4.33 problem 29

Internal problem ID [7902]
Book : Own collection of miscellaneous problems
Section : section 4.0
Problem number : 29
Date solved : Tuesday, October 22, 2024 at 02:53:30 PM
CAS classification : unknown

Solve

\begin{align*} \left (y-2 x y^{\prime }\right )^{2}&={y^{\prime }}^{3} \end{align*}

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

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

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

\begin{align*} y &= 2 x p +p^{{3}/{2}}\tag {1A}\\ y &= 2 x p -p^{{3}/{2}}\tag {2A} \end{align*}

This has the form

\begin{align*} y=xf(p)+g(p)\tag {*} \end{align*}

Where \(f,g\) are functions of \(p=y'(x)\). Each of the above ode’s is dAlembert ode which is now solved.

Solving ode 1A

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 &= 2 p\\ g &= p^{{3}/{2}} \end{align*}

Hence (2) becomes

\begin{align*} -p = \left (2 x +\frac {3 \sqrt {p}}{2}\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 = 0 \end{align*}

Solving the above for \(p\) results in

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

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

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

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 x \left (p \right )+\frac {3 \sqrt {p}}{2}}{p}\tag {4} \end{align*}

This ODE is now solved for \(x \left (p \right )\). The integrating factor is

\begin{align*} \mu &= {\mathrm e}^{\int \frac {2}{p}d p}\\ \mu &= p^{2}\\ \mu &= p^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (-\frac {3}{2 \sqrt {p}}\right ) \,dp} + c_1\right )\\ &= \frac {1}{\mu } \left (\frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}}+c_1\right ) \\ &= \frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}}\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).

Eliminating \(p\) from the following two equations

\begin{align*} x &= \frac {-3 p^{{5}/{2}}+5 c_1}{5 p^{2}} \\ y &= 2 x p +p^{{3}/{2}} \\ \end{align*}

results in

\begin{align*} p &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2} \\ \end{align*}

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} y &= 2 x \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}+{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

Solving ode 2A

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 &= 2 p\\ g &= -p^{{3}/{2}} \end{align*}

Hence (2) becomes

\begin{align*} -p = \left (2 x -\frac {3 \sqrt {p}}{2}\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 = 0 \end{align*}

Solving the above for \(p\) results in

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

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

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

Inverting the above ode gives

\begin{align*} \frac {d}{d p}x \left (p \right ) = -\frac {2 x \left (p \right )-\frac {3 \sqrt {p}}{2}}{p}\tag {4} \end{align*}

This ODE is now solved for \(x \left (p \right )\). The integrating factor is

\begin{align*} \mu &= {\mathrm e}^{\int \frac {2}{p}d p}\\ \mu &= p^{2}\\ \mu &= p^{2}\tag {5} \end{align*}

Integrating gives

\begin{align*} x \left (p \right )&= \frac {1}{\mu } \left ( \int { \mu \left (\frac {3}{2 \sqrt {p}}\right ) \,dp} + c_2\right )\\ &= \frac {1}{\mu } \left (\frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}}+c_2\right ) \\ &= \frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}}\tag {5} \end{align*}

Now we need to eliminate \(p\) between the above solution and (1A). The first method is to solve for \(p\) from Eq. (1A) and substitute the result into Eq. (5). The Second method is to solve for \(p\) from Eq. (5) and substitute the result into (1A).

Eliminating \(p\) from the following two equations

\begin{align*} x &= \frac {3 p^{{5}/{2}}+5 c_2}{5 p^{2}} \\ y &= 2 x p -p^{{3}/{2}} \\ \end{align*}

results in

\begin{align*} p &= \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2} \\ \end{align*}

Substituting the above into Eq (1A) and simplifying gives

\begin{align*} y &= 2 x \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}-{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}\right )}^{{3}/{2}} \\ \end{align*}

The solution

\[ y = 2 x \operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}+{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}+5 x \,\textit {\_Z}^{4}-5 c_1 \right )^{2}\right )}^{{3}/{2}} \]

was found not to satisfy the ode or the IC. Hence it is removed. The solution

\[ y = 2 x \operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}-{\left (\operatorname {RootOf}\left (3 \textit {\_Z}^{5}-5 x \,\textit {\_Z}^{4}+5 c_2 \right )^{2}\right )}^{{3}/{2}} \]

was found not to satisfy the ode or the IC. Hence it is removed.

4.33.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (y-2 x y^{\prime }\right )^{2}={y^{\prime }}^{3} \\ \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 }=\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}-\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}, y^{\prime }=-\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}\right )}{2}, y^{\prime }=-\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}\right )}{2}\right ] \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}-\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}-\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Solve the equation}\hspace {3pt} y^{\prime }=-\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{12}+\frac {3 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}+\frac {4 x^{2}}{3}+\frac {\mathrm {I} \sqrt {3}\, \left (\frac {\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}{6}+\frac {6 \left (\frac {4 y x}{3}-\frac {16 x^{4}}{9}\right )}{\left (-576 y x^{3}+108 y^{2}+512 x^{6}+12 \sqrt {-96 y^{3} x^{3}+81 y^{4}}\right )^{{1}/{3}}}\right )}{2} \\ \bullet & {} & \textrm {Set of solutions}\hspace {3pt} \\ {} & {} & \left \{\mathit {workingODE} , \mathit {workingODE} , \mathit {workingODE}\right \} \end {array} \]

4.33.2 Maple dsolve solution

Solving time : 11.450 (sec)
Leaf size : 73

dsolve((y(x)-2*x*diff(y(x),x))^2 = diff(y(x),x)^3, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ \left [x \left (\textit {\_T} \right ) &= \frac {3 \textit {\_T}^{{5}/{2}}+5 c_1}{5 \textit {\_T}^{2}}, y \left (\textit {\_T} \right ) &= \frac {\textit {\_T}^{{5}/{2}}+10 c_1}{5 \textit {\_T}}\right ] \\ \left [x \left (\textit {\_T} \right ) &= \frac {-3 \textit {\_T}^{{5}/{2}}+5 c_1}{5 \textit {\_T}^{2}}, y \left (\textit {\_T} \right ) &= \frac {-\textit {\_T}^{{5}/{2}}+10 c_1}{5 \textit {\_T}}\right ] \\ \end{align*}

4.33.3 Mathematica DSolve solution

Solving time : 0.0 (sec)
Leaf size : 0

DSolve[{(y[x]-2*x*D[y[x],x])^2== D[y[x],x]^3,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

Timed out