1.17 problem 17

Internal problem ID [7709]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 17
Date solved : Monday, October 21, 2024 at 03:57:49 PM
CAS classification : [[_1st_order, _with_linear_symmetries], _Clairaut]

Solve

\begin{align*} \frac {{y^{\prime }}^{2}}{4}-x y^{\prime }+y&=0 \end{align*}

1.17.1 Solved as first order Clairaut ode

Time used: 0.064 (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*} \frac {1}{4} p^{2}-x p +y = 0 \end{align*}

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

\begin{align*} y &= -\frac {1}{4} p^{2}+x p\tag {1A} \end{align*}

The above ode is a Clairaut ode which is now solved.

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

\begin{align*} y&=-\frac {1}{4} p^{2}+x p\\ &=-\frac {1}{4} p^{2}+x p \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&=-\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*}

1.17.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {{y^{\prime }}^{2}}{4}-x y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \left [y^{\prime }=2 x -2 \sqrt {x^{2}-y}, y^{\prime }=2 x +2 \sqrt {x^{2}-y}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=2 x -2 \sqrt {x^{2}-y} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=2 x +2 \sqrt {x^{2}-y} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

1.17.3 Maple trace

Methods for first order ODEs:
 

1.17.4 Maple dsolve solution

Solving time : 0.082 (sec)
Leaf size : 18

dsolve(1/4*diff(y(x),x)^2-x*diff(y(x),x)+y(x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= x^{2} \\ y &= -\frac {c_1 \left (-4 x +c_1 \right )}{4} \\ \end{align*}

1.17.5 Mathematica DSolve solution

Solving time : 0.008 (sec)
Leaf size : 25

DSolve[{(1/4)*(D[y[x],x])^2-x*D[y[x],x]+y[x]==0,{}}, 
       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*}