Answer to Question #128942 in Differential Equations for Sehrish

Question

Solve the legendres equation (1-x^2)y"-2xy'+n(n-1)y=0

Answer 1

$$Solution$$

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

Write the differential equation in standard form by dividing both sides of the equation by the coefficient of $y''$ : $(1-x^2)$

$$y''-\frac{2xy'}{(1-x^2)}+\frac{n(n-1)}{(1-x^2)}y=0$$

Now $1-x^2=0 \implies (1-x)(1+x)=0$

$\implies x=+1$ $or -1$

Thus $x=+1$ and $x=-1$ are two singular points of $(1)$

Let

$$y(x)=\Sigma ^{\infin}_{k=0}a_kx^{k+m}$$

$i.e$ $y=a_0x^m+a_1x^{m+1}+a_2x^{m+2}+a_3x^{m+3}+...+a_mx^{m+n}----->(2)$

$y'=ma_0x^{m-1}+(m-1)a_1x^{m}+(m+2)a_2x^{m+1}+(m+3)a_3x^{m+2}+...+(m+n)a_mx^{m+n-1}$

$y''=m(m-1)a_0x^{m-2}+(m+1)ma_1x^{m-1}+(m+2)(m-1)a_2x^{m}+(m+3)(m+2)a_3x^{m+1}+...+(m+n)(m+n-1)a_mx^{m+n-2}$

Put the values of $y,y',y''$ in $(1)$ to get;

$(1-x^2)[m(m-1)a_0x^{m-2}+(m+1)ma_1x^{m-1}+(m+2)(m-1)a_2x^{m}+(m+3)(m+2)a_3x^{m+1}+...]-2x[ma_0x^{m-1}+(m-1)a_1x^{m}+(m+2)a_2x^{m+1}+(m+3)a_3x^{m+2}+...]+n(n-1)[a_0x^m+a_1x^{m+1}+a_2x^{m+2}+a_3x^{m+3}+...+a_mx^{m+n}]=0$

Multiply the terms and simplify to get the Indicial equation

For Indicial equation, the least power of $x$ terms are equated to zero

Here, $x^{m-2}$ is the least power of $x$

Putting coefficients of $x^{m-2}$ equal to zero, we get;

$$\implies a_0m(m-1)=0$$

$$\implies m(m-1)=0, a_0 \ne0$$

Thus $m(m-1)=0$ is indicial equation$------->(3)$

Now we will find the coefficient $a_0, a_1, a_2, a_3, a_4....$ and then find the general solution.

Equating the coefficient of $x^{m-1}=0$

$$(m+1)ma_1=0 \implies a_1=\frac{0}{m(m+1)}---->(4)$$

Equating the coefficient of $x^m=0$

$$a_2(m+2)(m+1)-a_0(m-1)m-2a_0m+n(n-1)a_0=0$$

$$a_2(m+2)(m+1)-a_0[(m-1)m-2m+n(n-1)]=0----->(5)$$

Equating the coefficient of $x^{m+1}=0$

$$a_3(m+3)(m+1)-a_1[(m-1)m+2m-n(n-1)]=0----->(6)$$

Equating the coefficient of $x^{m+2}=0$

$a_4(m+4)(m+3)-a_2[(m+2)(m+1)+2(m+2)-n(n-1)]=0----->(7)$

To find the solution when $m=0$

$$a_2(0+2)(0+1)-a_0(0+1)*0-2a_0*0+n(n-1)a_0=0 (From (5))$$

$$2a_2+n(n-1)a_0=0$$

$$a_2=\frac{-n(n-1)}{2!}a_0$$

From $(6)$ Put $m=0$

$$a_3=\frac{-(n-2)(n+1)}{3!}a_1$$

From $(7)$ Put $m=0$

$$a_4=\frac{(n-3)(n-1)(n+2)n}{4!}a_0$$

Similarly

$$a_5=\frac{(n-4)(n-2)(n+1)(n+3)}{5!}a_1$$

Since given that $a_1=0$ Put values of $a_2, a_3, a_4, a_5$ in $(2)$ to get

$$y=a_0x^m+0+[\frac{-n(n-1)}{2!}a_1]x^{m+2}+0+[\frac{(n-3)(n-1)(n+2)n}{4!}]a_0x^{m+4}+0$$

$$y=a_0[1+\frac{n(n-1)}{2!}x^2+\frac{(n-3)(n-1)(n+2)n}{4!}x^4]x^{m}$$

Now Put $m=1$

From $(5)$

$$a_2(3)(2)-0-2a_0+n(n-1)a_0=0$$

$$a_2=(\frac{-n(n-1)-2}{6})a_0=-[\frac{n^2-n-2}{6}]a_0=-[\frac{(n+1)(n-2)}{6}]a_0$$

From $(6)$

$$a_3(4)(3)-a_1[2(1)+2(2)-n(n-1)]=0$$

Put $a_1=1$

$$a_3(4)(3)-a_1[6-n(n-1)]=0$$

$$\implies a_3=\frac{6-n(n-1)}{(4)(3)}a_1=-[\frac{n^2-n-6}{(4)(3)}]a_1=-[\frac{(n-3)(n+2)}{(4)(3)}]a_1$$

From $(7)$

$$a_4(5)(4)-a_2[(3)(2)+2(3)-n(n-1)]=0$$

$$a_4=\frac{-a_2}{(5)(4)}[(n-4)(n+3)]$$

Putting value of $a_2$ we get

$$a_4=\frac{(n+1)(n-2)}{(3)(2)}a_0(n-4)(n+3)=\frac{(n-4)(n+3)(n-2)(n+1)}{5!}a_0$$

Putting the values from $a_0,a_1,a_2,a_3$ in $(3)$ we get

$y=a_0x^m+a_1x^{m+1}+\frac{-(n-2)(n+1)}{3!}a_0^{m+2}-\frac{(n-3)(n+2)}{(4)(3)}a_1x^{m+3}+\frac{(n-4)(n+3)(n-2)(n+1)}{5!}a_0x^{m+1}$

Put $m=1$ and $a=1,$ we will get odd power terms as

$y=[x+\frac{-(n-2)(n+1)}{3!}x^2+\frac{(n-4)(n+3)(n-2)(n+1)}{5!}x^5+...]a_0----->Answer$