Internal problem ID [7440]
Internal file name [OUTPUT/6407_Sunday_June_05_2022_04_46_47_PM_1899037/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 51.
ODE order: 2.
ODE degree: 3.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {y {y^{\prime \prime }}^{3}+y^{3} y^{\prime }=0} \] The ode \begin {align*} y {y^{\prime \prime }}^{3}+y^{3} y^{\prime } = 0 \end {align*}
is factored to \begin {align*} y \left (y^{2} y^{\prime }+{y^{\prime \prime }}^{3}\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y = 0\tag {1} \\ y^{2} y^{\prime }+{y^{\prime \prime }}^{3} = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(y = 0\), is missing derivative in \(y\) then it is an algebraic equation. Solving for \(y\). \begin {align*} \end {align*}
Solving ODE (2) This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} y^{2} p \left (y \right )+p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} y^{2} p \left (y \right )+p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}
is factored to \begin {align*} p \left (y \right ) \left (p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{3}+y^{2}\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} p \left (y \right ) = 0\tag {1} \\ p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{3}+y^{2} = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(p \left (y \right ) = 0\), is missing derivative in \(p\) then it is an algebraic equation. Solving for \(p \left (y \right )\). \begin {align*} \end {align*}
Solving ODE (2) Let \(p=\frac {d}{d y}p \left (y \right )\) the ode becomes \begin {align*} p^{2} p^{3} = -y^{2} \end {align*}
Solving for \(p \left (y \right )\) from the above results in \begin {align*} p \left (y \right ) &= -\frac {y}{\sqrt {-p}\, p}\tag {1A}\\ p \left (y \right ) &= \frac {y}{\sqrt {-p}\, p}\tag {2A} \end {align*}
This has the form \begin {align*} p=yf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=p'(y)\). Each of the above ode’s is dAlembert ode which is now solved. Solving ode 1A Taking derivative of (*) w.r.t. \(y\) gives \begin {align*} p &= f+(y f'+g') \frac {dp}{dy}\\ p-f &= (y f'+g') \frac {dp}{dy}\tag {2} \end {align*}
Comparing the form \(p \left (y \right )=y f + g\) to (1A) shows that \begin {align*} f &= \frac {1}{\left (-p \right )^{\frac {3}{2}}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p -\frac {1}{\left (-p \right )^{\frac {3}{2}}} = \frac {3 y \left (\frac {d}{d y}p \left (y \right )\right )}{2 \left (-p \right )^{\frac {5}{2}}}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives \begin {align*} p -\frac {1}{\left (-p \right )^{\frac {3}{2}}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\\ p&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{3}+\textit {\_Z}^{2}-\textit {\_Z} +1, \operatorname {index} =4\right )^{2} \end {align*}
None of these values lead to defined solutions. Hence no singular solutions exist
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}\neq 0\). From eq. (2A). This results in \begin {align*} \frac {d}{d y}p \left (y \right ) = \frac {2 \left (p \left (y \right )-\frac {1}{\left (-p \left (y \right )\right )^{\frac {3}{2}}}\right ) \left (-p \left (y \right )\right )^{\frac {5}{2}}}{3 y}\tag {3} \end {align*}
This ODE is now solved for \(p \left (y \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}y \left (p \right ) = \frac {3 y \left (p \right )}{2 \left (-p \right )^{\frac {5}{2}} \left (p -\frac {1}{\left (-p \right )^{\frac {3}{2}}}\right )}\tag {4} \end {align*}
This ODE is now solved for \(y \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}y \left (p \right ) + p(p)y \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}y \left (p \right )+\frac {3 y \left (p \right )}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int \frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int \frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )}d p} y\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int \frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )}d p} y &= c_{3} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int \frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}-1\right )}d p}\) results in \begin {align*} y \left (p \right ) &= c_{3} {\mathrm e}^{\frac {3 \left (\int \frac {1}{p \left (\left (-p \right )^{\frac {5}{2}}+1\right )}d p \right )}{2}} \end {align*}
Since the solution \(y \left (p \right )\) has unresolved integral, unable to continue.
Solving ode 2A Taking derivative of (*) w.r.t. \(y\) gives \begin {align*} p &= f+(y f'+g') \frac {dp}{dy}\\ p-f &= (y f'+g') \frac {dp}{dy}\tag {2} \end {align*}
Comparing the form \(p \left (y \right )=y f + g\) to (1A) shows that \begin {align*} f &= -\frac {1}{\left (-p \right )^{\frac {3}{2}}}\\ g &= 0 \end {align*}
Hence (2) becomes \begin {align*} p +\frac {1}{\left (-p \right )^{\frac {3}{2}}} = -\frac {3 y \left (\frac {d}{d y}p \left (y \right )\right )}{2 \left (-p \right )^{\frac {5}{2}}}\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dy}=0\) in the above which gives \begin {align*} p +\frac {1}{\left (-p \right )^{\frac {3}{2}}} = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=-1\\ p&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =1\right )^{2}\\ p&=-\operatorname {RootOf}\left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+\textit {\_Z}^{2}+\textit {\_Z} +1, \operatorname {index} =4\right )^{2} \end {align*}
Removing solutions for \(p\) which leads to undefined results and substituting these in (1A) gives \begin {align*} p \left (y \right )&=-y \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}y}}\neq 0\). From eq. (2A). This results in \begin {align*} \frac {d}{d y}p \left (y \right ) = -\frac {2 \left (p \left (y \right )+\frac {1}{\left (-p \left (y \right )\right )^{\frac {3}{2}}}\right ) \left (-p \left (y \right )\right )^{\frac {5}{2}}}{3 y}\tag {3} \end {align*}
This ODE is now solved for \(p \left (y \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}y \left (p \right ) = -\frac {3 y \left (p \right )}{2 \left (-p \right )^{\frac {5}{2}} \left (p +\frac {1}{\left (-p \right )^{\frac {3}{2}}}\right )}\tag {4} \end {align*}
This ODE is now solved for \(y \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}y \left (p \right ) + p(p)y \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=-\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )}\\ q(p) &=0 \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}y \left (p \right )-\frac {3 y \left (p \right )}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )}d p} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left ({\mathrm e}^{\int -\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )}d p} y\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )}d p} y &= c_{5} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {3}{2 p \left (p \left (-p \right )^{\frac {3}{2}}+1\right )}d p}\) results in \begin {align*} y \left (p \right ) &= c_{5} {\mathrm e}^{-\frac {3 \left (\int \frac {1}{p \left (\left (-p \right )^{\frac {5}{2}}-1\right )}d p \right )}{2}} \end {align*}
Since the solution \(y \left (p \right )\) has unresolved integral, unable to continue.
For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} y^{\prime } = -y \end {align*}
Integrating both sides gives \begin {align*} \int -\frac {1}{y}d y &= x +c_{6}\\ -\ln \left (y \right )&=x +c_{6} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} y_1&={\mathrm e}^{-x -c_{6}}\\ &=\frac {{\mathrm e}^{-x}}{c_{6}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{6}} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{6}} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {{\mathrm e}^{-x}}{c_{6}} \\ \end{align*}
Verification of solutions
\[ y = \frac {{\mathrm e}^{-x}}{c_{6}} \] Verified OK.
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]
✓ Solution by Maple
Time used: 0.125 (sec). Leaf size: 128
dsolve(y(x)*diff(y(x),x$2)^3+y(x)^3*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -\left (\int _{}^{\textit {\_Z}}-\frac {1}{\textit {\_f}^{2}-\left (-\textit {\_f} \right )^{\frac {1}{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ y \left (x \right ) &= {\mathrm e}^{\int \operatorname {RootOf}\left (x +2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \sqrt {3}\, \left (-\textit {\_f} \right )^{\frac {1}{3}}+2 \textit {\_f}^{2}+\left (-\textit {\_f} \right )^{\frac {1}{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ y \left (x \right ) &= {\mathrm e}^{\int \operatorname {RootOf}\left (x -2 \left (\int _{}^{\textit {\_Z}}\frac {1}{i \sqrt {3}\, \left (-\textit {\_f} \right )^{\frac {1}{3}}-2 \textit {\_f}^{2}-\left (-\textit {\_f} \right )^{\frac {1}{3}}}d \textit {\_f} \right )+c_{1} \right )d x +c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 3.023 (sec). Leaf size: 800
DSolve[y[x]*y''[x]^3+y[x]^3*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5}{3} (-1) c_1\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 (-c_1)}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-\text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1+\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},-\frac {3 \sqrt [3]{-1} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (\sqrt [3]{-1} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \left (1-\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right ){}^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{5},\frac {8}{5},\frac {3 (-1)^{2/3} \text {$\#$1}^{5/3}}{5 c_1}\right )}{\left (-(-1)^{2/3} \text {$\#$1}^{5/3}+\frac {5 c_1}{3}\right ){}^{3/5}}\&\right ][x+c_2] \\ \end{align*}