2.3.25 Problem 25

Solved as second order missing y ode
Maple
Mathematica
Sympy

Internal problem ID [8883]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 25
Date solved : Wednesday, March 05, 2025 at 07:07:22 AM
CAS classification : [[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]

Solve

(x2+1)y+y2=0

Solved as second order missing y ode

Time used: 0.487 (sec)

This is second order ode with missing dependent variable y. Let

u(x)=y

Then

u(x)=y

Hence the ode becomes

(x2+1)u(x)+u(x)2=0

Which is now solved for u(x) as first order ode.

The ode

(1)u(x)=u(x)2x2+1

is separable as it can be written as

u(x)=u(x)2x2+1=f(x)g(u)

Where

f(x)=1x2+1g(u)=u2

Integrating gives

1g(u)du=f(x)dx1u2du=1x2+1dx
1u(x)=arctan(x)c1

In summary, these are the solution found for u(x)

1u(x)=arctan(x)c1

For solution 1u(x)=arctan(x)c1, since u=y then we now have a new first order ode to solve which is

1y=arctan(x)c1

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

dy=1arctan(x)c1dxy=1arctan(x)c1dx+c2
y=1arctan(x)c1dx+c2

In summary, these are the solution found for (y)

y=1arctan(x)c1dx+c2

Will add steps showing solving for IC soon.

Summary of solutions found

y=1arctan(x)c1dx+c2

Maple. Time used: 0.015 (sec). Leaf size: 14
ode:=(x^2+1)*diff(diff(y(x),x),x)+diff(y(x),x)^2 = 0; 
dsolve(ode,y(x), singsol=all);
 
y=1arctan(x)+c1dx+c2

Maple trace

`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 
`, `-> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE`, diff(_b(_a), _a) = -_b(_a)^2/(_a^2+1), _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

Maple step by step

Let’s solve(x2+1)(d2dx2y(x))+(ddxy(x))2=0Highest derivative means the order of the ODE is2d2dx2y(x)Make substitutionu=ddxy(x)to reduce order of ODE(x2+1)(ddxu(x))+u(x)2=0Solve for the highest derivativeddxu(x)=u(x)2x2+1Separate variablesddxu(x)u(x)2=1x2+1Integrate both sides with respect toxddxu(x)u(x)2dx=1x2+1dx+C1Evaluate integral1u(x)=arctan(x)+C1Solve foru(x)u(x)=1arctan(x)C1Solve 1st ODE foru(x)u(x)=1arctan(x)C1Make substitutionu=ddxy(x)ddxy(x)=1arctan(x)C1Integrate both sides to solve fory(x)(ddxy(x))dx=1arctan(x)C1dx+C2Compute integralsy(x)=1arctan(x)C1dx+C2
Mathematica. Time used: 60.287 (sec). Leaf size: 25
ode=(1+x^2)*D[y[x],{x,2}]+(D[y[x],x])^2==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 
y(x)1x1arctan(K[1])c1dK[1]+c2
Sympy. Time used: 1.464 (sec). Leaf size: 24
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq((x**2 + 1)*Derivative(y(x), (x, 2)) + Derivative(y(x), x)**2,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 
[y(x)=C11C2atan(x)dx, y(x)=C11C2atan(x)dx]