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How can I get the series solution for th...

psi 10 Maple 2022
ode homework series
June 08 2022
1 1
> 

NULL

Loading Student:-ODEs

> 

xi^2*(diff(psi[m](xi), xi, xi))+2*xi*(diff(psi[m](xi), xi))+(xi^2-m*(m+1))*psi[m](xi) = 0

xi^2*(diff(diff(psi[m](xi), xi), xi))+2*xi*(diff(psi[m](xi), xi))+(xi^2-m*(m+1))*psi[m](xi) = 0

 (1)
> 

Student:-ODEs:-ODESteps(xi^2*(diff(psi[m](xi), `$`(xi, 2)))+2*xi*(diff(psi[m](xi), xi))+(xi^2-m*(m+1))*psi[m](xi) = 0, psi[m](xi))

"[[,,"Let's solve"],[,,xi^2 (((ⅆ)^2)/(ⅆxi^2) psi[m](xi))+2 xi ((ⅆ)/(ⅆxi) psi[m](xi))+(xi^2-m (m+1)) psi[m](xi)=0],["•",,"Highest derivative means the order of the ODE is" 2],[,,((ⅆ)^2)/(ⅆxi^2) psi[m](xi)],["•",,"Isolate 2nd derivative"],[,,((ⅆ)^2)/(ⅆxi^2) psi[m](xi)=((m^2-xi^2+m) psi[m](xi))/(xi^2)-(2 ((ⅆ)/(ⅆxi) psi[m](xi)))/xi],["•",,"Group terms with" psi[m](xi) "on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear"],[,,((ⅆ)^2)/(ⅆxi^2) psi[m](xi)+(2 ((ⅆ)/(ⅆxi) psi[m](xi)))/xi-((m^2-xi^2+m) psi[m](xi))/(xi^2)=0],["▫",,"Check to see if" xi[0]=0 "is a regular singular point"],[,"?","Define functions"],[,,[P[2](xi)=2/xi,P[3](xi)=-(m^2-xi^2+m)/(xi^2)]],[,"?",xi*P[2](xi) "is analytic at" xi=0],[,,([]) ? ()|() ? (xi=0)=2],[,"?",xi^2*P[3](xi) "is analytic at" xi=0],[,,([]) ? ()|() ? (xi=0)=-m^2-m],[,"?",xi=0 "is a regular singular point"],[,,"Check to see if" xi[0]=0 "is a regular singular point"],[,,xi[0]=0],["•",,"Multiply by denominators"],[,,(-m^2+xi^2-m) psi[m](xi)+2 xi ((ⅆ)/(ⅆxi) psi[m](xi))+xi^2 (((ⅆ)^2)/(ⅆxi^2) psi[m](xi))=0],["•",,"Assume series solution for" psi[m](xi)],[,,psi[m](xi)=(∑)a[k] xi^(k+r)],["▫",,"Rewrite ODE with series expansions"],[,"?","Convert" xi^m*psi[m](xi) "to series expansion for" m=0..2],[,,[]=(∑)a[k] xi^(k+r+m)],[,"?","Shift index using" k "->" k-m],[,,[]=(∑)a[k-m] xi^(k+r)],[,"?","Convert" xi*((ⅆ)/(ⅆxi) psi[m](xi)) "to series expansion"],[,,[]=(∑)a[k] (k+r) xi^(k+r)],[,"?","Convert" xi^2*(((ⅆ)^2)/(ⅆxi^2) psi[m](xi)) "to series expansion"],[,,[]=(∑)a[k] (k+r) (k+r-1) xi^(k+r)],[,,"Rewrite ODE with series expansions"],[,,a[0] (r+1+m) (r-m) xi^r+a[1] (r+2+m) (r-m+1) xi^(1+r)+((∑)(a[k] (r+1+m+k) (r-m+k)+a[k-2]) xi^(k+r))=0],["•",,a[0] "cannot be 0 by assumption, giving the indicial equation"],[,,(r+1+m) (r-m)=0],["•",,"Values of r that satisfy the indicial equation"],[,,r in {m,-m-1}],["•",,"Each term must be 0"],[,,a[1] (r+2+m) (r-m+1)=0],["•",,"Solve for the dependent coefficient(s)"],[,,a[1]=0],["•",,"Each term in the series must be 0, giving the recursion relation"],[,,a[k] (r+1+m+k) (r-m+k)+a[k-2]=0],["•",,"Shift index using" k "->" k+2],[,,a[k+2] (r+3+m+k) (r-m+k+2)+a[k]=0],["•",,"Recursion relation that defines series solution to ODE"],[,,a[k+2]=-(a[k])/((r+3+m+k) (r-m+k+2))],["•",,"Recursion relation for" r=m],[,,a[k+2]=-(a[k])/((2 m+3+k) (k+2))],["•",,"Solution for" r=m],[,,[psi[m](xi)=(∑)a[k] xi^(k+m),a[k+2]=-(a[k])/((2 m+3+k) (k+2)),a[1]=0]],["•",,"Recursion relation for" r=-m-1],[,,a[k+2]=-(a[k])/((k+2) (-2 m+1+k))],["•",,"Solution for" r=-m-1],[,,[psi[m](xi)=(∑)a[k] xi^(k-m-1),a[k+2]=-(a[k])/((k+2) (-2 m+1+k)),a[1]=0]],["•",,"Combine solutions and rename parameters"],[,,[psi[m](xi)=((∑)a[k] xi^(k+m))+((∑)b[k] xi^(k-m-1)),a[k+2]=-(a[k])/((2 m+3+k) (k+2)),a[1]=0,b[k+2]=-(b[k])/((k+2) (-2 m+1+k)),b[1]=0]]]6""

 (2)

Solving for Sum(a[k]*xi^(k+m), k = 0 .. infinity)

 
> 

q := a(k+2) = -a(k)/((2*m+3+k)*(k+2))

a(k+2) = -a(k)/((2*m+3+k)*(k+2))

(1.1)
> 

``

When m =1

 
> 

NULL

> 

m[1] := eval(q, m = 1)

a(k+2) = -a(k)/((5+k)*(k+2))

(2.1)
> 

A := rsolve({m[1], a(0) = 1, a(1) = 0}, a(k))

3*(k+2)*cos((1/2)*k*Pi)/GAMMA(k+4)

(2.2)
> 

add(A*(eval(xi^(k+m), m = 1)), k = 0 .. 16)

xi-(1/10)*xi^3+(1/280)*xi^5-(1/15120)*xi^7+(1/1330560)*xi^9-(1/172972800)*xi^11+(1/31135104000)*xi^13-(1/7410154752000)*xi^15+(1/2252687044608000)*xi^17

 (2.3)
> 

``

When m =2

 

 

> 

NULL

> 

m[2] := eval(q, m = 2)

a(k+2) = -a(k)/((7+k)*(k+2))

(3.1)
> 

A := rsolve({m[2], a(0) = 1, a(1) = 0}, a(k))

105*(k+4)*(k+2)*(k+6)*cos((1/2)*k*Pi)/GAMMA(k+8)

(3.2)
> 

add(A*(eval(xi^(k+m), m = 2)), k = 0 .. 16)

xi^2-(1/18)*xi^4+(1/792)*xi^6-(1/61776)*xi^8+(1/7413120)*xi^10-(1/1260230400)*xi^12+(1/287332531200)*xi^14-(1/84475764172800)*xi^16+(1/31087081215590400)*xi^18

 (3.3)
> 

NULL

When m=3

 
> 

NULL

> 

m[3] := eval(q, m = 3)

a(k+2) = -a(k)/((9+k)*(k+2))

(4.1)
> 

A := rsolve({m[3], a(0) = 1, a(1) = 0}, a(k))

105*(k+4)*(k+2)*(k+6)*cos((1/2)*k*Pi)/GAMMA(k+8)

(4.2)
> 

add(A*(eval(xi^(k+m), m = 3)), k = 0 .. 16)

xi^3-(1/18)*xi^5+(1/792)*xi^7-(1/61776)*xi^9+(1/7413120)*xi^11-(1/1260230400)*xi^13+(1/287332531200)*xi^15-(1/84475764172800)*xi^17+(1/31087081215590400)*xi^19

 (4.3)
> 

NULL

When m=4

 
> 

NULL

> 

m[4] := eval(q, m = 4)

a(k+2) = -a(k)/((11+k)*(k+2))

(5.1)
> 

A := rsolve({m[4], a(0) = 1, a(1) = 0}, a(k))

945*(k+4)*(k+8)*(k+2)*(k+6)*cos((1/2)*k*Pi)/GAMMA(k+10)

(5.2)
> 

add(A*(eval(xi^(k+m), m = 4)), k = 0 .. 16)

xi^4-(1/22)*xi^6+(1/1144)*xi^8-(1/102960)*xi^10+(1/14002560)*xi^12-(1/2660486400)*xi^14+(1/670442572800)*xi^16-(1/215882508441600)*xi^18+(1/86353003376640000)*xi^20

 (5.3)
> 

``

When m=5

 
> 

NULL

> 

m[5] := eval(q, m = 5)

a(k+2) = -a(k)/((13+k)*(k+2))

(6.1)
> 

A := rsolve({m[5], a(0) = 1, a(1) = 0}, a(k))

10395*(k+10)*(k+4)*(k+8)*(k+2)*(k+6)*cos((1/2)*k*Pi)/GAMMA(k+12)

(6.2)
> 

add(A*(eval(xi^(k+m), m = 5)), k = 0 .. 16)

xi^5-(1/26)*xi^7+(1/1560)*xi^9-(1/159120)*xi^11+(1/24186240)*xi^13-(1/5079110400)*xi^15+(1/1401834470400)*xi^17-(1/490642064640000)*xi^19+(1/211957371924480000)*xi^21

 (6.3)
> 

NULL

Now solving  for Sum(b[k]*xi^(k-m-1), k = 0 .. infinity)

 
> 

s := b(k+2) = -b(k)/((k+2)*(-2*m+1+k))

b(k+2) = -b(k)/((k+2)*(-2*m+1+k))

(7.1)
> 

NULL

The final solution is the sum of 2 terms and I am doing it individually for 1st term. I think I am doing it wrong because answer did not match when comparing with textbook answer. Can anyone teach me or hint me compute the final series for m=1 to 10. An example of final series for m=1 would be helpful. 

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