrel{(*)}{=}x\int \frac{e^{-In\left|1-x^2\right|}}{x^2}dx$

$=x\int \frac{1}{x^2-x^4}dx$

$=x\left(\frac{1}{x^2}dx+\frac{1}{2}\int \frac{1}{1-x}dx+\frac{1}{2}\int \frac{1}{1+x}dx\right)$

$=x\left(-\frac{1}{x}-\frac{1}{2}In(1-x)+\frac{1}{2}In(1+x)\right)$

$=1+\frac{x}{2}In\left(\frac{1+x}{1-x}\right)$

A graph of the above equation will be shown bellow