2.3.25 Problem 25

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

$$\begin{array}{rl}(x^2+1)y^{\prime \prime }+{y^{\prime }}^2& =0\end{array}$$

Solved as second order missing y ode

Time used: 0.487 (sec)

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

$$\begin{array}{rl}u(x)& =y^{\prime }\end{array}$$

Then

$$\begin{array}{rl}{u}^{\prime }(x)& =y^{\prime \prime }\end{array}$$

Hence the ode becomes

$$\begin{array}{r}(x^2+1)u^{\prime }\left(x\right)+u{\left(x\right)}^2=0\end{array}$$

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

The ode

$$\begin{array}{}\text{(1)}& u^{\prime }\left(x\right)=-\frac{u{\left(x\right)}^2}{x^2+1}\end{array}$$

is separable as it can be written as

$$\begin{array}{rl}{u}^{\prime }\left(x\right)& =-\frac{u{\left(x\right)}^2}{x^2+1}\\ & =f(x)g(u)\end{array}$$

Where

$$\begin{array}{rl}f(x)& =-\frac{1}{x^2+1}\\ g(u)& =u^2\end{array}$$

Integrating gives

$$\begin{array}{rl}\int \frac{1}{g(u)}du& =\int f(x)dx\\ \int \frac{1}{u^2}du& =\int -\frac{1}{x^2+1}dx\end{array}$$

$$\frac{1}{u\left(x\right)}=\arctan \left(x\right)-c_1$$

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

$$\begin{array}{rl}\frac{1}{u\left(x\right)}& =\arctan \left(x\right)-c_1\end{array}$$

For solution $\frac{1}{u\left(x\right)}=\arctan \left(x\right)-c_1$, since $u=y^{\prime }$ then we now have a new first order ode to solve which is

$$\begin{array}{r}\frac{1}{y^{\prime }}=\arctan \left(x\right)-c_1\end{array}$$

Since the ode has the form $y^{\prime }=f(x)$, then we only need to integrate $f(x)$.

$$\begin{array}{rl}\int dy& =\int \frac{1}{\arctan \left(x\right)-c_1}dx\\ y& =\int \frac{1}{\arctan \left(x\right)-c_1}dx+c_2\end{array}$$

$$\begin{array}{rl}y& =\int \frac{1}{\arctan \left(x\right)-c_1}dx+c_2\end{array}$$

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

$$\begin{array}{rl}y& =\int \frac{1}{\arctan \left(x\right)-c_1}dx+c_2\end{array}$$

Will add steps showing solving for IC soon.

Summary of solutions found

$$\begin{array}{rl}y& =\int \frac{1}{\arctan \left(x\right)-c_1}dx+c_2\end{array}$$

✓ 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);
 

$$y=\int \frac{1}{\arctan \left(x\right)+c_1}dx+c_2$$

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

$$\begin{array}{lll}& & \text{Let’s solve}\\ & & (x^2+1)\left(\frac{d^2}{d{x}^2}y\left(x\right)\right)+{\left(\frac{d}{dx}y\left(x\right)\right)}^2=0\\ \bullet & & \text{Highest derivative means the order of the ODE is}2\\ & & \frac{d^2}{d{x}^2}y\left(x\right)\\ \bullet & & \text{Make substitution}u=\frac{d}{dx}y\left(x\right)\text{to reduce order of ODE}\\ & & (x^2+1)\left(\frac{d}{dx}u\left(x\right)\right)+u{\left(x\right)}^2=0\\ \bullet & & \text{Solve for the highest derivative}\\ & & \frac{d}{dx}u\left(x\right)=-\frac{u{\left(x\right)}^2}{x^2+1}\\ \bullet & & \text{Separate variables}\\ & & \frac{\frac{d}{dx}u\left(x\right)}{u{\left(x\right)}^2}=-\frac{1}{x^2+1}\\ \bullet & & \text{Integrate both sides with respect to}x\\ & & \int \frac{\frac{d}{dx}u\left(x\right)}{u{\left(x\right)}^2}dx=\int -\frac{1}{x^2+1}dx+\mathit{C}\mathit{1}\\ \bullet & & \text{Evaluate integral}\\ & & -\frac{1}{u\left(x\right)}=-\arctan \left(x\right)+\mathit{C}\mathit{1}\\ \bullet & & \text{Solve for}u\left(x\right)\\ & & u\left(x\right)=\frac{1}{\arctan \left(x\right)-\mathit{C}\mathit{1}}\\ \bullet & & \text{Solve 1st ODE for}u\left(x\right)\\ & & u\left(x\right)=\frac{1}{\arctan \left(x\right)-\mathit{C}\mathit{1}}\\ \bullet & & \text{Make substitution}u=\frac{d}{dx}y\left(x\right)\\ & & \frac{d}{dx}y\left(x\right)=\frac{1}{\arctan \left(x\right)-\mathit{C}\mathit{1}}\\ \bullet & & \text{Integrate both sides to solve for}y\left(x\right)\\ & & \int \left(\frac{d}{dx}y\left(x\right)\right)dx=\int \frac{1}{\arctan \left(x\right)-\mathit{C}\mathit{1}}dx+\mathit{C}\mathit{2}\\ \bullet & & \text{Compute integrals}\\ & & y\left(x\right)=\int \frac{1}{\arctan \left(x\right)-\mathit{C}\mathit{1}}dx+\mathit{C}\mathit{2}\end{array}$$

✓ 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]
 

$$y(x)\to {\int }_1^x\frac{1}{\arctan (K[1])-c_1}dK[1]+c_2$$

✓ 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)
 

$$[y\left(x\right)=C_1-\int \frac{1}{C_2-\mathrm{atan}\left(x\right)}dx,y\left(x\right)=C_1-\int \frac{1}{C_2-\mathrm{atan}\left(x\right)}dx]$$