2.1.59 Problem 59

Solved using first_order_nonlinear_p_but_separable
Maple
Mathematica
Sympy

Internal problem ID [9042]
Book : First order enumerated odes
Section : section 1
Problem number : 59
Date solved : Friday, April 25, 2025 at 05:36:08 PM
CAS classification : [_separable]

Solved using first_order_nonlinear_p_but_separable

Time used: 0.425 (sec)

Solve

y2=1x3y4

The ode has the form

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

Where n=2,m=1,f=1x3,g=1y4. Hence the ode is

(y)2=1x3y4

Solving for y from (1) gives

y=fgy=fg

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

1x3>01y4>0

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

y=fgy=fg

Therefore

1gdy=(f)dx1gdy=(f)dx

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

11y4dy=(1x3)dx11y4dy=(1x3)dx

Integrating now gives the following solutions

11y4dy=1x3dx+c1y51y43=2x1x311y4dy=1x3dx+c1y51y43=2x1x3

Therefore

y51y43=2x1x3+c1y51y43=2x1x3+c1

Summary of solutions found

y51y43=2x1x3+c1y51y43=2x1x3+c1
Maple. Time used: 0.036 (sec). Leaf size: 133
ode:=diff(y(x),x)^2 = 1/y(x)^4/x^3; 
dsolve(ode,y(x), singsol=all);
 
y=(c1x6x)1/3y=(c1x6x)1/3(1+i3)2y=(c1x6x)1/3(1+i3)2y=(c1x+6x)1/3y=(c1x+6x)1/3(1+i3)2y=(c1x+6x)1/3(1+i3)2

Maple trace

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
trying simple symmetries for implicit equations 
Successful isolation of dy/dx: 2 solutions were found. Trying to solve each res\ 
ulting ODE. 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
------------------- 
* Tackling next ODE. 
   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful
 

Maple step by step

Let’s solve(ddxy(x))2=1x3y(x)4Highest derivative means the order of the ODE is1ddxy(x)Solve for the highest derivative[ddxy(x)=1x3/2y(x)2,ddxy(x)=1x3/2y(x)2]Solve the equationddxy(x)=1x3/2y(x)2Separate variables(ddxy(x))y(x)2=1x3/2Integrate both sides with respect tox(ddxy(x))y(x)2dx=1x3/2dx+_C1Evaluate integraly(x)33=2x+_C1Solve fory(x){y(x)=(3_C1x6x)1/3,y(x)=(3_C1x6x)1/32I3(3_C1x6x)1/32,y(x)=(3_C1x6x)1/32+I3(3_C1x6x)1/32}Simplify{y(x)=(3_C1x6x)1/3,y(x)=(_C1x2x)1/3(31/3+I35/6)2,y(x)=(_C1x2x)1/3(31/3+I35/6)2}Redefine the integration constant(s){y(x)=(_C1x6x)1/3,y(x)=(_C1x2x)1/3(31/3+I35/6)2,y(x)=(_C1x2x)1/3(31/3+I35/6)2}Solve the equationddxy(x)=1x3/2y(x)2Separate variables(ddxy(x))y(x)2=1x3/2Integrate both sides with respect tox(ddxy(x))y(x)2dx=1x3/2dx+_C1Evaluate integraly(x)33=2x+_C1Solve fory(x){y(x)=(3_C1x+6x)1/3,y(x)=(3_C1x+6x)1/32I3(3_C1x+6x)1/32,y(x)=(3_C1x+6x)1/32+I3(3_C1x+6x)1/32}Simplify{y(x)=(3_C1x+6x)1/3,y(x)=(_C1x+2x)1/3(31/3+I35/6)2,y(x)=(_C1x+2x)1/3(31/3+I35/6)2}Redefine the integration constant(s){y(x)=(_C1x+6x)1/3,y(x)=(_C1x+2x)1/3(31/3+I35/6)2,y(x)=(_C1x+2x)1/3(31/3+I35/6)2}Set of solutions{y(x)=(C1x6x)1/3,y(x)=(C1x+6x)1/3,y(x)=(C1x2x)1/3(31/3+I35/6)2,y(x)=(C1x2x)1/3(31/3+I35/6)2,y(x)=(C1x+2x)1/3(31/3+I35/6)2,y(x)=(C1x+2x)1/3(31/3+I35/6)2}
Mathematica. Time used: 3.402 (sec). Leaf size: 157
ode=(D[y[x],x])^2==1/(x^3*y[x]^4); 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)332x+c13y(x)332x+c13y(x)(1)2/3332x+c13y(x)332x+c13y(x)332x+c13y(x)(1)2/3332x+c13
Sympy. Time used: 4.459 (sec). Leaf size: 170
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(Derivative(y(x), x)**2 - 1/(x**3*y(x)**4),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
[y(x)=C1+6x1x33, y(x)=(33356i)C1+2x1x332, y(x)=(33+356i)C1+2x1x332, y(x)=C16x1x33, y(x)=(33356i)C12x1x332, y(x)=(33+356i)C12x1x332]