Answer to Question #128942 in Differential Equations for Sehrish

Question #128942
Solve the legendres equation (1-x^2)y"-2xy'+n(n-1)y=0
1
Expert's answer
2020-08-10T18:12:41-0400
"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"










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