2.4.30 Problem 26

Maple
Mathematica
Sympy

Internal problem ID [8919]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 26
Date solved : Wednesday, March 05, 2025 at 07:08:13 AM
CAS classification : [[_2nd_order, _exact, _linear, _homogeneous]]

Solve

x2(3+x)y+5x(1+x)y(14x)y=0

Using series expansion around x=0

The type of the expansion point is first determined. This is done on the homogeneous part of the ODE.

(x3+3x2)y+(5x2+5x)y+(4x1)y=0

The following is summary of singularities for the above ode. Writing the ode as

y+p(x)y+q(x)y=0

Where

p(x)=5+5xx(3+x)q(x)=4x1x2(3+x)
Table 2.79: Table p(x),q(x) singularites.
p(x)=5+5xx(3+x)
singularity type
x=3 “regular”
x=0 “regular”
q(x)=4x1x2(3+x)
singularity type
x=3 “regular”
x=0 “regular”

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : [3,0,]

Irregular singular points : []

Since x=0 is regular singular point, then Frobenius power series is used. The ode is normalized to be

x2(3+x)y+(5x2+5x)y+(4x1)y=0

Let the solution be represented as Frobenius power series of the form

y=n=0anxn+r

Then

y=n=0(n+r)anxn+r1y=n=0(n+r)(n+r1)anxn+r2

Substituting the above back into the ode gives

(1)x2(3+x)(n=0(n+r)(n+r1)anxn+r2)+(5x2+5x)(n=0(n+r)anxn+r1)+(4x1)(n=0anxn+r)=0

Which simplifies to

(2A)(n=0x1+n+ran(n+r)(n+r1))+(n=03xn+ran(n+r)(n+r1))+(n=05x1+n+ran(n+r))+(n=05xn+ran(n+r))+(n=04x1+n+ran)+n=0(anxn+r)=0

The next step is to make all powers of x be n+r in each summation term. Going over each summation term above with power of x in it which is not already xn+r and adjusting the power and the corresponding index gives

n=0x1+n+ran(n+r)(n+r1)=n=1an1(n+r1)(n+r2)xn+rn=05x1+n+ran(n+r)=n=15an1(n+r1)xn+rn=04x1+n+ran=n=14an1xn+r

Substituting all the above in Eq (2A) gives the following equation where now all powers of x are the same and equal to n+r.

(2B)(n=1an1(n+r1)(n+r2)xn+r)+(n=03xn+ran(n+r)(n+r1))+(n=15an1(n+r1)xn+r)+(n=05xn+ran(n+r))+(n=14an1xn+r)+n=0(anxn+r)=0

The indicial equation is obtained from n=0. From Eq (2B) this gives

3xn+ran(n+r)(n+r1)+5xn+ran(n+r)anxn+r=0

When n=0 the above becomes

3xra0r(1+r)+5xra0ra0xr=0

Or

(3xrr(1+r)+5xrrxr)a0=0

Since a00 then the above simplifies to

(3r2+2r1)xr=0

Since the above is true for all x then the indicial equation becomes

3r2+2r1=0

Solving for r gives the roots of the indicial equation as

r1=13r2=1

Since a00 then the indicial equation becomes

(3r2+2r1)xr=0

Solving for r gives the roots of the indicial equation as [13,1].

Since r1r2=43 is not an integer, then we can construct two linearly independent solutions

y1(x)=xr1(n=0anxn)y2(x)=xr2(n=0bnxn)

Or

y1(x)=n=0anxn+13y2(x)=n=0bnxn1

We start by finding y1(x). Eq (2B) derived above is now used to find all an coefficients. The case n=0 is skipped since it was used to find the roots of the indicial equation. a0 is arbitrary and taken as a0=1. For 1n the recursive equation is

(3)an1(n+r1)(n+r2)+3an(n+r)(n+r1)+5an1(n+r1)+5an(n+r)+4an1an=0

Solving for an from recursive equation (4) gives

(4)an=(1+n+r)an13n+3r1

Which for the root r=13 becomes

(5)an=(4+3n)an19n

At this point, it is a good idea to keep track of an in a table both before substituting r=13 and after as more terms are found using the above recursive equation.

n an,r an
a0 1 1

For n=1, using the above recursive equation gives

a1=2r2+3r

Which for the root r=13 becomes

a1=79

And the table now becomes

n an,r an
a0 1 1
a1 2r2+3r 79

For n=2, using the above recursive equation gives

a2=r2+5r+69r2+21r+10

Which for the root r=13 becomes

a2=3581

And the table now becomes

n an,r an
a0 1 1
a1 2r2+3r 79
a2 r2+5r+69r2+21r+10 3581

For n=3, using the above recursive equation gives

a3=r39r226r2427r3+135r2+198r+80

Which for the root r=13 becomes

a3=4552187

And the table now becomes

n an,r an
a0 1 1
a1 2r2+3r 79
a2 r2+5r+69r2+21r+10 3581
a3 r39r226r2427r3+135r2+198r+80 4552187

For n=4, using the above recursive equation gives

a4=r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880

Which for the root r=13 becomes

a4=182019683

And the table now becomes

n an,r an
a0 1 1
a1 2r2+3r 79
a2 r2+5r+69r2+21r+10 3581
a3 r39r226r2427r3+135r2+198r+80 4552187
a4 r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880 182019683

For n=5, using the above recursive equation gives

a5=r520r4155r3580r21044r720243r5+3240r4+16065r3+36360r2+36492r+12320

Which for the root r=13 becomes

a5=6916177147

And the table now becomes

n an,r an
a0 1 1
a1 2r2+3r 79
a2 r2+5r+69r2+21r+10 3581
a3 r39r226r2427r3+135r2+198r+80 4552187
a4 r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880 182019683
a5 r520r4155r3580r21044r720243r5+3240r4+16065r3+36360r2+36492r+12320 6916177147

Using the above table, then the solution y1(x) is

y1(x)=x1/3(a0+a1x+a2x2+a3x3+a4x4+a5x5+a6x6)=x1/3(17x9+35x281455x32187+1820x4196836916x5177147+O(x6))

Now the second solution y2(x) is found. Eq (2B) derived above is now used to find all bn coefficients. The case n=0 is skipped since it was used to find the roots of the indicial equation. b0 is arbitrary and taken as b0=1. For 1n the recursive equation is

(3)bn1(n+r1)(n+r2)+3bn(n+r)(n+r1)+5bn1(n+r1)+5bn(n+r)+4bn1bn=0

Solving for bn from recursive equation (4) gives

(4)bn=(1+n+r)bn13n+3r1

Which for the root r=1 becomes

(5)bn=nbn13n4

At this point, it is a good idea to keep track of bn in a table both before substituting r=1 and after as more terms are found using the above recursive equation.

n bn,r bn
b0 1 1

For n=1, using the above recursive equation gives

b1=2r2+3r

Which for the root r=1 becomes

b1=1

And the table now becomes

n bn,r bn
b0 1 1
b1 2r2+3r 1

For n=2, using the above recursive equation gives

b2=r2+5r+69r2+21r+10

Which for the root r=1 becomes

b2=1

And the table now becomes

n bn,r bn
b0 1 1
b1 2r2+3r 1
b2 r2+5r+69r2+21r+10 1

For n=3, using the above recursive equation gives

b3=r39r226r2427r3+135r2+198r+80

Which for the root r=1 becomes

b3=35

And the table now becomes

n bn,r bn
b0 1 1
b1 2r2+3r 1
b2 r2+5r+69r2+21r+10 1
b3 r39r226r2427r3+135r2+198r+80 35

For n=4, using the above recursive equation gives

b4=r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880

Which for the root r=1 becomes

b4=310

And the table now becomes

n bn,r bn
b0 1 1
b1 2r2+3r 1
b2 r2+5r+69r2+21r+10 1
b3 r39r226r2427r3+135r2+198r+80 35
b4 r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880 310

For n=5, using the above recursive equation gives

b5=r520r4155r3580r21044r720243r5+3240r4+16065r3+36360r2+36492r+12320

Which for the root r=1 becomes

b5=322

And the table now becomes

n bn,r bn
b0 1 1
b1 2r2+3r 1
b2 r2+5r+69r2+21r+10 1
b3 r39r226r2427r3+135r2+198r+80 35
b4 r4+14r3+71r2+154r+12081r4+702r3+2079r2+2418r+880 310
b5 r520r4155r3580r21044r720243r5+3240r4+16065r3+36360r2+36492r+12320 322

Using the above table, then the solution y2(x) is

y2(x)=x1/3(b0+b1x+b2x2+b3x3+b4x4+b5x5+b6x6)=1+xx2+3x353x410+3x522+O(x6)x

Therefore the homogeneous solution is

yh(x)=c1y1(x)+c2y2(x)=c1x1/3(17x9+35x281455x32187+1820x4196836916x5177147+O(x6))+c2(1+xx2+3x353x410+3x522+O(x6))x

Hence the final solution is

y=yh=c1x1/3(17x9+35x281455x32187+1820x4196836916x5177147+O(x6))+c2(1+xx2+3x353x410+3x522+O(x6))x
Maple. Time used: 0.015 (sec). Leaf size: 48
Order:=6; 
ode:=x^2*(x+3)*diff(diff(y(x),x),x)+5*x*(x+1)*diff(y(x),x)-(1-4*x)*y(x) = 0; 
dsolve(ode,y(x),type='series',x=0);
 
y=c2x4/3(179x+3581x24552187x3+182019683x46916177147x5+O(x6))+c1(1+xx2+35x3310x4+322x5+O(x6))x

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
trying a symmetry of the form [xi=0, eta=F(x)] 
   One independent solution has integrals. Trying a hypergeometric solution free of integrals... 
   -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
   <- hyper3 successful: received ODE is equivalent to the 2F1 ODE 
   -> Trying to convert hypergeometric functions to elementary form... 
   <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals 
<- linear_1 successful`
 

Maple step by step

Let’s solvex2(x+3)(d2dx2y(x))+5x(x+1)(ddxy(x))(14x)y(x)=0Highest derivative means the order of the ODE is2d2dx2y(x)Isolate 2nd derivatived2dx2y(x)=(4x1)y(x)x2(x+3)5(x+1)(ddxy(x))x(x+3)Group terms withy(x)on the lhs of the ODE and the rest on the rhs of the ODE; ODE is lineard2dx2y(x)+5(x+1)(ddxy(x))x(x+3)+(4x1)y(x)x2(x+3)=0Check to see ifx0is a regular singular pointDefine functions[P2(x)=5(x+1)x(x+3),P3(x)=4x1x2(x+3)](x+3)P2(x)is analytic atx=3((x+3)P2(x))|x=3=103(x+3)2P3(x)is analytic atx=3((x+3)2P3(x))|x=3=0x=3is a regular singular pointCheck to see ifx0is a regular singular pointx0=3Multiply by denominatorsx2(x+3)(d2dx2y(x))+5x(x+1)(ddxy(x))+(4x1)y(x)=0Change variables usingx=u3so that the regular singular point is atu=0(u36u2+9u)(d2du2y(u))+(5u225u+30)(dduy(u))+(4u13)y(u)=0Assume series solution fory(u)y(u)=k=0akuk+rRewrite ODE with series expansionsConvertumy(u)to series expansion form=0..1umy(u)=k=0akuk+r+mShift index usingk>kmumy(u)=k=makmuk+rConvertum(dduy(u))to series expansion form=0..2um(dduy(u))=k=0ak(k+r)uk+r1+mShift index usingk>k+1mum(dduy(u))=k=1+mak+1m(k+1m+r)uk+rConvertum(d2du2y(u))to series expansion form=1..3um(d2du2y(u))=k=0ak(k+r)(k+r1)uk+r2+mShift index usingk>k+2mum(d2du2y(u))=k=2+mak+2m(k+2m+r)(k+1m+r)uk+rRewrite ODE with series expansions3a0r(7+3r)u1+r+(3a1(1+r)(10+3r)a0(13+6r)(1+r))ur+(k=1(3ak+1(k+r+1)(3k+10+3r)ak(6k+6r+13)(k+r+1)+ak1(k+r+1)2)uk+r)=0a0cannot be 0 by assumption, giving the indicial equation3r(7+3r)=0Values of r that satisfy the indicial equationr{0,73}Each term must be 03a1(1+r)(10+3r)a0(13+6r)(1+r)=0Each term in the series must be 0, giving the recursion relation6((akak163ak+12)k+(akak163ak+12)r+13ak6ak165ak+1)(k+r+1)=0Shift index usingk>k+16((ak+1ak63ak+22)(k+1)+(ak+1ak63ak+22)r+13ak+16ak65ak+2)(k+r+2)=0Recursion relation that defines series solution to ODEak+2=kak6kak+1+rak6rak+1+2ak19ak+13(3k+13+3r)Recursion relation forr=0ak+2=kak6kak+1+2ak19ak+13(3k+13)Solution forr=0[y(u)=k=0akuk,ak+2=kak6kak+1+2ak19ak+13(3k+13),30a113a0=0]Revert the change of variablesu=x+3[y(x)=k=0ak(x+3)k,ak+2=kak6kak+1+2ak19ak+13(3k+13),30a113a0=0]Recursion relation forr=73ak+2=kak6kak+113ak5ak+13(3k+6)Solution forr=73[y(u)=k=0akuk73,ak+2=kak6kak+113ak5ak+13(3k+6),12a14a03=0]Revert the change of variablesu=x+3[y(x)=k=0ak(x+3)k73,ak+2=kak6kak+113ak5ak+13(3k+6),12a14a03=0]Combine solutions and rename parameters[y(x)=(k=0ak(x+3)k)+(k=0bk(x+3)k73),ak+2=kak6kak+1+2ak19ak+13(3k+13),30a113a0=0,bk+2=kbk6kbk+113bk5bk+13(3k+6),12b14b03=0]
Mathematica. Time used: 0.016 (sec). Leaf size: 82
ode=x^2*(3+x)*D[y[x],{x,2}] + 5*x*(1+x)*D[y[x],x] - (1-4*x)*y[x] == 0; 
ic={}; 
AsymptoticDSolveValue[{ode,ic},y[x],{x,0,5}]
 
y(x)c1x3(6916x5177147+1820x419683455x32187+35x2817x9+1)+c2(3x5223x410+3x35x2+x+1)x
Sympy. Time used: 1.118 (sec). Leaf size: 15
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**2*(x + 3)*Derivative(y(x), (x, 2)) + 5*x*(x + 1)*Derivative(y(x), x) - (1 - 4*x)*y(x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics,hint="2nd_power_series_regular",x0=0,n=6)
 
y(x)=C2x3+C1x+O(x6)