2.1.14 Problem 14

Internal problem ID [9085]
Book : Second order enumerated odes
Section : section 1
Problem number : 14
Date solved : Sunday, March 30, 2025 at 02:06:35 PM
CAS classification : [[_2nd_order, _missing_x], _Liouville, [_2nd_order, _reducible, _mu_xy]]

Solved as second order missing x ode

Time used: 0.219 (sec)

Solve

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

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 \(p(y)\).

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*}

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

\begin{align*} -\ln \left (y^{\prime }\right ) = 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_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 } = 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_3 \\ y &= c_3 \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&=c_3\\ y&=-c_1 +\ln \left (x +c_2 \right ) \end{align*}

Summary of solutions found

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

Solved as second order missing y ode

Time used: 0.158 (sec)

Solve

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

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 \(u\). 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 = 0 \end{align*}

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

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

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

\begin{align*} \frac {1}{y^{\prime }\left (x \right )} = 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_2 \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_3 \\ y &= c_3 \end{align*}

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

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

Will add steps showing solving for IC soon.

Summary of solutions found

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

✓ Maple. Time used: 0.019 (sec). Leaf size: 10

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

\[ y = \ln \left (c_1 x +c_2 \right ) \]

Maple trace

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

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y \left (x \right )+\left (\frac {d}{d x}y \left (x \right )\right )^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}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}=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 )}=-x +\mathit {C1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {1}{-x +\mathit {C1}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & u \left (x \right )=\frac {1}{-\mathit {C1} +x} \\ \bullet & {} & \textrm {Redefine the integration constant(s)}\hspace {3pt} \\ {} & {} & 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} \]

✓ Mathematica. Time used: 0.191 (sec). Leaf size: 15

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

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

✓ Sympy. Time used: 0.513 (sec). Leaf size: 8

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

\[ y{\left (x \right )} = C_{1} + \log {\left (C_{2} + x \right )} \]