2.1.64 Problem 64

Solved as first order homogeneous class C ode
Solved using Lie symmetry for first order ode
Solved as first order ode of type dAlembert
Maple
Mathematica
Sympy

Internal problem ID [9048]
Book : First order enumerated odes
Section : section 1
Problem number : 64
Date solved : Wednesday, March 05, 2025 at 07:18:07 AM
CAS classification : [[_homogeneous, `class C`], _dAlembert]

Solve

y=(π+x+7y)7/2

Solved as first order homogeneous class C ode

Time used: 0.567 (sec)

Let

(1)z=π+x+7y

Then

z(x)=1+7y

Therefore

y=z(x)717

Hence the given ode can now be written as

z(x)717=z7/2

This is separable first order ode. Integrating

dx=17z7/2+1dzx+c1=(_R=RootOf(49_Z71)ln(z_R)_R6)343+(_R=RootOf(7_Z7+1)ln(z_R)_R5)49+(_R=RootOf(7_Z71)ln(z_R)_R5)49

Replacing z back by its value from (1) then the above gives the solution as

Summary of solutions found

\begin{align*} -\frac {\left (\munderset {\textit {\_R} &=\operatorname {RootOf}\left (49 \textit {\_Z}^{7}-1\right )}{\sum }\frac {\ln \left (\pi +x +7 y-\textit {\_R} \right )}{\textit {\_R}^{6}}\right )}{343}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (7 \textit {\_Z}^{7}+1\right )}{\sum }\frac {\ln \left (\sqrt {\pi +x +7 y}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{49}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (7 \textit {\_Z}^{7}-1\right )}{\sum }\frac {\ln \left (\sqrt {\pi +x +7 y}-\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{49} = x +c_1 \\ \end{align*}
Solved using Lie symmetry for first order ode

Time used: 2.763 (sec)

Writing the ode as

y=(π+x+7y)7/2y=ω(x,y)

The condition of Lie symmetry is the linearized PDE given by

(A)ηx+ω(ηyξx)ω2ξyωxξωyη=0

To determine ξ,η then (A) is solved using ansatz. Making bivariate polynomials of degree 1 to use as anstaz gives

(1E)ξ=xa2+ya3+a1(2E)η=xb2+yb3+b1

Where the unknown coefficients are

{a1,a2,a3,b1,b2,b3}

Substituting equations (1E,2E) and ω into (A) gives

(5E)b2+(π+x+7y)7/2(b3a2)(π+x+7y)7a37(π+x+7y)5/2(xa2+ya3+a1)249(π+x+7y)5/2(xb2+yb3+b1)2=0

Putting the above in normal form gives

294πx5ya35145πx4y2a348020πx3y3a3252105πx2y4a3705894πxy5a3294π5xya3735π4x2ya35145π4xy2a3980π3x3ya310290π3x2y2a348020π3xy3a3735π2x4ya310290π2x3y2a372030π2x2y3a3252105π2xy4a3x7a3823543y7a3+(π+x+7y)7/2b3(π+x+7y)7/2a2π7a37(π+x+7y)5/2a1249(π+x+7y)5/2b1249x6ya31029x5y2a312005x4y3a384035x3y4a3352947x2y5a3823543xy6a37π6xa349π6ya321π5x2a31029π5y2a335π4x3a312005π4y3a335π3x4a384035π3y4a321π2x5a3352947π2y5a37πx6a3823543πy6a37(π+x+7y)5/2xa227(π+x+7y)5/2ya3249(π+x+7y)5/2xb2249(π+x+7y)5/2yb32+b2=0

Setting the numerator to zero gives

(6E)588πx5ya310290πx4y2a396040πx3y3a3504210πx2y4a31411788πxy5a3588π5xya31470π4x2ya310290π4xy2a31960π3x3ya320580π3x2y2a396040π3xy3a31470π2x4ya320580π2x3y2a3144060π2x2y3a3504210π2xy4a32x7a31647086y7a3+2(π+x+7y)7/2b32(π+x+7y)7/2a22π7a37(π+x+7y)5/2a149(π+x+7y)5/2b198x6ya32058x5y2a324010x4y3a3168070x3y4a3705894x2y5a31647086xy6a314π6xa398π6ya342π5x2a32058π5y2a370π4x3a324010π4y3a370π3x4a3168070π3y4a342π2x5a3705894π2y5a314πx6a31647086πy6a37(π+x+7y)5/2xa27(π+x+7y)5/2ya349(π+x+7y)5/2xb249(π+x+7y)5/2yb3+2b2=0

Since the PDE has radicals, simplifying gives

7π2π+x+7yyb320πx2π+x+7ya298πx2π+x+7yb2+6πx2π+x+7yb3294ππ+x+7yy2a298ππ+x+7yy2a3392ππ+x+7yy2b3140x2π+x+7yya27x2π+x+7yya3686x2π+x+7yyb27x2π+x+7yyb3637xπ+x+7yy2a298xπ+x+7yy2a32401xπ+x+7yy2b2392xπ+x+7yy2b314πxπ+x+7ya198πxπ+x+7yb198ππ+x+7yya1686ππ+x+7yyb198xπ+x+7yya1686xπ+x+7yyb113π2xπ+x+7ya249π2xπ+x+7yb2+6π2xπ+x+7yb342π2π+x+7yya27π2π+x+7yya3182πxπ+x+7yya214πxπ+x+7yya3686πxπ+x+7yyb214πxπ+x+7yyb3588πx5ya310290πx4y2a396040πx3y3a3504210πx2y4a31411788πxy5a3588π5xya31470π4x2ya310290π4xy2a31960π3x3ya320580π3x2y2a396040π3xy3a31470π2x4ya320580π2x3y2a3144060π2x2y3a3504210π2xy4a32x7a31647086y7a32π7a39x3π+x+7ya249x3π+x+7yb2+2x3π+x+7yb3686π+x+7yy3a2343π+x+7yy3a31715π+x+7yy3b37π2π+x+7ya149π2π+x+7yb17x2π+x+7ya149x2π+x+7yb1343π+x+7yy2a12401π+x+7yy2b12π3π+x+7ya2+2π3π+x+7yb398x6ya32058x5y2a324010x4y3a3168070x3y4a3705894x2y5a31647086xy6a314π6xa398π6ya342π5x2a32058π5y2a370π4x3a324010π4y3a370π3x4a3168070π3y4a342π2x5a3705894π2y5a314πx6a31647086πy6a3+2b2=0

Looking at the above PDE shows the following are all the terms with {x,y} in them.

{x,y,π+x+7y}

The following substitution is now made to be able to collect on all terms with {x,y} in them

{x=v1,y=v2,π+x+7y=v3}

The above PDE (6E) now becomes

(7E)2π7a314π6v1a398π6v2a342π5v12a3588π5v1v2a32058π5v22a370π4v13a31470π4v12v2a310290π4v1v22a324010π4v23a370π3v14a31960π3v13v2a320580π3v12v22a396040π3v1v23a3168070π3v24a342π2v15a31470π2v14v2a320580π2v13v22a3144060π2v12v23a3504210π2v1v24a3705894π2v25a314πv16a3588πv15v2a310290πv14v22a396040πv13v23a3504210πv12v24a31411788πv1v25a31647086πv26a32v17a398v16v2a32058v15v22a324010v14v23a3168070v13v24a3705894v12v25a31647086v1v26a31647086v27a32π3v3a2+2π3v3b313π2v1v3a242π2v3v2a27π2v3v2a349π2v1v3b2+6π2v1v3b37π2v3v2b320πv12v3a2182πv1v3v2a2294πv3v22a214πv1v3v2a398πv3v22a398πv12v3b2686πv1v3v2b2+6πv12v3b314πv1v3v2b3392πv3v22b39v13v3a2140v12v3v2a2637v1v3v22a2686v3v23a27v12v3v2a398v1v3v22a3343v3v23a349v13v3b2686v12v3v2b22401v1v3v22b2+2v13v3b37v12v3v2b3392v1v3v22b31715v3v23b37π2v3a149π2v3b114πv1v3a198πv3v2a198πv1v3b1686πv3v2b17v12v3a198v1v3v2a1343v3v22a149v12v3b1686v1v3v2b12401v3v22b1+2b2=0

Collecting the above on the terms vi introduced, and these are

{v1,v2,v3}

Equation (7E) now becomes

(8E)2π7a3+(2π3a2+2π3b37π2a149π2b1)v398v16v2a32058v15v22a324010v14v23a3168070v13v24a3705894v12v25a31647086v1v26a314π6v1a398π6v2a342π5v12a32058π5v22a370π4v13a324010π4v23a370π3v14a3168070π3v24a342π2v15a3705894π2v25a314πv16a31647086πv26a3+(140a27a3686b27b3)v12v2v3+(637a298a32401b2392b3)v1v22v3+(182πa214πa3686πb214πb398a1686b1)v1v2v3+2b2+(686a2343a31715b3)v23v3+(294πa298πa3392πb3343a12401b1)v22v3+(42π2a27π2a37π2b398πa1686πb1)v2v3+(9a249b2+2b3)v13v3+(20πa298πb2+6πb37a149b1)v12v3+(13π2a249π2b2+6π2b314πa198πb1)v1v3588πv15v2a310290πv14v22a396040πv13v23a3504210πv12v24a31411788πv1v25a3588π5v1v2a31470π4v12v2a310290π4v1v22a31960π3v13v2a320580π3v12v22a396040π3v1v23a31470π2v14v2a320580π2v13v22a3144060π2v12v23a3504210π2v1v24a32v17a31647086v27a3=0

Setting each coefficients in (8E) to zero gives the following equations to solve

1647086a3=0705894a3=0168070a3=024010a3=02058a3=098a3=02a3=01647086πa3=01411788πa3=0504210πa3=096040πa3=010290πa3=0588πa3=014πa3=0705894π2a3=0504210π2a3=0144060π2a3=020580π2a3=01470π2a3=042π2a3=0168070π3a3=096040π3a3=020580π3a3=01960π3a3=070π3a3=024010π4a3=010290π4a3=01470π4a3=070π4a3=02058π5a3=0588π5a3=042π5a3=098π6a3=014π6a3=0686a2343a31715b3=09a249b2+2b3=0637a298a32401b2392b3=0140a27a3686b27b3=02π7a3+2b2=02π3a2+2π3b37π2a149π2b1=0294πa298πa3392πb3343a12401b1=042π2a27π2a37π2b398πa1686πb1=020πa298πb2+6πb37a149b1=013π2a249π2b2+6π2b314πa198πb1=0182πa214πa3686πb214πb398a1686b1=0

Solving the above equations for the unknowns gives

a1=7b1a2=0a3=0b1=b1b2=0b3=0

Substituting the above solution in the anstaz (1E,2E) (using 1 as arbitrary value for any unknown in the RHS) gives

ξ=7η=1

The next step is to determine the canonical coordinates R,S. The canonical coordinates map (x,y)(R,S) where (R,S) are the canonical coordinates which make the original ode become a quadrature and hence solved by integration.

The characteristic pde which is used to find the canonical coordinates is

(1)dxξ=dyη=dS

The above comes from the requirements that (ξx+ηy)S(x,y)=1. Starting with the first pair of ode’s in (1) gives an ode to solve for the independent variable R in the canonical coordinates, where S(R). Therefore

dydx=ηξ=17=17

This is easily solved to give

y=x7+c1

Where now the coordinate R is taken as the constant of integration. Hence

R=x7+y

And S is found from

dS=dxξ=dx7

Integrating gives

S=dxT=x7

Where the constant of integration is set to zero as we just need one solution. Now that R,S are found, we need to setup the ode in these coordinates. This is done by evaluating

(2)dSdR=Sx+ω(x,y)SyRx+ω(x,y)Ry

Where in the above Rx,Ry,Sx,Sy are all partial derivatives and ω(x,y) is the right hand side of the original ode given by

ω(x,y)=(π+x+7y)7/2

Evaluating all the partial derivatives gives

Rx=17Ry=1Sx=17Sy=0

Substituting all the above in (2) and simplifying gives the ode in canonical coordinates.

(2A)dSdR=11+7(π+x+7y)7/2

We now need to express the RHS as function of R only. This is done by solving for x,y in terms of R,S from the result obtained earlier and simplifying. This gives

dSdR=11+7(π+7R)7/2

The above is a quadrature ode. This is the whole point of Lie symmetry method. It converts an ode, no matter how complicated it is, to one that can be solved by integration when the ode is in the canonical coordiates R,S.

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

dS=11+7(π+7R)7/2dRS(R)=2(_R=RootOf(7_Z7+1)ln(π+7R_R)_R5)343+c2
S(R)=11+7(π+7R)7/2dR+c2

This results in

x7=y11+7(π+x+7_a)7/2d_a+c2

Summary of solutions found

x7=y11+7(π+x+7_a)7/2d_a+c2
Solved as first order ode of type dAlembert

Time used: 13.437 (sec)

Let p=y the ode becomes

p=(π+x+7y)7/2

Solving for y from the above results in

(1)y=p2/77π7x7(2)y=(cos(2π7)+icos(3π14))2p2/77π7x7(3)y=(cos(3π7)+icos(π14))2p2/77π7x7(4)y=(cos(π7)+icos(5π14))2p2/77π7x7(5)y=(cos(π7)icos(5π14))2p2/77π7x7(6)y=(cos(3π7)icos(π14))2p2/77π7x7(7)y=(cos(2π7)icos(3π14))2p2/77π7x7

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=17g=p2/77π7

Hence (2) becomes

(2A)p+17=2p(x)49p5/7

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

p+17=0

Solving the above for p results in

p1=17

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

y=(1)2/775/749π7x7

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

(3)p(x)=49(p(x)+17)p(x)5/72

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)27(7τ+1)τ5/7dτ=x+c1

Singular solutions are found by solving

7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=RootOf(_Z27(7τ+1)τ5/7dτ+x+c1)2/77π7x7y=π7x7y=(1)2/775/749π7x7

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=17g=p2/7(cos(3π7)+isin(3π7))7π7

Hence (2) becomes

(2A)p+17=(2cos(3π7)49p5/7+2isin(3π7)49p5/7)p(x)

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

p+17=0

Solving the above for p results in

p1=17

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

y=75/7(1)6/749π7x7

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

(3)p(x)=p(x)+172cos(3π7)49p(x)5/7+2isin(3π7)49p(x)5/7

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)2(1)4/77(7τ+1)τ5/7dτ=x+c2

Singular solutions are found by solving

7(1)3/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+RootOf(_Z2(1)4/77(7τ+1)τ5/7dτ+x+c2)2/7(cos(3π7)+isin(3π7))7π7y=π7x7y=x7cos(3π7)(1)2/775/749+sin(3π7)(1)11/1475/749π7

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=17g=p2/7(isin(π7)cos(π7))7π7

Hence (2) becomes

(2A)p+17=(2isin(π7)49p5/72cos(π7)49p5/7)p(x)

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

p+17=0

No valid singular solutions found.

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

(3)p(x)=p(x)+172isin(π7)49p(x)5/72cos(π7)49p(x)5/7

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)2(1)1/77(7τ+1)τ5/7dτ=x+c3

Singular solutions are found by solving

7(1)6/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+RootOf(_Z2(1)1/77(7τ+1)τ5/7dτ+x+c3)2/7(isin(π7)cos(π7))7π7y=π7x7y=x7(1)11/1475/7sin(π7)49(1)2/775/7cos(π7)49π7

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=17g=(isin(2π7)+cos(2π7))p2/77π7

Hence (2) becomes

(2A)p+17=(2isin(2π7)49p5/7+2cos(2π7)49p5/7)p(x)

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

p+17=0

No valid singular solutions found.

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

(3)p(x)=p(x)+172isin(2π7)49p(x)5/7+2cos(2π7)49p(x)5/7

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)2(1)5/77(7τ+1)τ5/7dτ=x+c4

Singular solutions are found by solving

7(1)2/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+(isin(2π7)+cos(2π7))RootOf(_Z2(1)5/77(7τ+1)τ5/7dτ+x+c4)2/77π7y=π7x7y=x7(1)11/1475/7sin(2π7)49+(1)2/775/7cos(2π7)49π7

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=17g=p2/7(isin(2π7)+cos(2π7))7π7

Hence (2) becomes

(2A)p+17=(2isin(2π7)49p5/7+2cos(2π7)49p5/7)p(x)

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

p+17=0

No valid singular solutions found.

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

(3)p(x)=p(x)+172isin(2π7)49p(x)5/7+2cos(2π7)49p(x)5/7

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)2(1)2/77(7τ+1)τ5/7dτ=x+c5

Singular solutions are found by solving

7(1)5/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+RootOf(_Z2(1)2/77(7τ+1)τ5/7dτ+x+c5)2/7(isin(2π7)+cos(2π7))7π7y=π7x7y=x7+(1)11/1475/7sin(2π7)49+(1)2/775/7cos(2π7)49π7

Solving ode 6A

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=17g=p2/7(isin(π7)cos(π7))7π7

Hence (2) becomes

(2A)p+17=(2isin(π7)49p5/72cos(π7)49p5/7)p(x)

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

p+17=0

Solving the above for p results in

p1=17

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

y=75/7(1)1/749π7x7

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

(3)p(x)=p(x)+172isin(π7)49p(x)5/72cos(π7)49p(x)5/7

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)2(1)6/77(7τ+1)τ5/7dτ=x+c6

Singular solutions are found by solving

7(1)1/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+RootOf(_Z2(1)6/77(7τ+1)τ5/7dτ+x+c6)2/7(isin(π7)cos(π7))7π7y=π7x7y=x7+(1)11/1475/7sin(π7)49(1)2/775/7cos(π7)49π7

Solving ode 7A

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=17g=p2/7(cos(3π7)isin(3π7))7π7

Hence (2) becomes

(2A)p+17=(2cos(3π7)49p5/72isin(3π7)49p5/7)p(x)

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

p+17=0

Solving the above for p results in

p1=17

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

y=(7)5/749π7x7

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

(3)p(x)=p(x)+172cos(3π7)49p(x)5/72isin(3π7)49p(x)5/7

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)2(1)3/77(7τ+1)τ5/7dτ=x+c7

Singular solutions are found by solving

7(1)4/7(7p+1)p5/72=0

for p(x). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

p(x)=0p(x)=17

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

y=x7+RootOf(_Z2(1)3/77(7τ+1)τ5/7dτ+x+c7)2/7(cos(3π7)isin(3π7))7π7y=π7x7y=x7cos(3π7)(1)2/775/749sin(3π7)(1)11/1475/749π7

The solution

y=π7x7

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

y=x7+(isin(2π7)+cos(2π7))RootOf(_Z2(1)5/77(7τ+1)τ5/7dτ+x+c4)2/77π7

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

y=x7+RootOf(_Z2(1)1/77(7τ+1)τ5/7dτ+x+c3)2/7(isin(π7)cos(π7))7π7

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

y=x7+RootOf(_Z2(1)2/77(7τ+1)τ5/7dτ+x+c5)2/7(isin(2π7)+cos(2π7))7π7

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

y=x7+RootOf(_Z2(1)3/77(7τ+1)τ5/7dτ+x+c7)2/7(cos(3π7)isin(3π7))7π7

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

y=x7+RootOf(_Z2(1)4/77(7τ+1)τ5/7dτ+x+c2)2/7(cos(3π7)+isin(3π7))7π7

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

y=x7+RootOf(_Z2(1)6/77(7τ+1)τ5/7dτ+x+c6)2/7(isin(π7)cos(π7))7π7

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

y=x7(1)11/1475/7sin(π7)49(1)2/775/7cos(π7)49π7

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

y=x7(1)11/1475/7sin(2π7)49+(1)2/775/7cos(2π7)49π7

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

y=x7+(1)11/1475/7sin(2π7)49+(1)2/775/7cos(2π7)49π7

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

Summary of solutions found

y=(7)5/749π7x7y=RootOf(_Z27(7τ+1)τ5/7dτ+x+c1)2/77π7x7y=(1)2/775/749π7x7y=75/7(1)1/749π7x7y=75/7(1)6/749π7x7y=x7+(1)11/1475/7sin(π7)49(1)2/775/7cos(π7)49π7y=x7cos(3π7)(1)2/775/749sin(3π7)(1)11/1475/749π7y=x7cos(3π7)(1)2/775/749+sin(3π7)(1)11/1475/749π7
Maple. Time used: 0.039 (sec). Leaf size: 33
ode:=diff(y(x),x) = (Pi+x+7*y(x))^(7/2); 
dsolve(ode,y(x), singsol=all);
 
y=x7+RootOf(x+7(_Z11+7(π+7_a)7/2d_a)+c1)

Maple trace

`Methods for first order ODEs: 
--- Trying classification methods --- 
trying homogeneous types: 
trying homogeneous C 
1st order, trying the canonical coordinates of the invariance group 
   -> Calling odsolve with the ODE`, diff(y(x), x) = -1/7, y(x)`      *** Sublevel 2 *** 
      Methods for first order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      trying 1st order linear 
      <- 1st order linear successful 
<- 1st order, canonical coordinates successful 
<- homogeneous successful`
 

Maple step by step

Let’s solveddxy(x)=(π+x+7y(x))7/2Highest derivative means the order of the ODE is1ddxy(x)Solve for the highest derivativeddxy(x)=(π+x+7y(x))7/2
Mathematica. Time used: 0.532 (sec). Leaf size: 43
ode=D[y[x],x]==(Pi+x+7*y[x])^(7/2); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
Solve[(7y(x)+x+π)(Hypergeometric2F1(27,1,97,7(x+7y(x)+π)7/2)1)7y(x)=c1,y(x)]
Sympy. Time used: 67.756 (sec). Leaf size: 400
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-(x + 7*y(x) + pi)**(7/2) + Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
y(x)=C1+7x+7π3C2x7π7r2401r3π7r1029r2ππ7r+147rπ2π7r7π3π7r1dr+1029πC2x7r2π7r2401r3π7r1029r2ππ7r+147rπ2π7r7π3π7r1dr2401C2x7r3π7r2401r3π7r1029r2ππ7r+147rπ2π7r7π3π7r1dr147π2C2x7rπ7r2401r3π7r1029r2ππ7r+147rπ2π7r7π3π7r1dr49C2x712401r3π7r1029r2ππ7r+147rπ2π7r7π3π7r1dr