1.35 problem 36

Internal problem ID [7727]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 36
Date solved : Monday, October 21, 2024 at 03:59:23 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*}

1.35.1 Solved as first order Clairaut ode

Time used: 0.079 (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.

1.35.2 Maple step by step solution

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

1.35.3 Maple trace

Methods for first order ODEs:
 

1.35.4 Maple dsolve solution

Solving time : 0.136 (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 \,{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda } \lambda +{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }-1\right )}{4 \left ({\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda } \lambda +{\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z}}\right ) \lambda }-1\right )^{2} \left ({\mathrm e}^{\operatorname {RootOf}\left (2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}+2 \,{\mathrm e}^{\textit {\_Z} \left (1+2 \lambda \right )}-2 \lambda \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \right )}-x \,\lambda ^{2}-4 \,{\mathrm e}^{\textit {\_Z} \left (1+\lambda \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*}

1.35.5 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*}