2.1.51 problem 51

Internal problem ID [8534]
Book : Second order enumerated odes
Section : section 1
Problem number : 51
Date solved : Sunday, November 10, 2024 at 04:00:55 AM
CAS classification : [[_2nd_order, _missing_x]]

Solve

\begin{align*} y {y^{\prime \prime }}^{3}+y^{3} y^{\prime }&=0 \end{align*}

Factoring the ode gives these factors

\begin{align*} \tag{1} y &= 0 \\ \tag{2} y^{\prime } y^{2} &= 0 \\ \end{align*}

Now each of the above equations is solved in turn.

Solving equation (1)

Solving for \(y\) from

\begin{align*} y = 0 \end{align*}

Solving gives \(y = 0\)

Solving equation (2)

Solved as first order quadrature ode

Time used: 0.017 (sec)

Since the ode has the form \(y^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {dy} &= \int {0\, dx} + c_1 \\ y &= c_1 \end{align*}

Figure 2.155: Slope field plot
\(y^{\prime } y^{2} = 0\)

Summary of solutions found

\begin{align*} y &= c_1 \\ \end{align*}

Solved as first order homogeneous class D2 ode

Time used: 0.148 (sec)

Applying change of variables \(y = u \left (x \right ) x\), then the ode becomes

\begin{align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) u \left (x \right )^{2} x^{2} = 0 \end{align*}

Which is now solved The ode \(u^{\prime }\left (x \right ) = -\frac {u \left (x \right )}{x}\) is separable as it can be written as

\begin{align*} u^{\prime }\left (x \right )&= -\frac {u \left (x \right )}{x}\\ &= f(x) g(u) \end{align*}

Where

\begin{align*} f(x) &= -\frac {1}{x}\\ g(u) &= u \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(u)} \,du} &= \int { f(x) \,dx}\\ \int { \frac {1}{u}\,du} &= \int { -\frac {1}{x} \,dx}\\ \ln \left (u \left (x \right )\right )&=\ln \left (\frac {1}{x}\right )+c_1 \end{align*}

We now need to find the singular solutions, these are found by finding for what values \(g(u)\) is zero, since we had to divide by this above. Solving \(g(u)=0\) or \(u=0\) for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right )&=0 \end{align*}

Now we go over each such singular solution and check if it verifies the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (u \left (x \right )\right ) = \ln \left (\frac {1}{x}\right )+c_1\\ u \left (x \right ) = 0 \end{align*}

Solving for \(u \left (x \right )\) gives

\begin{align*} u \left (x \right ) &= \frac {{\mathrm e}^{c_1}}{x} \\ \end{align*}

Converting \(u \left (x \right ) = 0\) back to \(y\) gives

\begin{align*} y = 0 \end{align*}

Converting \(u \left (x \right ) = \frac {{\mathrm e}^{c_1}}{x}\) back to \(y\) gives

\begin{align*} y = {\mathrm e}^{c_1} \end{align*}

Figure 2.156: Slope field plot
\(y^{\prime } y^{2} = 0\)

Summary of solutions found

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{c_1} \\ \end{align*}

Solved as first order ode of type differential

Time used: 0.010 (sec)

Writing the ode as

\begin{align*} y^{\prime }&=0\tag {1} \end{align*}

Which becomes

\begin{align*} \left (1\right ) dy &= \left (0\right ) dx\tag {2} \end{align*}

But the RHS is complete differential because

\begin{align*} \left (0\right ) dx &= d\left (0\right ) \end{align*}

Hence (2) becomes

\begin{align*} \left (1\right ) dy &= d\left (0\right ) \end{align*}

Integrating gives

\begin{align*} y = c_1 \end{align*}

Figure 2.157: Slope field plot
\(y^{\prime } y^{2} = 0\)

Summary of solutions found

\begin{align*} y &= c_1 \\ \end{align*}

Maple step by step solution

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying 2nd order Liouville 
      trying 2nd order WeierstrassP 
      trying 2nd order JacobiSN 
      differential order: 2; trying a linearization to 3rd order 
      trying 2nd order ODE linearizable_by_differentiation 
      trying 2nd order, 2 integrating factors of the form mu(x,y) 
      trying differential order: 2; missing variables 
      `, `-> Computing symmetries using: way = 3 
      Try integration with the canonical coordinates of the symmetry [0, y] 
      -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -_b(_a)^2+(-_b(_a))^(1/3), _b(_a), explicit, HINT = [[1, 0]]`         *** 
         symmetry methods on request 
      `, `1st order, trying reduction of order with given symmetries:`[1, 0]
 

Maple dsolve solution

Solving time : 0.283 (sec)
Leaf size : 126

dsolve(y(x)*diff(diff(y(x),x),x)^3+y(x)^3*diff(y(x),x) = 0, 
       y(x),singsol=all)
 

\begin{align*} y &= 0 \\ y &= c_{1} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -\left (\int _{}^{\textit {\_Z}}-\frac {1}{\textit {\_f}^{2}-\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x +2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \sqrt {3}\, \left (-\textit {\_f} \right )^{{1}/{3}}+2 \textit {\_f}^{2}+\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ y &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \sqrt {3}\, \left (-\textit {\_f} \right )^{{1}/{3}}-2 \textit {\_f}^{2}-\left (-\textit {\_f} \right )^{{1}/{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ \end{align*}

Mathematica DSolve solution

Solving time : 4.228 (sec)
Leaf size : 800

DSolve[{y[x]*D[y[x],{x,2}]^3+y[x]^3*D[y[x],x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5}{3} (-1) c_1\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ \end{align*}