Internal problem ID [7435]
Internal file name [OUTPUT/6402_Sunday_June_05_2022_04_43_22_PM_4687353/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 46.
ODE order: 2.
ODE degree: 2.
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]]
\[ \boxed {y {y^{\prime \prime }}^{2}+{y^{\prime }}^{3}=0} \]
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 p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right )^{3} = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} y p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right )^{3} = 0 \end {align*}
is factored to \begin {align*} p \left (y \right )^{2} \left (\left (\frac {d}{d y}p \left (y \right )\right )^{2} y +p \left (y \right )\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} p \left (y \right )^{2} = 0\tag {1} \\ \left (\frac {d}{d y}p \left (y \right )\right )^{2} y +p \left (y \right ) = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(p \left (y \right )^{2} = 0\), is missing derivative in \(p\) then it is an algebraic equation. Solving for \(p \left (y \right )\). \begin {align*} \end {align*}
Solving ODE (2) The ode has the form \begin {align*} (p')^{\frac {n}{m}} &= f(y) g(p)\tag {1} \end {align*}
Where \(n=2, m=1, f=-\frac {1}{y} , g=p\). Hence the ode is \begin {align*} (p')^{2} &= -\frac {p}{y} \end {align*}
Solving for \(\frac {d}{d y}p \left (y \right )\) from (1) gives \begin {align*} \frac {d}{d y}p \left (y \right ) &=\sqrt {f g}\\ \frac {d}{d y}p \left (y \right ) &=-\sqrt {f g} \end {align*}
To be able to solve as separable ode, we have to now assume that \(f>0,g>0\). \begin {align*} -\frac {1}{y} &> 0\\ p &> 0 \end {align*}
Under the above assumption the differential equations become separable and can be written as \begin {align*} \frac {d}{d y}p \left (y \right ) &=\sqrt {f}\, \sqrt {g}\\ \frac {d}{d y}p \left (y \right ) &=-\sqrt {f}\, \sqrt {g} \end {align*}
Therefore \begin {align*} \frac {1}{\sqrt {g}} \, dp &= \left (\sqrt {f}\right )\,dy\\ -\frac {1}{\sqrt {g}} \, dp &= \left (\sqrt {f}\right )\,dy \end {align*}
Replacing \(f(y),g(p)\) by their values gives \begin {align*} \frac {1}{\sqrt {p}} \, dp &= \left (\sqrt {-\frac {1}{y}}\right )\,dy\\ -\frac {1}{\sqrt {p}} \, dp &= \left (\sqrt {-\frac {1}{y}}\right )\,dy \end {align*}
Integrating now gives the solutions. \begin {align*} \int \frac {1}{\sqrt {p}}d p &= \int \sqrt {-\frac {1}{y}}d y +c_{1}\\ \int -\frac {1}{\sqrt {p}}d p &= \int \sqrt {-\frac {1}{y}}d y +c_{1} \end {align*}
Integrating gives \begin {align*} 2 \sqrt {p \left (y \right )} &= 2 \sqrt {-\frac {1}{y}}\, y +c_{1}\\ -2 \sqrt {p \left (y \right )} &= 2 \sqrt {-\frac {1}{y}}\, y +c_{1} \end {align*}
Therefore \begin{align*} p \left (y \right ) &= \sqrt {-\frac {1}{y}}\, y c_{1} +\frac {c_{1}^{2}}{4}-y \\ p \left (y \right ) &= \sqrt {-\frac {1}{y}}\, y c_{1} +\frac {c_{1}^{2}}{4}-y \\ \end{align*}
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 } = \sqrt {-\frac {1}{y}}\, y c_{1} +\frac {c_{1}^{2}}{4}-y \end {align*}
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {-\frac {1}{y}}\, y c_{1} +\frac {c_{1}^{2}}{4}-y}d y &= \int d x \\ -\frac {2 c_{1}^{2}}{c_{1}^{2}+4 y}-\ln \left (c_{1}^{2}+4 y\right )-\frac {2 \sqrt {-\frac {1}{y}}\, \sqrt {y}\, \left (\arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) c_{1}^{2}+4 \arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) y-2 \sqrt {y}\, c_{1} \right )}{c_{1}^{2}+4 y}&=x +c_{2} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} -\frac {2 c_{1}^{2}}{c_{1}^{2}+4 y}-\ln \left (c_{1}^{2}+4 y\right )-\frac {2 \sqrt {-\frac {1}{y}}\, \sqrt {y}\, \left (\arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) c_{1}^{2}+4 \arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) y-2 \sqrt {y}\, c_{1} \right )}{c_{1}^{2}+4 y} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ -\frac {2 c_{1}^{2}}{c_{1}^{2}+4 y}-\ln \left (c_{1}^{2}+4 y\right )-\frac {2 \sqrt {-\frac {1}{y}}\, \sqrt {y}\, \left (\arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) c_{1}^{2}+4 \arctan \left (\frac {2 \sqrt {y}}{c_{1}}\right ) y-2 \sqrt {y}\, c_{1} \right )}{c_{1}^{2}+4 y} = x +c_{2} \] Verified OK.
Maple trace
`Methods for second order ODEs: *** Sublevel 2 *** Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 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))^(3/2)-_b(_a)^2, _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: 0.093 (sec). Leaf size: 166
dsolve(y(x)*diff(y(x),x$2)^2+diff(y(x),x)^3=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {c_{2} {\left (\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (c_{1} {\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ y \left (x \right ) &= \frac {c_{2} {\left (\operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-1+\frac {x}{2}}\right )+1\right )}^{2}}{\operatorname {LambertW}\left (-c_{1} {\mathrm e}^{-1+\frac {x}{2}}\right )^{2}} \\ y \left (x \right ) &= {\mathrm e}^{-\left (\int {\mathrm e}^{2 \operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}+1\right )^{2}\right )+c_{1} {\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}+\ln \left (\left ({\mathrm e}^{\textit {\_Z}}+1\right )^{2}\right )+c_{1} -2 \textit {\_Z} +x -2\right )}d x \right )-2 \left (\int {\mathrm e}^{\operatorname {RootOf}\left ({\mathrm e}^{\textit {\_Z}} \ln \left (\left ({\mathrm e}^{\textit {\_Z}}+1\right )^{2}\right )+c_{1} {\mathrm e}^{\textit {\_Z}}-2 \,{\mathrm e}^{\textit {\_Z}} \textit {\_Z} +x \,{\mathrm e}^{\textit {\_Z}}+\ln \left (\left ({\mathrm e}^{\textit {\_Z}}+1\right )^{2}\right )+c_{1} -2 \textit {\_Z} +x -2\right )}d x \right )-x +c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.165 (sec). Leaf size: 361
DSolve[y[x]*y''[x]^2+y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )-\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}-i (-c_1)\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i (-c_1)}{2 \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i (-1) c_1\right )\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}-i c_1\right )-\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}-i c_1\right )}\right )\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-4 \left (\frac {i c_1}{2 \left (2 \sqrt {\text {$\#$1}}+i c_1\right )}+\frac {1}{2} \log \left (2 \sqrt {\text {$\#$1}}+i c_1\right )\right )\&\right ][x+c_2] \\ \end{align*}