2.1.2 problem 2

Internal problem ID [4088]
Book : Applied Differential equations, Newby Curle. Van Nostrand Reinhold. 1972
Section : Examples, page 35
Problem number : 2
Date solved : Tuesday, December 17, 2024 at 06:22:01 AM
CAS classification : [[_1st_order, _with_linear_symmetries], _rational, _Clairaut]

Solve

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

Solved as first order Clairaut ode

Time used: 0.196 (sec)

This is Clairaut ODE. It has the form

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

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

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

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

\begin{align*} y &= x p +\sqrt {p^{2}+1}\tag {1A} \\ y &= x p -\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 We start by replacing \(y^{\prime }\) by \(p\) which gives

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

Writing the ode as

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

Then we see that

\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 We start by replacing \(y^{\prime }\) by \(p\) which gives

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

Writing the ode as

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

Then we see that

\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 step by step solution

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

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 dsolve solution

Solving time : 0.077 (sec)
Leaf size : 57

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

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

Mathematica DSolve solution

Solving time : 0.174 (sec)
Leaf size : 73

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