2.1.14 Problem 14

Internal problem ID [9085]
Book : Second order enumerated odes
Section : section 1
Problem number : 14
Date solved : Monday, January 27, 2025 at 05:32:21 PM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]

Solve

\begin{align*} y^{\prime \prime }+{y^{\prime }}^{2}&=0 \end{align*}

Solved as second order missing x ode

Time used: 0.197 (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} = 0 \end{align*}

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

Factoring the ode gives these factors

\begin{align*} \tag{1} p &= 0 \\ \tag{2} p^{\prime }+p &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(p\) from

\begin{align*} p = 0 \end{align*}

Solving gives \(p = 0\)

Solving equation (2)

Integrating gives

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

Singular solutions are found by solving

\begin{align*} -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 = 0 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= 0 \\ p &= {\mathrm e}^{-y -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 } = 0 \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 {0\, dx} + c_2 \\ y &= 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 } = {\mathrm e}^{-y-c_1} \end{align*}

Integrating gives

\begin{align*} \int {\mathrm e}^{y +c_1}d y &= dx\\ {\mathrm e}^{y +c_1}&= x +c_3 \end{align*}

Solving for \(y\) gives

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Solved as second order missing y ode

Time used: 0.076 (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} = 0 \end{align*}

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

Integrating gives

\begin{align*} \int -\frac {1}{u^{2}}d u &= dx\\ \frac {1}{u}&= x +c_1 \end{align*}

Singular solutions are found by solving

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

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

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

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

\begin{align*} u \left (x \right ) &= 0 \\ u \left (x \right ) &= \frac {1}{x +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*}

Since 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*}

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

\begin{align*} y^{\prime } = \frac {1}{x +c_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 {\frac {1}{x +c_1}\, dx}\\ y &= \ln \left (x +c_1 \right ) + c_3 \end{align*}

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y \left (x \right )\right )^{2}+\frac {d^{2}}{d x^{2}}y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right ) \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (x \right )^{2}+\frac {d}{d x}u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}u \left (x \right )=-u \left (x \right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{2}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}u \left (x \right )}{u \left (x \right )^{2}}d x =\int \left (-1\right )d x +\mathit {C1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{u \left (x \right )}=\mathit {C1} -x \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\mathit {C1} -x} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{\mathit {C1} -x} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =\frac {d}{d x}y \left (x \right ) \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {1}{\mathit {C1} -x} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \left (x \right ) \\ {} & {} & \int \left (\frac {d}{d x}y \left (x \right )\right )d x =\int -\frac {1}{\mathit {C1} -x}d x +\mathit {C2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y \left (x \right )=\ln \left (\mathit {C1} -x \right )+\mathit {C2} \end {array} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
<- 2nd_order Liouville successful`
 

Maple dsolve solution

Solving time : 0.011 (sec)
Leaf size : 10

dsolve(diff(diff(y(x),x),x)+diff(y(x),x)^2 = 0,y(x),singsol=all)
 

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

Mathematica DSolve solution

Solving time : 0.191 (sec)
Leaf size : 15

DSolve[{D[y[x],{x,2}]+(D[y[x],x])^2==0,{}},y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \log (x-c_1)+c_2 \]