2.1.2 Problem 2

Internal problem ID [4087]
Book : Applied Differential equations, Newby Curle. Van Nostrand Reinhold. 1972
Section : Examples, page 35
Problem number : 2
Date solved : Wednesday, November 26, 2025 at 03:33:44 PM
CAS classification : [[_1st_order, _with_linear_symmetries], _rational, _Clairaut]

Solved using first_order_ode_clairaut

Time used: 0.089 (sec)

Solve

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

This is Clairaut ODE. It has the form

\[ y=y^{\prime } x+g\left (y^{\prime }\right ) \]

Where \(g\) is function of \(y'(x)\). Let \(p=y^{\prime }\) the ode becomes

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

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

\begin{align*} y &= p x +\sqrt {p^{2}+1}\tag {1A} \\ y &= p x -\sqrt {p^{2}+1}\tag {2A} \end{align*}

Each of the above ode’s is a Clairaut ode which is now solved.

Solving ode 1A Writing the equation (1A) as

\begin{align*} y&= p x +g \left (p \right ) \end{align*}

We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes

\begin{align*} y = p x +g\tag {1} \end{align*}

Where

\begin{align*} g&=\sqrt {p^{2}+1} \end{align*}

Taking derivative of (1) w.r.t. \(x\) gives

\begin{align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end{align*}

Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\).

The general solution is given by

\begin{align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end{align*}

Substituting this in (1) gives the general solution as

\begin{align*} y = c_1 x +\sqrt {c_1^{2}+1} \end{align*}

The singular solution is found from solving for \(p\) from

\begin{align*} x+g'\left ( p\right ) &=0 \end{align*}

And substituting the result back in (1). Since we found above that \(g=\sqrt {p^{2}+1}\), then the above equation becomes

\begin{align*} x+g'\left ( p\right ) &= x +\frac {p}{\sqrt {p^{2}+1}}\\ &= 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=y = \left (-x^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}} \end{align*}

Substituting the above back in (1) results in

\begin{align*} y = \left (-x^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}} \end{align*}

Solving ode 2A Writing the equation (1A) as

\begin{align*} y&= p x +g \left (p \right ) \end{align*}

We now write \(g\equiv g\left ( p\right ) \) to make notation simpler but we should always remember that \(g\) is function of \(p\) which in turn is function of \(x\). Hence the above becomes

\begin{align*} y = p x +g\tag {1} \end{align*}

Where

\begin{align*} g&=-\sqrt {p^{2}+1} \end{align*}

Taking derivative of (1) w.r.t. \(x\) gives

\begin{align*} p &=\frac {d}{dx}\left (x p+g\right ) \\ p & =\left ( p+x\frac {dp}{dx}\right ) +\left ( g' \frac {dp}{dx}\right ) \\ p & =p+\left ( x+g'\right ) \frac {dp}{dx}\\ 0 & =\left ( x+g'\right ) \frac {dp}{dx} \end{align*}

Where \(g'\) is derivative of \(g\left ( p\right ) \) w.r.t. \(p\).

The general solution is given by

\begin{align*} \frac {dp}{dx} & =0\\ p &=c_{1} \end{align*}

Substituting this in (1) gives the general solution as

\begin{align*} y = c_2 x -\sqrt {c_2^{2}+1} \end{align*}

The singular solution is found from solving for \(p\) from

\begin{align*} x+g'\left ( p\right ) &=0 \end{align*}

And substituting the result back in (1). Since we found above that \(g=-\sqrt {p^{2}+1}\), then the above equation becomes

\begin{align*} x+g'\left ( p\right ) &= x -\frac {p}{\sqrt {p^{2}+1}}\\ &= 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=y = \sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right ) \end{align*}

Substituting the above back in (1) results in

\begin{align*} y = \sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right ) \end{align*}

Summary of solutions found

\begin{align*} y &= \sqrt {-\frac {1}{x^{2}-1}}\, \left (x^{2}-1\right ) \\ y &= \left (-x^{2}+1\right ) \sqrt {-\frac {1}{x^{2}-1}} \\ y &= c_1 x +\sqrt {c_1^{2}+1} \\ y &= c_2 x -\sqrt {c_2^{2}+1} \\ \end{align*}

✓ Maple. Time used: 0.042 (sec). Leaf size: 57

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

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

Maple trace

Methods for first order ODEs: 
   *** Sublevel 2 *** 
   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 
   trying dAlembert 
   <- dAlembert successful 
   <- dAlembert successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y \left (x \right )-x \left (\frac {d}{d x}y \left (x \right )\right )\right )^{2}=1+\left (\frac {d}{d x}y \left (x \right )\right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) x -\sqrt {y \left (x \right )^{2}+x^{2}-1}}{x^{2}-1}, \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) x +\sqrt {y \left (x \right )^{2}+x^{2}-1}}{x^{2}-1}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) x -\sqrt {y \left (x \right )^{2}+x^{2}-1}}{x^{2}-1} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=\frac {y \left (x \right ) x +\sqrt {y \left (x \right )^{2}+x^{2}-1}}{x^{2}-1} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

✓ Mathematica. Time used: 0.076 (sec). Leaf size: 73

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

\begin{align*} y(x)&\to c_1 x-\sqrt {1+c_1{}^2}\\ y(x)&\to c_1 x+\sqrt {1+c_1{}^2}\\ y(x)&\to -\sqrt {1-x^2}\\ y(x)&\to \sqrt {1-x^2} \end{align*}

✗ Sympy

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

Timed Out