2.1.52 Problem 52

Solved as second order missing x ode
Maple
Mathematica
Sympy

Internal problem ID [9123]
Book : Second order enumerated odes
Section : section 1
Problem number : 52
Date solved : Wednesday, March 05, 2025 at 07:26:52 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

yy3+y3y5=0

Factoring the ode gives these factors

(1)y=0(2)y2y5+y3=0

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for y from

y=0

Solving gives y=0

Solving equation (2)

Solved as second order missing x ode

Time used: 226.095 (sec)

This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable y an independent variable. Using

y=p

Then

y=dpdx=dpdydydx=pdpdy

Hence the ode becomes

yp(y)3(ddyp(y))3+y3p(y)5=0

Which is now solved as first order ode for p(y).

Factoring the ode gives these factors

(1)p3=0(2)p2y2+p3=0

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for p from

p3=0

Solving gives p=0

Solving equation (2)

The ode has the form

(1)(p)nm=f(y)g(p)

Where n=3,m=1,f=y2,g=p2. Hence the ode is

(p)3=p2y2

Solving for p from (1) gives

p=(fg)1/3p=(fg)1/32+i3(fg)1/32p=(fg)1/32i3(fg)1/32

To be able to solve as separable ode, we have to now assume that f>0,g>0.

y2>0p2>0

Under the above assumption the differential equations become separable and can be written as

p=f1/3g1/3p=f1/3g1/3(1+i3)2p=f1/3g1/3(1+i3)2

Therefore

1g1/3dp=(f1/3)dy2g1/3(1+i3)dp=(f1/3)dy2g1/3(1+i3)dp=(f1/3)dy

Replacing f(y),g(p) by their values gives

1(p2)1/3dp=((y2)1/3)dy2(p2)1/3(1+i3)dp=((y2)1/3)dy2(p2)1/3(1+i3)dp=((y2)1/3)dy

Integrating now gives the following solutions

1(p2)1/3dp=(y2)1/3dy+c13(p2)2/3p=3y(y2)1/352(p2)1/3(1+i3)dp=(y2)1/3dy+c13(p2)2/3(1+i3)2p=3y(y2)1/352(p2)1/3(1+i3)dp=(y2)1/3dy+c13(p2)2/3(1+i3)2p=3y(y2)1/35

Therefore

3(p2)2/3p=3y(y2)1/35+c1p=y5125+(y2)2/3c1y225+(y2)1/3c12y15+c1327p=y5125+(y2)2/3c1y225+(y2)1/3c12y15+c1327

For solution (1) found earlier, since p=y then we now have a new first order ode to solve which is

3(y2)2/3y=3y(y2)1/35+c1

Let p=y the ode becomes

3(p2)2/3p=3y(y2)1/35+c1

Solving for y from the above results in

(1)y=53/532/5((27(p2)4/3c19(p2)2/3c12p+c13p227p3)p3)1/53p(2)y=(5414+i25+54)53/532/5((27(p2)4/3c19(p2)2/3c12p+c13p227p3)p3)1/53p(3)y=(5414+i2554)53/532/5((27(p2)4/3c19(p2)2/3c12p+c13p227p3)p3)1/53p(4)y=(5414i2554)53/532/5((27(p2)4/3c19(p2)2/3c12p+c13p227p3)p3)1/53p(5)y=(5414i25+54)53/532/5((27(p2)4/3c19(p2)2/3c12p+c13p227p3)p3)1/53p

This has the form

(*)y=xf(p)+g(p)

Where f,g are functions of p=y(x). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

Taking derivative of (*) w.r.t. x gives

p=f+(xf+g)dpdx(2)pf=(xf+g)dpdx

Comparing the form y=xf+g to (1A) shows that

f=0g=53/532/5(27(c12(p2)2/33+p(c1327c1(p2)1/3+p))p4)1/53p

Hence (2) becomes

(2A)p=(453/532/5p4c125(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/5(p2)1/3+53/532/5p3c133(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/5+953/532/5p3c1(p2)1/3(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/55453/532/5p45(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/5+653/532/5p5c15(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/5(p2)2/31253/532/5p2c12(p2)2/35(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)4/553/532/5(9p4c12(p2)2/3+p5c13+27p5c1(p2)1/327p6)1/53p2)p(x)

The singular solution is found by setting dpdx=0 in the above which gives

p=0

Solving the above for p results in

p1=0

Substituting these in (1A) and keeping singular solution that verifies the ode gives

y=(c13)1/553/532/53

The general solution is found when dpdx0. From eq. (2A). This results in

(3)p(x)=p(x)453/532/5p(x)4c125(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/5(p(x)2)1/3+53/532/5p(x)3c133(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/5+953/532/5p(x)3c1(p(x)2)1/3(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/55453/532/5p(x)45(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/5+653/532/5p(x)5c15(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/5(p(x)2)2/31253/532/5p(x)2c12(p(x)2)2/35(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)4/553/532/5(9p(x)4c12(p(x)2)2/3+p(x)5c13+27p(x)5c1(p(x)2)1/327p(x)6)1/53p(x)2

This ODE is now solved for p(x). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

p(x)53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3+τ(c13+27c1(τ2)1/327τ))τ4)4/5(τ2)2/3dτ=x+c2

Substituing the above solution for p in (2A) gives

y=53/532/5(27(c12(RootOf(_Zτ3(c12(τ2)1/3+9(τ2)2/36c1τ)6754/5225(τ2)2/3((9c12(τ2)2/3+c13τ+27(τ2)1/3c1τ27τ2)4τ16)1/5dτ+x+c2)2)2/33+RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)(c1327c1(RootOf(_Zτ3(c12(τ2)1/3+9(τ2)2/36c1τ)6754/5225(τ2)2/3((9c12(τ2)2/3+c13τ+27(τ2)1/3c1τ27τ2)4τ16)1/5dτ+x+c2)2)1/3+RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)))RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)4)1/53RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)

Solving ode 2A

Taking derivative of (*) w.r.t. x gives

p=f+(xf+g)dpdx(2)pf=(xf+g)dpdx

Comparing the form y=xf+g to (1A) shows that

f=0g=(i25+5+51)53/532/5(27(c12(p2)2/33+p(c1327c1(p2)1/3+p))p4)1/512p

Hence (2) becomes

(2A)Expression too large to display

The singular solution is found by setting dpdx=0 in the above which gives

p=0

Solving the above for p results in

p1=0

Substituting these in (1A) and keeping singular solution that verifies the ode gives

y=(c13)1/532/5(i53/525+553/5+551/10)12

The general solution is found when dpdx0. From eq. (2A). This results in

(3)Expression too large to display

This ODE is now solved for p(x). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

p(x)32/5τ3(c12(τ2)1/39(τ2)2/3+6c1τ)(i53/525+553/5+551/10)20((9c12(τ2)2/3+τ(c13+27c1(τ2)1/327τ))τ4)4/5(τ2)2/3dτ=x+c3

Substituing the above solution for p in (2A) gives

Expression too large to display

Solving ode 3A

Taking derivative of (*) w.r.t. x gives

p=f+(xf+g)dpdx(2)pf=(xf+g)dpdx

Comparing the form y=xf+g to (1A) shows that

f=0g=(i25551)53/532/5(27(c12(p2)2/33+p(c1327c1(p2)1/3+p))p4)1/512p

Hence (2) becomes

(2A)Expression too large to display

The singular solution is found by setting dpdx=0 in the above which gives

p=0

Solving the above for p results in

p1=0

Substituting these in (1A) and keeping singular solution that verifies the ode gives

y=(c13)1/532/5(i53/525553/5551/10)12

The general solution is found when dpdx0. From eq. (2A). This results in

(3)Expression too large to display

This ODE is now solved for p(x). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

p(x)32/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)(i53/525553/5551/10)20((9c12(τ2)2/3+τ(c13+27c1(τ2)1/327τ))τ4)4/5(τ2)2/3dτ=x+c4

Substituing the above solution for p in (2A) gives

Expression too large to display

Solving ode 4A

Taking derivative of (*) w.r.t. x gives

p=f+(xf+g)dpdx(2)pf=(xf+g)dpdx

Comparing the form y=xf+g to (1A) shows that

f=0g=(i255+5+1)53/532/5(27(c12(p2)2/33+p(c1327c1(p2)1/3+p))p4)1/512p

Hence (2) becomes

(2A)Expression too large to display

The singular solution is found by setting dpdx=0 in the above which gives

p=0

Solving the above for p results in

p1=0

Substituting these in (1A) and keeping singular solution that verifies the ode gives

y=(c13)1/532/5(i53/5255+53/5+551/10)12

The general solution is found when dpdx0. From eq. (2A). This results in

(3)Expression too large to display

This ODE is now solved for p(x). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

p(x)32/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)(i53/5255+53/5+551/10)20((9c12(τ2)2/3+τ(c13+27c1(τ2)1/327τ))τ4)4/5(τ2)2/3dτ=x+c5

Substituing the above solution for p in (2A) gives

Expression too large to display

Solving ode 5A

Taking derivative of (*) w.r.t. x gives

p=f+(xf+g)dpdx(2)pf=(xf+g)dpdx

Comparing the form y=xf+g to (1A) shows that

f=0g=(i25+5+51)53/532/5(27(c12(p2)2/33+p(c1327c1(p2)1/3+p))p4)1/512p

Hence (2) becomes

(2A)Expression too large to display

The singular solution is found by setting dpdx=0 in the above which gives

p=0

Solving the above for p results in

p1=0

Substituting these in (1A) and keeping singular solution that verifies the ode gives

y=(c13)1/532/5(i53/525+5+53/5551/10)12

The general solution is found when dpdx0. From eq. (2A). This results in

(3)Expression too large to display

This ODE is now solved for p(x). No inversion is needed.

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

p(x)32/5τ3(c12(τ2)1/39(τ2)2/3+6c1τ)(i53/525+5+53/5551/10)20((9c12(τ2)2/3+τ(c13+27c1(τ2)1/327τ))τ4)4/5(τ2)2/3dτ=x+c6

Substituing the above solution for p in (2A) gives

Expression too large to display

For solution (2) found earlier, since p=y then we now have a new first order ode to solve which is

y=0

Since the ode has the form y=f(x), then we only need to integrate f(x).

dy=0dx+c7y=c7

For solution (3) found earlier, since p=y then we now have a new first order ode to solve which is

y=y5125+y2(y2)2/3c125+y(y2)1/3c1215+c1327

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

y1τ5125+(τ2)2/3c1τ225+(τ2)1/3c12τ15+c1327dτ=x+c8

Will add steps showing solving for IC soon.

The solution

Expression too large to display

was found not to satisfy the ode or the IC. Hence it is removed. The solution

Expression too large to display

was found not to satisfy the ode or the IC. Hence it is removed. The solution

Expression too large to display

was found not to satisfy the ode or the IC. Hence it is removed. The solution

Expression too large to display

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

y1τ5125+(τ2)2/3c1τ225+(τ2)1/3c12τ15+c1327dτ=x+c8y=c7y=(c13)1/532/5(i53/525553/5551/10)12y=(c13)1/532/5(i53/5255+53/5+551/10)12y=(c13)1/532/5(i53/525+553/5+551/10)12y=(c13)1/532/5(i53/525+5+53/5551/10)12y=(c13)1/553/532/53y=53/532/5(27(c12(RootOf(_Zτ3(c12(τ2)1/3+9(τ2)2/36c1τ)6754/5225(τ2)2/3((9c12(τ2)2/3+c13τ+27(τ2)1/3c1τ27τ2)4τ16)1/5dτ+x+c2)2)2/33+RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)(c1327c1(RootOf(_Zτ3(c12(τ2)1/3+9(τ2)2/36c1τ)6754/5225(τ2)2/3((9c12(τ2)2/3+c13τ+27(τ2)1/3c1τ27τ2)4τ16)1/5dτ+x+c2)2)1/3+RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)))RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)4)1/53RootOf(_Z53/532/5τ3(c12(τ2)1/3+9(τ2)2/36c1τ)5((9c12(τ2)2/3c13τ27(τ2)1/3c1τ+27τ2)τ4)4/5(τ2)2/3dτ+x+c2)

Maple. Time used: 0.214 (sec). Leaf size: 208
ode:=y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x)^5 = 0; 
dsolve(ode,y(x), singsol=all);
 
y=0y=c1y1RootOf(5(_g_Z1_a(_a2_f2)1/35_fd_f)ln(_a5+125)+5c1)d_axc2=0y1RootOf(3ln(_a5+125)iln(_a5+125)+20(_g_Z12i_a(_a2_f2)1/3+5i_f+53_fd_f)20c1)d_axc2=0y1RootOf(3ln(_a5+125)+iln(_a5+125)+20(_g_Z12i_a(_a2_f2)1/3+53_f5i_fd_f)20c1)d_axc2=0

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
      `, `-> Computing symmetries using: way = 3 
      -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(-_a^2*_b(_a)^2)^(1/3)*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]]` 
         symmetry methods on request 
      `, `1st order, trying reduction of order with given symmetries:`[_a, 5*_b]
 

Mathematica. Time used: 24.151 (sec). Leaf size: 449
ode=y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]^5==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
Solution too large to show

Sympy. Time used: 13.171 (sec). Leaf size: 3
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(y(x)**3*Derivative(y(x), x)**5 + y(x)*Derivative(y(x), (x, 2))**3,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
y(x)=0