2.1.35 problem 36

Internal problem ID [8423]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 36
Date solved : Tuesday, December 17, 2024 at 12:51:12 PM
CAS classification : [_Clairaut]

Solve

\begin{align*} x f^{\prime }-f&=\frac {{f^{\prime }}^{2} \left (1-{f^{\prime }}^{\lambda }\right )^{2}}{\lambda ^{2}} \end{align*}

Solved as first order Clairaut ode

Time used: 0.075 (sec)

This is Clairaut ODE. It has the form

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

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

\begin{align*} x p -f = \frac {p^{2} \left (1-p^{\lambda }\right )^{2}}{\lambda ^{2}} \end{align*}

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

\begin{align*} f &= -\frac {p \left (p^{2 \lambda } p -x \,\lambda ^{2}-2 p^{\lambda } p +p \right )}{\lambda ^{2}}\tag {1A} \end{align*}

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

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

\begin{align*} f&=x p -\frac {p^{2} \left (p^{2 \lambda }-2 p^{\lambda }+1\right )}{\lambda ^{2}}\\ &=x p -\frac {p^{2} \left (p^{2 \lambda }-2 p^{\lambda }+1\right )}{\lambda ^{2}} \end{align*}

Writing the ode as

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

Then we see that

\begin{align*} g&=-\frac {p^{2} \left (p^{2 \lambda }-2 p^{\lambda }+1\right )}{\lambda ^{2}} \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*} f = c_2 x -\frac {c_2^{2} \left (c_2^{2 \lambda }-2 c_2^{\lambda }+1\right )}{\lambda ^{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 \(g=-\frac {p^{2} \left (p^{2 \lambda }-2 p^{\lambda }+1\right )}{\lambda ^{2}}\), then the above equation becomes

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

Unable to solve for \(p\). No singular solutions can be found.

Summary of solutions found

\begin{align*} f &= c_2 x -\frac {c_2^{2} \left (c_2^{2 \lambda }-2 c_2^{\lambda }+1\right )}{\lambda ^{2}} \\ \end{align*}

Maple step by step solution

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

Maple trace

`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 
<- 1st order, parametric methods successful 
<- dAlembert successful`
 

Maple dsolve solution

Solving time : 0.677 (sec)
Leaf size : 318

dsolve(x*diff(f(x),x)-f(x) = diff(f(x),x)^2/lambda^2*(1-diff(f(x),x)^lambda)^2, 
       f(x),singsol=all)
 

\begin{align*} f &= 0 \\ f &= \frac {\lambda ^{2} x^{2} \left (2 \lambda \,{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }+{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }-1\right )}{4 \left (\lambda \,{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }+{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }-1\right )^{2} \left ({\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (2 \lambda +1\right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (\lambda +1\right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }-1\right )} \\ f &= c_{1} x -\frac {c_{1}^{2} \left (-1+c_{1}^{\lambda }\right )^{2}}{\lambda ^{2}} \\ \end{align*}

Mathematica DSolve solution

Solving time : 14.489 (sec)
Leaf size : 30

DSolve[{x*D[ f[x],x]-f[x]==D[ f[x],x]^2/\[Lambda]^2*(1-D[ f[x],x]^\[Lambda])^2,{}}, 
       f[x],x,IncludeSingularSolutions->True]
 

\begin{align*} f(x)\to c_1 \left (x-\frac {c_1 \left (-1+c_1{}^{\lambda }\right ){}^2}{\lambda ^2}\right ) \\ f(x)\to 0 \\ \end{align*}