2.1.17 Problem 17

Internal problem ID [9088]
Book : Second order enumerated odes
Section : section 1
Problem number : 17
Date solved : Monday, January 27, 2025 at 05:37:05 PM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_xy]]

Solve

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

Solved as second order missing x ode

Time used: 6.675 (sec)

This 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 \(p(y)\).

Integrating 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*}

Applying the exponential to both sides gives

\begin{align*} e^{\ln \left (\frac {1}{\sqrt {p^{2}-1}}\right )} &= e^{y +c_1}\\ \frac {1}{\sqrt {p^{2}-1}}&=c_1 \,{\mathrm e}^{y} \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*}

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

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

Solving for the derivative gives these ODE’s to solve

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

Now each of the above is solved separately.

Solving Eq. (1)

Integrating gives

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

Singular solutions are found by solving

\begin{align*} \frac {\sqrt {1+{\mathrm e}^{2 y} c_1^{2}}\, {\mathrm e}^{-y}}{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 {\ln \left (-\frac {1}{c_1^{2}}\right )}{2} \end{align*}

Solving for \(y\) gives

\begin{align*} y &= \frac {\ln \left (-\frac {1}{c_1^{2}}\right )}{2} \\ y &= -x -c_2 +\frac {\ln \left (\frac {\left ({\mathrm e}^{2 x +2 c_2}-1\right )^{2}}{4 c_1^{2}}\right )}{2} \\ \end{align*}

Solving Eq. (2)

Integrating gives

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

Singular solutions are found by solving

\begin{align*} -\frac {\sqrt {1+{\mathrm e}^{2 y} c_1^{2}}\, {\mathrm e}^{-y}}{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 {\ln \left (-\frac {1}{c_1^{2}}\right )}{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*}

Since 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_4 \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 } = 1 \end{align*}

Since 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_5 \end{align*}

Will add steps showing solving for IC soon.

Solving for \(y\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\begin{align*} y&=\frac {\ln \left (-\frac {1}{c_1^{2}}\right )}{2}\\ y&=-x +c_4\\ y&=x +c_5\\ y&=-x -c_2 +\frac {\ln \left (\frac {\left ({\mathrm e}^{2 x +2 c_2}-1\right )^{2}}{4 c_1^{2}}\right )}{2}\\ y&=-x -c_3 +\frac {\ln \left (\frac {\left ({\mathrm e}^{2 x +2 c_3}-1\right )^{2}}{4 c_1^{2}}\right )}{2} \end{align*}

The solution

\[ y = \frac {\ln \left (-\frac {1}{c_1^{2}}\right )}{2} \]

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

Summary of solutions found

\begin{align*} y &= -x +c_4 \\ y &= x +c_5 \\ y &= -x -c_2 +\frac {\ln \left (\frac {\left ({\mathrm e}^{2 x +2 c_2}-1\right )^{2}}{4 c_1^{2}}\right )}{2} \\ y &= -x -c_3 +\frac {\ln \left (\frac {\left ({\mathrm e}^{2 x +2 c_3}-1\right )^{2}}{4 c_1^{2}}\right )}{2} \\ \end{align*}

Solved as second order missing y ode

Time used: 0.259 (sec)

This 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.

Integrating 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 \(u(x)\)

\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*}

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

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

Since 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 \(u \left (x \right ) = \tanh \left (x +c_1 \right )\), 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*}

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}+\frac {d^{2}}{d x^{2}}y \left (x \right )=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} \\ {} & {} & u \left (x \right )^{2}+\frac {d}{d x}u \left (x \right )=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} \]

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) 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful`
 

Maple dsolve solution

Solving time : 0.010 (sec)
Leaf size : 21

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

\[ y = x -\ln \left (2\right )+\ln \left (c_{1} {\mathrm e}^{-2 x}-c_{2} \right ) \]

Mathematica DSolve solution

Solving time : 0.354 (sec)
Leaf size : 48

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

\begin{align*} y(x)\to -\frac {1}{2} \log \left (e^{2 x}\right )+\log \left (e^{2 x}+e^{2 c_1}\right )+c_2 \\ y(x)\to \frac {1}{2} \log \left (e^{2 x}\right )+c_2 \\ \end{align*}