2.1.1 Problem 1

Internal problem ID [4086]
Book : Applied Differential equations, Newby Curle. Van Nostrand Reinhold. 1972
Section : Examples, page 35
Problem number : 1
Date solved : Saturday, December 06, 2025 at 04:17:25 PM
CAS classification : [_quadrature]

2.1.1.1 Solved using first_order_ode_dAlembert

0.098 (sec)

Entering first order ode dAlembert solver

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

Let \(p=y^{\prime }\) the ode becomes

\begin{align*} y = p +\frac {1}{2} p^{2} \end{align*}

Solving for \(y\) from the above results in

\begin{align*} \tag{1} y &= p +\frac {1}{2} p^{2} \\ \end{align*}

This has the form

\begin{align*} y=x f(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). The above ode is dAlembert ode which is now solved.

Taking derivative of (*) w.r.t. \(x\) gives

\begin{align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end{align*}

Comparing the form \(y=x f + g\) to (1A) shows that

\begin{align*} f &= 0\\ g &= p +\frac {1}{2} p^{2} \end{align*}

Hence (2) becomes

\begin{equation} \tag{2A} p = \left (1+p \right ) p^{\prime }\left (x \right ) \end{equation}

The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives

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

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0 \end{align*}

Substituting these in (1A) and keeping singular solution that verifies the ode gives

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

The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in

\begin{equation} \tag{3} p^{\prime }\left (x \right ) = \frac {p \left (x \right )}{1+p \left (x \right )} \end{equation}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed.

Integrating gives

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

Singular solutions are found by solving

\begin{align*} \frac {p}{1+p}&= 0 \end{align*}

for \(p \left (x \right )\). 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*} p \left (x \right ) = 0 \end{align*}

Substituing the above solution for \(p\) in (2A) gives

\begin{align*} y &= {\mathrm e}^{-\operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right )+x +c_1}+\frac {{\mathrm e}^{-2 \operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right )+2 x +2 c_1}}{2} \\ y &= 0 \\ \end{align*}

Simplifying the above gives

\begin{align*} y &= 0 \\ y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right ) \left (2+\operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right )\right )}{2} \\ y &= 0 \\ \end{align*}

Summary of solutions found

\begin{align*} y &= 0 \\ y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right ) \left (2+\operatorname {LambertW}\left ({\mathrm e}^{x +c_1}\right )\right )}{2} \\ \end{align*}

2.1.1.2 Solved using first_order_ode_parametric method

0.178 (sec)

Entering first order ode parametric solver

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

Let \(y^{\prime }\) be a parameter \(\lambda \). The ode becomes

\begin{align*} y -\lambda -\frac {1}{2} \lambda ^{2} = 0 \end{align*}

Isolating \(y\) gives

\begin{align*} y&=\lambda +\frac {1}{2} \lambda ^{2}\\ &=\lambda +\frac {1}{2} \lambda ^{2}\\ &=F \left (x , \lambda \right ) \end{align*}

Now we generate an ode in \(x \left (\lambda \right )\) using

\begin{align*} \frac {d}{d \lambda }x \left (\lambda \right ) &= \frac { \frac {\partial F}{\partial \lambda }} { \lambda -\frac {\partial F}{\partial x} } \\ &= \frac {1+\lambda }{\lambda }\\ &= \frac {1+\lambda }{\lambda } \end{align*}

Which is now solved for \(x\).

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

\begin{align*} \int {dx} &= \int {\frac {1+\lambda }{\lambda }\, d\lambda }\\ x \left (\lambda \right ) &= \lambda +\ln \left (\lambda \right ) + c_1 \end{align*}

Now that we found solution \(x\) we have two equations with parameter \(\lambda \). They are

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

Eliminating \(\lambda \) gives the solution for \(y\). Simplifying the above gives

\begin{align*} y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right ) \left (\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right )+2\right )}{2} \\ y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right ) \left (\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right )+2\right )}{2} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right ) \left (\operatorname {LambertW}\left ({\mathrm e}^{x -c_1}\right )+2\right )}{2} \\ \end{align*}

2.1.1.3 ✓ Maple. Time used: 0.014 (sec). Leaf size: 106

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

\begin{align*} y &= \frac {\operatorname {LambertW}\left (-\sqrt {2}\, {\mathrm e}^{-1+x -c_1}\right ) \left (\operatorname {LambertW}\left (-\sqrt {2}\, {\mathrm e}^{-1+x -c_1}\right )+2\right )}{2} \\ y &= \frac {{\mathrm e}^{2 \operatorname {RootOf}\left (-\textit {\_Z} -2 x +2 \,{\mathrm e}^{\textit {\_Z}}-2+2 c_1 -\ln \left (2\right )+\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )}}{2}-{\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} -2 x +2 \,{\mathrm e}^{\textit {\_Z}}-2+2 c_1 -\ln \left (2\right )+\ln \left ({\mathrm e}^{\textit {\_Z}} \left ({\mathrm e}^{\textit {\_Z}}-2\right )^{2}\right )\right )} \\ \end{align*}

Maple trace

Methods for first order ODEs: 
-> Solving 1st order ODE of high degree, 1st attempt 
trying 1st order WeierstrassP solution for high degree ODE 
trying 1st order WeierstrassPPrime solution for high degree ODE 
trying 1st order JacobiSN solution for high degree ODE 
trying 1st order ODE linearizable_by_differentiation 
trying differential order: 1; missing variables 
<- differential order: 1; missing  x  successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {d}{d x}y \left (x \right )+\frac {\left (\frac {d}{d x}y \left (x \right )\right )^{2}}{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=-1-\sqrt {1+2 y \left (x \right )}, \frac {d}{d x}y \left (x \right )=-1+\sqrt {1+2 y \left (x \right )}\right ] \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-1-\sqrt {1+2 y \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-1-\sqrt {1+2 y \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-1-\sqrt {1+2 y \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+2 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\textit {\_C1} \\ \square & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=-1+\sqrt {1+2 y \left (x \right )} \\ {} & \circ & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d x}y \left (x \right )}{-1+\sqrt {1+2 y \left (x \right )}}=1 \\ {} & \circ & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {\frac {d}{d x}y \left (x \right )}{-1+\sqrt {1+2 y \left (x \right )}}d x =\int 1d x +\textit {\_C1} \\ {} & \circ & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+2 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\textit {\_C1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\frac {\ln \left (y \left (x \right )\right )}{2}-\sqrt {1+2 y \left (x \right )}-\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}+\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\mathit {C1} , \frac {\ln \left (y \left (x \right )\right )}{2}+\sqrt {1+2 y \left (x \right )}+\frac {\ln \left (-1+\sqrt {1+2 y \left (x \right )}\right )}{2}-\frac {\ln \left (1+\sqrt {1+2 y \left (x \right )}\right )}{2}=x +\mathit {C1} \right \} \end {array} \]

2.1.1.4 ✓ Mathematica. Time used: 11.291 (sec). Leaf size: 66

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

\begin{align*} y(x)&\to \frac {1}{2} W\left (-e^{x-1-c_1}\right ) \left (2+W\left (-e^{x-1-c_1}\right )\right )\\ y(x)&\to \frac {1}{2} W\left (e^{x-1+c_1}\right ) \left (2+W\left (e^{x-1+c_1}\right )\right )\\ y(x)&\to 0 \end{align*}

2.1.1.5 ✓ Sympy. Time used: 0.540 (sec). Leaf size: 51

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

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