2.3.23 Problem 23

Internal problem ID [10155]
Book : Own collection of miscellaneous problems
Section : section 3.0
Problem number : 23
Date solved : Monday, December 08, 2025 at 07:45:16 PM
CAS classification : [[_2nd_order, _missing_y], [_2nd_order, _reducible, _mu_y_y1]]

2.3.23.1 second order ode missing y

1.544 (sec)

\begin{align*} \left (x^{2}+1\right ) y^{\prime \prime }+1+x {y^{\prime }}^{2}&=1 \\ \end{align*}

Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y\). Let

\begin{align*} u(x) &= y^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= y^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} \left (x^{2}+1\right ) u^{\prime }\left (x \right )+x u \left (x \right )^{2} = 0 \end{align*}

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

Entering first order ode separable solverThe ode

\begin{equation} u^{\prime }\left (x \right ) = -\frac {x u \left (x \right )^{2}}{x^{2}+1} \end{equation}

is separable as it can be written as

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

Where

\begin{align*} f(x) &= -\frac {x}{x^{2}+1}\\ g(u) &= u^{2} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx} \\ \int { \frac {1}{u^{2}}\,du} &= \int { -\frac {x}{x^{2}+1} \,dx} \\ \end{align*}

\[ -\frac {1}{u \left (x \right )}=\ln \left (\frac {1}{\sqrt {x^{2}+1}}\right )+c_1 \]

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or

\[ u^{2}=0 \]

for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} -\frac {1}{u \left (x \right )} &= \ln \left (\frac {1}{\sqrt {x^{2}+1}}\right )+c_1 \\ u \left (x \right ) &= 0 \\ \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {2}{\ln \left (x^{2}+1\right )-2 c_1} \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {2}{\ln \left (x^{2}+1\right )-2 c_1} \\ \end{align*}

For solution \(u \left (x \right ) = 0\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 0 \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_2 \\ y &= c_2 \end{align*}

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

\begin{align*} y &= c_2 \\ \end{align*}

For solution \(u \left (x \right ) = \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}\), since \(u=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \frac {2}{\ln \left (x^{2}+1\right )-2 c_1} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {\frac {2}{\ln \left (x^{2}+1\right )-2 c_1}\, dx}\\ y &= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x + c_3 \end{align*}

\begin{align*} y&= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x +c_3 \end{align*}

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

\begin{align*} y &= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x +c_3 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_2 \\ y &= \int \frac {2}{\ln \left (x^{2}+1\right )-2 c_1}d x +c_3 \\ \end{align*}

2.3.23.2 SSolved using second order ode arccos transformation

1.555 (sec)

\begin{align*} \left (x^{2}+1\right ) y^{\prime \prime }+1+x {y^{\prime }}^{2}&=1 \\ \end{align*}

Applying change of variable \(x = \arccos \left (\tau \right )\) to the above ode results in the following new ode

\[ 1-\frac {\left (\tau -1\right ) \left (\tau +1\right ) \left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \left (\frac {d^{2}}{d \tau ^{2}}y \left (\tau \right )\right )}{4}+\arccos \left (\tau \right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )^{2} \left (-\tau ^{2}+1\right )-\frac {\tau \left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \left (\frac {d}{d \tau }y \left (\tau \right )\right )}{4} = 1 \]

Which is now solved for \(y \left (\tau \right )\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(y \left (\tau \right )\). Let

\begin{align*} u(\tau ) &= \frac {d}{d \tau }y \left (\tau \right ) \end{align*}

Then

\begin{align*} u'(\tau ) &= \frac {d^{2}}{d \tau ^{2}}y \left (\tau \right ) \end{align*}

Hence the ode becomes

\begin{align*} -\frac {\left (\tau -1\right ) \left (\tau +1\right ) \left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \left (\frac {d}{d \tau }u \left (\tau \right )\right )}{4}+\arccos \left (\tau \right ) u \left (\tau \right )^{2} \left (-\tau ^{2}+1\right )-\frac {\tau \left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) u \left (\tau \right )}{4} = 0 \end{align*}

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

Entering first order ode bernoulli solver In canonical form, the ODE is

\begin{align*} u' &= F(\tau ,u)\\ &= -\frac {u \left (4 \arccos \left (\tau \right ) \tau ^{2} u +\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau -4 \arccos \left (\tau \right ) u +4 \tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4} \end{align*}

This is a Bernoulli ODE.

\[ u' = \left (-\frac {\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau }{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\right ) u \left (\tau \right ) + \left (-\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\right )u^{2} \tag {1} \]

The standard Bernoulli ODE has the form

\[ u' = f_0(\tau )u+f_1(\tau )u^n \tag {2} \]

Comparing this to (1) shows that

\begin{align*} f_0 &=-\frac {\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau }{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\\ f_1 &=-\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4} \end{align*}

The first step is to divide the above equation by \(u^n \) which gives

\[ \frac {u'}{u^n} = f_0(\tau ) u^{1-n} +f_1(\tau ) \tag {3} \]

The next step is use the substitution \(v = u^{1-n}\) in equation (3) which generates a new ODE in \(v \left (\tau \right )\) which will be linear and can be easily solved using an integrating factor. Backsubstitution then gives the solution \(u(\tau )\) which is what we want.

This method is now applied to the ODE at hand. Comparing the ODE (1) With (2) Shows that

\begin{align*} f_0(\tau )&=-\frac {\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau }{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\\ f_1(\tau )&=-\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\\ n &=2 \end{align*}

Dividing both sides of ODE (1) by \(u^n=u^{2}\) gives

\begin{align*} u'\frac {1}{u^{2}} &= -\frac {\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau }{\left (\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4\right ) u} -\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4} \tag {4} \end{align*}

Let

\begin{align*} v &= u^{1-n} \\ &= \frac {1}{u} \tag {5} \end{align*}

Taking derivative of equation (5) w.r.t \(\tau \) gives

\begin{align*} v' &= -\frac {1}{u^{2}}u' \tag {6} \end{align*}

Substituting equations (5) and (6) into equation (4) gives

\begin{align*} -\frac {d}{d \tau }v \left (\tau \right )&= -\frac {\left (\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau \right ) v \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}-\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}\\ v' &= \frac {\left (\pi ^{2} \tau -4 \pi \arcsin \left (\tau \right ) \tau +4 \arcsin \left (\tau \right )^{2} \tau +4 \tau \right ) v}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4}+\frac {4 \arccos \left (\tau \right ) \tau ^{2}-4 \arccos \left (\tau \right )}{\pi ^{2} \tau ^{2}-\pi ^{2}-4 \arcsin \left (\tau \right ) \pi \,\tau ^{2}+4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2} \tau ^{2}-4 \arcsin \left (\tau \right )^{2}+4 \tau ^{2}-4} \tag {7} \end{align*}

The above now is a linear ODE in \(v \left (\tau \right )\) which is now solved.

In canonical form a linear first order is

\begin{align*} \frac {d}{d \tau }v \left (\tau \right ) + q(\tau )v \left (\tau \right ) &= p(\tau ) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(\tau ) &=-\frac {\tau }{\tau ^{2}-1}\\ p(\tau ) &=\frac {4 \arccos \left (\tau \right )}{\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,d\tau }}\\ &= {\mathrm e}^{\int -\frac {\tau }{\tau ^{2}-1}d \tau }\\ &= \frac {1}{\sqrt {\tau ^{2}-1}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }}\left ( \mu v\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }}\left ( \mu v\right ) &= \left (\mu \right ) \left (\frac {4 \arccos \left (\tau \right )}{\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}\tau }} \left (\frac {v}{\sqrt {\tau ^{2}-1}}\right ) &= \left (\frac {1}{\sqrt {\tau ^{2}-1}}\right ) \left (\frac {4 \arccos \left (\tau \right )}{\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4}\right ) \\ \mathrm {d} \left (\frac {v}{\sqrt {\tau ^{2}-1}}\right ) &= \left (\frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}\right )\, \mathrm {d} \tau \\ \end{align*}

Integrating gives

\begin{align*} \frac {v}{\sqrt {\tau ^{2}-1}}&= \int {\frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}} \,d\tau } \\ &=\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau + c_1 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{\sqrt {\tau ^{2}-1}}\) gives the final solution

\[ v \left (\tau \right ) = \sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right ) \]

The substitution \(v = u^{1-n}\) is now used to convert the above solution back to \(u \left (\tau \right )\) which results in

\[ \frac {1}{u \left (\tau \right )} = \sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right ) \]

Solving for \(u \left (\tau \right )\) gives

\begin{align*} u \left (\tau \right ) &= \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )} \\ \end{align*}

In summary, these are the solution found for \(y \left (\tau \right )\)

\begin{align*} u \left (\tau \right ) &= \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )} \\ \end{align*}

For solution \(u \left (\tau \right ) = \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )}\), since \(u=\frac {d}{d \tau }y \left (\tau \right )\) then we now have a new first order ode to solve which is

\begin{align*} \frac {d}{d \tau }y \left (\tau \right ) = \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )} \end{align*}

Entering first order ode quadrature solverSince the ode has the form \(\frac {d}{d \tau }y \left (\tau \right )=f(\tau )\), then we only need to integrate \(f(\tau )\).

\begin{align*} \int {dy} &= \int {\frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )}\, d\tau }\\ y \left (\tau \right ) &= \int \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )}d \tau + c_2 \end{align*}

\begin{align*} y \left (\tau \right )&= \int \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )}d \tau +c_2 \end{align*}

In summary, these are the solution found for \((y \left (\tau \right ))\)

\begin{align*} y \left (\tau \right ) &= \int \frac {1}{\sqrt {\tau ^{2}-1}\, \left (\int \frac {4 \arccos \left (\tau \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\tau \right )+4 \arcsin \left (\tau \right )^{2}+4\right ) \sqrt {\tau ^{2}-1}}d \tau +c_1 \right )}d \tau +c_2 \\ \end{align*}

Applying change of variable \(\tau = \cos \left (x \right )\) to the solutions above gives

\begin{align*} y &= \int -\frac {\sin \left (x \right )}{\sqrt {\cos \left (x \right )^{2}-1}\, \left (\int -\frac {4 \arccos \left (\cos \left (x \right )\right ) \sin \left (x \right )}{\left (\pi ^{2}-4 \pi \arcsin \left (\cos \left (x \right )\right )+4 \arcsin \left (\cos \left (x \right )\right )^{2}+4\right ) \sqrt {\cos \left (x \right )^{2}-1}}d x +c_1 \right )}d x +c_2 \\ \end{align*}

2.3.23.3 ✓ Maple. Time used: 0.010 (sec). Leaf size: 22

ode:=(x^2+1)*diff(diff(y(x),x),x)+1+x*diff(y(x),x)^2 = 1; 
dsolve(ode,y(x), singsol=all);
 

\[ y = 2 \int \frac {1}{\ln \left (x^{2}+1\right )+2 c_1}d x +c_2 \]

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/(_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} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+1+x \left (\frac {d}{d x}y \left (x \right )\right )^{2}=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (x^{2}+1\right ) \left (\frac {d}{d x}u \left (x \right )\right )+1+x u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-\frac {x u \left (x \right )^{2}}{x^{2}+1} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{2}}=-\frac {x}{x^{2}+1} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{2}}d x =\int -\frac {x}{x^{2}+1}d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=-\frac {\ln \left (x^{2}+1\right )}{2}+\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )-2 \mathit {C1}} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )+\mathit {C1}} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )+\mathit {C1}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {2}{\ln \left (x^{2}+1\right )+\mathit {C1}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int \frac {2}{\ln \left (x^{2}+1\right )+\mathit {C1}}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\int \frac {2}{\ln \left (x^{2}+1\right )+\mathit {C1}}d x +\mathit {C2} \end {array} \]

2.3.23.4 ✓ Mathematica. Time used: 60.084 (sec). Leaf size: 33

ode=(1+x^2)*D[y[x],{x,2}]+1+x*(D[y[x],x])^2==1; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \int _1^x-\frac {2}{2 c_1-\log \left (K[1]^2+1\right )}dK[1]+c_2 \end{align*}

2.3.23.5 ✓ Sympy. Time used: 1.780 (sec). Leaf size: 34

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x*Derivative(y(x), x)**2 + (x**2 + 1)*Derivative(y(x), (x, 2)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

\[ \left [ y{\left (x \right )} = C_{1} - 2 \int \frac {1}{C_{2} - \log {\left (x^{2} + 1 \right )}}\, dx, \ y{\left (x \right )} = C_{1} - 2 \int \frac {1}{C_{2} - \log {\left (x^{2} + 1 \right )}}\, dx\right ] \]