2.1.29 problem 30

Internal problem ID [8417]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 30
Date solved : Tuesday, December 17, 2024 at 12:50:59 PM
CAS classification : [[_homogeneous, `class C`], _rational, _dAlembert]

Solve

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

Solved as first order ode of type dAlembert

Time used: 0.179 (sec)

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

\begin{align*} y = x \,p^{2}+p^{2} \end{align*}

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

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

This has the form

\begin{align*} y=xf(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 &= p^{2}\\ g &= p^{2} \end{align*}

Hence (2) becomes

\begin{align*} -p^{2}+p = \left (2 x p +2 p \right ) p^{\prime }\left (x \right )\tag {2A} \end{align*}

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

\begin{align*} -p^{2}+p = 0 \end{align*}

Solving the above for \(p\) results in

\begin{align*} p_{1} &=0\\ p_{2} &=1 \end{align*}

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

\begin{align*} y = 0\\ y = x +1 \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{align*} p^{\prime }\left (x \right ) = \frac {-p \left (x \right )^{2}+p \left (x \right )}{2 p \left (x \right ) x +2 p \left (x \right )}\tag {3} \end{align*}

This ODE is now solved for \(p \left (x \right )\). No inversion is needed. In canonical form a linear first order is

\begin{align*} p^{\prime }\left (x \right ) + q(x)p \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {1}{2 x +2}\\ p(x) &=\frac {1}{2 x +2} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {1}{2 x +2}d x}\\ &= \sqrt {x +1} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu p\right ) &= \left (\mu \right ) \left (\frac {1}{2 x +2}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (p \sqrt {x +1}\right ) &= \left (\sqrt {x +1}\right ) \left (\frac {1}{2 x +2}\right ) \\ \mathrm {d} \left (p \sqrt {x +1}\right ) &= \left (\frac {\sqrt {x +1}}{2 x +2}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} p \sqrt {x +1}&= \int {\frac {\sqrt {x +1}}{2 x +2} \,dx} \\ &=\sqrt {x +1} + c_1 \end{align*}

Dividing throughout by the integrating factor \(\sqrt {x +1}\) gives the final solution

\[ p \left (x \right ) = \frac {\sqrt {x +1}+c_1}{\sqrt {x +1}} \]

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

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

Summary of solutions found

\begin{align*} y &= 0 \\ y &= x +1 \\ y &= \frac {x \left (\sqrt {x +1}+c_1 \right )^{2}}{x +1}+\frac {\left (\sqrt {x +1}+c_1 \right )^{2}}{x +1} \\ \end{align*}

Maple step by step solution

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

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`
 

Maple dsolve solution

Solving time : 0.048 (sec)
Leaf size : 53

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

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

Mathematica DSolve solution

Solving time : 0.065 (sec)
Leaf size : 57

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

\begin{align*} y(x)\to x-c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\ y(x)\to x+c_1 \sqrt {x+1}+1+\frac {c_1{}^2}{4} \\ y(x)\to 0 \\ \end{align*}