2.1.17 Problem 17

Internal problem ID [10376]
Book : Second order enumerated odes
Section : section 1
Problem number : 17
Date solved : Monday, December 08, 2025 at 08:23:26 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]

2.1.17.1 second order ode missing x

4.677 (sec)

\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=1 \\ \end{align*}

Entering second order ode missing \(x\) solverThis is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using

\begin{align*} y' &= p \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= p \frac {dp}{dy} \end{align*}

Hence the ode becomes

\begin{align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+p \left (y \right )^{2} = 1 \end{align*}

Which is now solved as first order ode for \(p(y)\).

Entering first order ode autonomous solverIntegrating gives

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

Singular solutions are found by solving

\begin{align*} -\frac {p^{2}-1}{p}&= 0 \end{align*}

for \(p\). 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.

\begin{align*} p = -1\\ p = 1 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= -1 \\ p &= 1 \\ p &= \sqrt {1+{\mathrm e}^{-2 y -2 c_1}} \\ p &= -\sqrt {1+{\mathrm e}^{-2 y -2 c_1}} \\ \end{align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -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 {-1\, dx}\\ y &= -x + c_2 \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = 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 {1\, dx}\\ y &= x + c_3 \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = \sqrt {1+{\mathrm e}^{-2 y-2 c_1}} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{\sqrt {1+{\mathrm e}^{-2 y -2 c_1}}}d y &= dx\\ \operatorname {arctanh}\left (\sqrt {1+{\mathrm e}^{-2 y -2 c_1}}\right )&= x +c_4 \end{align*}

Singular solutions are found by solving

\begin{align*} \sqrt {1+{\mathrm e}^{-2 y -2 c_1}}&= 0 \end{align*}

for \(y\). 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.

\begin{align*} y = -\frac {i \pi }{2}-c_1 \end{align*}

Simplifying the above gives

\begin{align*} \operatorname {arctanh}\left (\sqrt {{\mathrm e}^{-2 y-2 c_1} \left ({\mathrm e}^{2 y+2 c_1}+1\right )}\right ) &= x +c_4 \\ y &= -\frac {i \pi }{2}-c_1 \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_4 \right )^{2}-1\right )}{2} \\ y &= -\frac {i \pi }{2}-c_1 \\ \end{align*}

For solution (4) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime } = -\sqrt {1+{\mathrm e}^{-2 y-2 c_1}} \end{align*}

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int -\frac {1}{\sqrt {1+{\mathrm e}^{-2 y -2 c_1}}}d y &= dx\\ -\operatorname {arctanh}\left (\sqrt {1+{\mathrm e}^{-2 y -2 c_1}}\right )&= x +c_5 \end{align*}

Singular solutions are found by solving

\begin{align*} -\sqrt {1+{\mathrm e}^{-2 y -2 c_1}}&= 0 \end{align*}

for \(y\). 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.

\begin{align*} y = -\frac {i \pi }{2}-c_1 \end{align*}

Simplifying the above gives

\begin{align*} -\operatorname {arctanh}\left (\sqrt {{\mathrm e}^{-2 y-2 c_1} \left ({\mathrm e}^{2 y+2 c_1}+1\right )}\right ) &= x +c_5 \\ y &= -\frac {i \pi }{2}-c_1 \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_5 \right )^{2}-1\right )}{2} \\ y &= -\frac {i \pi }{2}-c_1 \\ \end{align*}

The solution

\[ y = -\frac {i \pi }{2}-c_1 \]

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

Summary of solutions found

\begin{align*} y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_4 \right )^{2}-1\right )}{2} \\ y &= -c_1 -\frac {\ln \left (\tanh \left (x +c_5 \right )^{2}-1\right )}{2} \\ y &= c_2 -x \\ y &= x +c_3 \\ \end{align*}

2.1.17.2 second order ode missing y

0.610 (sec)

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

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

Entering first order ode autonomous solverIntegrating gives

\begin{align*} \int \frac {1}{-u^{2}+1}d u &= dx\\ \operatorname {arctanh}\left (u \right )&= x +c_1 \end{align*}

Singular solutions are found by solving

\begin{align*} -u^{2}+1&= 0 \end{align*}

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

\begin{align*} u \left (x \right ) = -1\\ u \left (x \right ) = 1 \end{align*}

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

\begin{align*} u \left (x \right ) &= -1 \\ u \left (x \right ) &= 1 \\ u \left (x \right ) &= \tanh \left (x +c_1 \right ) \\ \end{align*}

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

\begin{align*} u \left (x \right ) &= -1 \\ u \left (x \right ) &= 1 \\ u \left (x \right ) &= \tanh \left (x +c_1 \right ) \\ \end{align*}

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

\begin{align*} y^{\prime } = -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 {-1\, dx}\\ y &= -x + c_2 \end{align*}

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

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

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

\begin{align*} y^{\prime } = 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 {1\, dx}\\ y &= x + c_3 \end{align*}

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

\begin{align*} y &= x +c_3 \\ \end{align*}

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

\begin{align*} y^{\prime } = \tanh \left (x +c_1 \right ) \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 {\tanh \left (x +c_1 \right )\, dx}\\ y &= \ln \left (\cosh \left (x +c_1 \right )\right ) + c_4 \end{align*}

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

\begin{align*} y &= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_2 -x \\ y &= x +c_3 \\ y &= \ln \left (\cosh \left (x +c_1 \right )\right )+c_4 \\ \end{align*}

2.1.17.3 SSolved using second order ode arccos transformation

7.648 (sec)

\begin{align*} y^{\prime \prime }+{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

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

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

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

Entering first order ode riccati solverIn canonical form the ODE is

\begin{align*} u' &= F(\tau ,u)\\ &= -\frac {\tau ^{2} u^{2}+u \tau -u^{2}+1}{\tau ^{2}-1} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ u' = -\frac {\tau ^{2} u^{2}}{\tau ^{2}-1}-\frac {u \tau }{\tau ^{2}-1}+\frac {u^{2}}{\tau ^{2}-1}-\frac {1}{\tau ^{2}-1} \]

With Riccati ODE standard form

\[ u' = f_0(\tau )+ f_1(\tau )u+f_2(\tau )u^{2} \]

Shows that \(f_0(\tau )=-\frac {1}{\tau ^{2}-1}\), \(f_1(\tau )=-\frac {\tau }{\tau ^{2}-1}\) and \(f_2(\tau )=-1\). Let

\begin{align*} u &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-u} \tag {1} \end{align*}

Using the above substitution in the given ODE results (after some simplification) in a second order ODE to solve for \(u(x)\) which is

\begin{align*} f_2 u''(\tau ) -\left ( f_2' + f_1 f_2 \right ) u'(\tau ) + f_2^2 f_0 u(\tau ) &= 0 \tag {2} \end{align*}

But

\begin{align*} f_2' &=0\\ f_1 f_2 &=\frac {\tau }{\tau ^{2}-1}\\ f_2^2 f_0 &=-\frac {1}{\tau ^{2}-1} \end{align*}

Substituting the above terms back in equation (2) gives

\[ -\frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )-\frac {\tau \left (\frac {d}{d \tau }u \left (\tau \right )\right )}{\tau ^{2}-1}-\frac {u \left (\tau \right )}{\tau ^{2}-1} = 0 \]

Entering second order change of variable on \(x\) method 2 solverIn normal form the ode

\begin{align*} -\frac {d^{2}u}{d \tau ^{2}}-\frac {\tau \left (\frac {d u}{d \tau }\right )}{\tau ^{2}-1}-\frac {u}{\tau ^{2}-1} = 0\tag {1} \end{align*}

Becomes

\begin{align*} \frac {d^{2}u}{d \tau ^{2}}+p \left (\tau \right ) \left (\frac {d u}{d \tau }\right )+q \left (\tau \right ) u&=0 \tag {2} \end{align*}

Where

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

Applying change of variables \(\tau = g \left (\tau \right )\) to (2) gives

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+p_{1} \left (\frac {d}{d \tau }u \left (\tau \right )\right )+q_{1} u \left (\tau \right )&=0 \tag {3} \end{align*}

Where \(\tau \) is the new independent variable, and

\begin{align*} p_{1} \left (\tau \right ) &=\frac {\frac {d^{2}}{d \tau ^{2}}\tau \left (\tau \right )+p \left (\tau \right ) \left (\frac {d}{d \tau }\tau \left (\tau \right )\right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\tag {4} \\ q_{1} \left (\tau \right ) &=\frac {q \left (\tau \right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\tag {5} \end{align*}

Let \(p_{1} = 0\). Eq (4) simplifies to

\begin{align*} \frac {d^{2}}{d \tau ^{2}}\tau \left (\tau \right )+p \left (\tau \right ) \left (\frac {d}{d \tau }\tau \left (\tau \right )\right )&=0 \end{align*}

This ode is solved resulting in

\begin{align*} \tau &= \int {\mathrm e}^{-\int p \left (\tau \right )d \tau }d \tau \\ &= \int {\mathrm e}^{-\int \frac {\tau }{\tau ^{2}-1}d \tau }d \tau \\ &= \int e^{-\frac {\ln \left (\tau -1\right )}{2}-\frac {\ln \left (\tau +1\right )}{2}} \,d\tau \\ &= \int \frac {1}{\sqrt {\tau -1}\, \sqrt {\tau +1}}d \tau \\ &= \frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\tag {6} \end{align*}

Using (6) to evaluate \(q_{1}\) from (5) gives

\begin{align*} q_{1} \left (\tau \right ) &= \frac {q \left (\tau \right )}{\left (\frac {d}{d \tau }\tau \left (\tau \right )\right )^{2}}\\ &= \frac {\frac {1}{\tau ^{2}-1}}{\frac {1}{\left (\tau +1\right ) \left (\tau -1\right )}}\\ &= 1\tag {7} \end{align*}

Substituting the above in (3) and noting that now \(p_{1} = 0\) results in

\begin{align*} \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+q_{1} u \left (\tau \right )&=0 \\ \frac {d^{2}}{d \tau ^{2}}u \left (\tau \right )+u \left (\tau \right )&=0 \end{align*}

The above ode is now solved for \(u \left (\tau \right )\).Entering second order linear constant coefficient ode solver

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A u''(\tau ) + B u'(\tau ) + C u(\tau ) = 0 \]

Where in the above \(A=1, B=0, C=1\). Let the solution be \(u \left (\tau \right )=e^{\lambda \tau }\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\tau \lambda }+{\mathrm e}^{\tau \lambda } = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda \tau }\) gives

\[ \lambda ^{2}+1 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form.Using the quadratic formula

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=1\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (1\right )}\\ &= \pm i \end{align*}

Hence

\begin{align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end{align*}

Which simplifies to

\begin{align*} \lambda _1 &= i \\ \lambda _2 &= -i \\ \end{align*}

Since roots are complex conjugate of each others, then let the roots be

\[ \lambda _{1,2} = \alpha \pm i \beta \]

Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution, when using Euler relation, can be written as

\[ u \left (\tau \right ) = e^{\alpha \tau } \left ( c_1 \cos (\beta \tau ) + c_2 \sin (\beta \tau ) \right ) \]

Which becomes

\[ u \left (\tau \right ) = e^{0}\left (c_1 \cos \left (\tau \right )+c_2 \sin \left (\tau \right )\right ) \]

Or

\[ u \left (\tau \right ) = c_1 \cos \left (\tau \right )+c_2 \sin \left (\tau \right ) \]

The above solution is now transformed back to \(u\) using (6) which results in

\[ u = c_1 \cos \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_2 \sin \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right ) \]

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (\tau \right ) = -c_1 \left (\frac {\ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) \tau }{\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \sqrt {\tau -1}\, \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \left (\tau -1\right )^{{3}/{2}} \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \sqrt {\tau -1}\, \left (\tau +1\right )^{{3}/{2}}}+\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \left (1+\frac {\tau }{\sqrt {\tau ^{2}-1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\tau +\sqrt {\tau ^{2}-1}\right )}\right ) \sin \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_2 \left (\frac {\ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) \tau }{\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \sqrt {\tau -1}\, \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \left (\tau -1\right )^{{3}/{2}} \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \sqrt {\tau -1}\, \left (\tau +1\right )^{{3}/{2}}}+\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \left (1+\frac {\tau }{\sqrt {\tau ^{2}-1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\tau +\sqrt {\tau ^{2}-1}\right )}\right ) \cos \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right ) \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} u &= \frac {-u'}{f_2 u} \\ u &= \frac {-u'}{-u} \\ u &= \frac {-c_1 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_2 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (c_1 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_2 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ u = \frac {-\left (\frac {\ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) \tau }{\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \sqrt {\tau -1}\, \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \left (\tau -1\right )^{{3}/{2}} \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \sqrt {\tau -1}\, \left (\tau +1\right )^{{3}/{2}}}+\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \left (1+\frac {\tau }{\sqrt {\tau ^{2}-1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\tau +\sqrt {\tau ^{2}-1}\right )}\right ) \sin \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \left (\frac {\ln \left (\tau +\sqrt {\tau ^{2}-1}\right ) \tau }{\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \sqrt {\tau -1}\, \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \left (\tau -1\right )^{{3}/{2}} \sqrt {\tau +1}}-\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{2 \sqrt {\tau -1}\, \left (\tau +1\right )^{{3}/{2}}}+\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \left (1+\frac {\tau }{\sqrt {\tau ^{2}-1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\tau +\sqrt {\tau ^{2}-1}\right )}\right ) \cos \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\cos \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\left (\tau -1\right ) \left (\tau +1\right )}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )} \]

Simplifying the above gives

\begin{align*} u &= \frac {-\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )} \\ \end{align*}

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

\begin{align*} u &= \frac {-\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )} \\ \end{align*}

For solution \(u = \frac {-\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )}\), since \(u=y^{\prime }\left (\tau \right )\) then we now have a new first order ode to solve which is

\begin{align*} y^{\prime }\left (\tau \right ) = \frac {-\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )} \end{align*}

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

\begin{align*} \int {dy} &= \int {-\frac {\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )-c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )}\, d\tau }\\ y \left (\tau \right ) &= \int -\frac {\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )-c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )}d \tau + c_4 \end{align*}

\begin{align*} y \left (\tau \right )&= \int -\frac {\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )-c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )}d \tau +c_4 \end{align*}

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

\begin{align*} y \left (\tau \right ) &= \int -\frac {\sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )-c_3 \cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}\, \left (\cos \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )+c_3 \sin \left (\frac {\sqrt {\tau ^{2}-1}\, \ln \left (\tau +\sqrt {\tau ^{2}-1}\right )}{\sqrt {\tau -1}\, \sqrt {\tau +1}}\right )\right )}d \tau +c_4 \\ \end{align*}

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

\begin{align*} y \left (x \right ) &= \int \frac {\left (\sin \left (\frac {\sqrt {\cos \left (x \right )^{2}-1}\, \ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}\right )-c_3 \cos \left (\frac {\sqrt {\cos \left (x \right )^{2}-1}\, \ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}\right )\right ) \sin \left (x \right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}\, \left (\cos \left (\frac {\sqrt {\cos \left (x \right )^{2}-1}\, \ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}\right )+c_3 \sin \left (\frac {\sqrt {\cos \left (x \right )^{2}-1}\, \ln \left (\cos \left (x \right )+\sqrt {\cos \left (x \right )^{2}-1}\right )}{\sqrt {\cos \left (x \right )-1}\, \sqrt {\cos \left (x \right )+1}}\right )\right )}d x +c_4 \\ \end{align*}

2.1.17.4 ✓ Maple. Time used: 0.008 (sec). Leaf size: 22

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

\[ y = -x -\ln \left (2\right )+\ln \left (-c_1 \,{\mathrm e}^{2 x}+c_2 \right ) \]

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 a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )+\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} \\ {} & {} & \frac {d}{d x}u \left (x \right )+u \left (x \right )^{2}=1 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2}+1 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{-u \left (x \right )^{2}+1}=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}+1}d x =\int 1d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \mathrm {arctanh}\left (u \left (x \right )\right )=x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=\tanh \left (x +\mathit {C1} \right ) \\ \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 \tanh \left (x +\mathit {C1} \right )d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (\cosh \left (x +\mathit {C1} \right )\right )+\mathit {C2} \end {array} \]

2.1.17.5 ✓ Mathematica. Time used: 0.426 (sec). Leaf size: 45

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

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

2.1.17.6 ✗ Sympy

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

Timed Out