2.1.17 Problem 17
Internal
problem
ID
[10003]
Book
:
Own
collection
of
miscellaneous
problems
Section
:
section
1.0
Problem
number
:
17
Date
solved
:
Thursday, November 27, 2025 at 10:04:40 AM
CAS
classification
:
[[_1st_order, _with_linear_symmetries], _Clairaut]
Solved using first_order_ode_clairaut
Time used: 0.029 (sec)
Solve
\begin{align*}
\frac {{y^{\prime }}^{2}}{4}-y^{\prime } x +y&=0 \\
\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*} \frac {1}{4} p^{2}-p x +y = 0 \end{align*}
Solving for \(y\) from the above results in
\begin{align*} y &= -\frac {1}{4} p^{2}+p x\tag {1A} \end{align*}
The above ode is a Clairaut ode which is now solved.
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&=-\frac {p^{2}}{4} \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 -\frac {1}{4} c_1^{2} \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 \(g=-\frac {p^{2}}{4}\), then the above equation
becomes
\begin{align*} x+g'\left ( p\right ) &= x -\frac {p}{2}\\ &= 0 \end{align*}
Solving the above for \(p\) results in
\begin{align*} p_{1} &=y = x^{2} \end{align*}
Substituting the above back in (1) results in
\begin{align*} y = x^{2} \end{align*}
Simplifying the above gives
\begin{align*}
y &= -\frac {c_1 \left (-4 x +c_1 \right )}{4} \\
y &= x^{2} \\
\end{align*}
Summary of solutions found
\begin{align*}
y &= x^{2} \\
y &= -\frac {c_1 \left (-4 x +c_1 \right )}{4} \\
\end{align*}
✓ Maple. Time used: 0.017 (sec). Leaf size: 18
ode:=1/4*diff(y(x),x)^2-diff(y(x),x)*x+y(x) = 0;
dsolve(ode,y(x), singsol=all);
\begin{align*}
y &= x^{2} \\
y &= -\frac {c_1 \left (c_1 -4 x \right )}{4} \\
\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
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d x}y \left (x \right )\right )^{2}}{4}-x \left (\frac {d}{d x}y \left (x \right )\right )+y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [\frac {d}{d x}y \left (x \right )=2 x -2 \sqrt {x^{2}-y \left (x \right )}, \frac {d}{d x}y \left (x \right )=2 x +2 \sqrt {x^{2}-y \left (x \right )}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=2 x -2 \sqrt {x^{2}-y \left (x \right )} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} \frac {d}{d x}y \left (x \right )=2 x +2 \sqrt {x^{2}-y \left (x \right )} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]
✓ Mathematica. Time used: 0.005 (sec). Leaf size: 25
ode=(1/4)*(D[y[x],x])^2-x*D[y[x],x]+y[x]==0;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to c_1 x-\frac {c_1{}^2}{4}\\ y(x)&\to x^2 \end{align*}
✓ Sympy. Time used: 1.091 (sec). Leaf size: 10
from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x*Derivative(y(x), x) + y(x) + Derivative(y(x), x)**2/4,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
y{\left (x \right )} = x^{2} - \left (C_{1} + x\right )^{2}
\]