Internal problem ID [7438]
Internal file name [OUTPUT/6405_Sunday_June_05_2022_04_46_32_PM_41712938/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 49.
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], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{3} {y^{\prime \prime }}^{2}+y y^{\prime }=0} \] The ode \begin {align*} y^{3} {y^{\prime \prime }}^{2}+y y^{\prime } = 0 \end {align*}
is factored to \begin {align*} y \left (y^{2} {y^{\prime \prime }}^{2}+y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y = 0\tag {1} \\ y^{2} {y^{\prime \prime }}^{2}+y^{\prime } = 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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} y^{2} p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right )^{2}+p \left (y \right ) = 0 \end {align*}
is factored to \begin {align*} p \left (y \right ) \left (p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )^{2} y^{2}+1\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} p \left (y \right ) = 0\tag {1} \\ p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )^{2} y^{2}+1 = 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) Solving the given ode for \(\frac {d}{d y}p \left (y \right )\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} \frac {d}{d y}p \left (y \right )&=-\frac {1}{\sqrt {-p \left (y \right )}\, y} \tag {1} \\ \frac {d}{d y}p \left (y \right )&=\frac {1}{\sqrt {-p \left (y \right )}\, y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {1}{\sqrt {-p}\, y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {1}{\sqrt {-p}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{\sqrt {-p}}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {1}{\sqrt {-p}}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ -\frac {2 \left (-p \right )^{\frac {3}{2}}}{3}&=-\ln \left (y \right )+c_{1} \\ \end{align*} The solution is \[ -\frac {2 \left (-p \left (y \right )\right )^{\frac {3}{2}}}{3}+\ln \left (y \right )-c_{1} = 0 \] Solving equation (2)
In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= \frac {1}{\sqrt {-p}\, y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=\frac {1}{\sqrt {-p}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{\sqrt {-p}}} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {1}{\sqrt {-p}}} \,dp} &= \int {\frac {1}{y} \,d y} \\ -\frac {2 \left (-p \right )^{\frac {3}{2}}}{3}&=\ln \left (y \right )+c_{2} \\ \end{align*} The solution is \[ -\frac {2 \left (-p \left (y \right )\right )^{\frac {3}{2}}}{3}-\ln \left (y \right )-c_{2} = 0 \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {2 \left (-y^{\prime }\right )^{\frac {3}{2}}}{3}+\ln \left (y\right )-c_{1} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {\left (12 \ln \left (y\right )-12 c_{1} \right )^{\frac {2}{3}}}{4} \tag {1} \\ y^{\prime }&=-\left (-\frac {\left (12 \ln \left (y\right )-12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (12 \ln \left (y\right )-12 c_{1} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {2} \\ y^{\prime }&=-\left (-\frac {\left (12 \ln \left (y\right )-12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (12 \ln \left (y\right )-12 c_{1} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {4}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{3} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{4} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{5} \end {align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} -\frac {2 \left (-y^{\prime }\right )^{\frac {3}{2}}}{3}-\ln \left (y\right )-c_{2} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {\left (-12 \ln \left (y\right )-12 c_{2} \right )^{\frac {2}{3}}}{4} \tag {1} \\ y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (y\right )-12 c_{2} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 \ln \left (y\right )-12 c_{2} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {2} \\ y^{\prime }&=-\left (-\frac {\left (-12 \ln \left (y\right )-12 c_{2} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 \ln \left (y\right )-12 c_{2} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{6} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{7} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{8} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}-\frac {4}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} &= x +c_{3} \\ \tag{2} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} &= x +c_{4} \\ \tag{3} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} &= x +c_{5} \\ \tag{4} \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}}}d \textit {\_a} &= x +c_{6} \\ \tag{5} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} &= x +c_{7} \\ \tag{6} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} &= x +c_{8} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}-\frac {4}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{4} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{5} \] Verified OK.
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{6} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{7} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{8} \] Verified OK.
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}-\frac {4}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} &= x +c_{3} \\ \tag{2} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} &= x +c_{4} \\ \tag{3} \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} &= x +c_{5} \\ \tag{4} \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}}}d \textit {\_a} &= x +c_{6} \\ \tag{5} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} &= x +c_{7} \\ \tag{6} \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} &= x +c_{8} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}-\frac {4}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{4} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (12 \ln \left (\textit {\_a} \right )-12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{5} \] Verified OK.
\[ \int _{}^{y}-\frac {4}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}}}d \textit {\_a} = x +c_{6} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d \textit {\_a} = x +c_{7} \] Verified OK.
\[ \int _{}^{y}-\frac {16}{\left (-12 \ln \left (\textit {\_a} \right )-12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d \textit {\_a} = x +c_{8} \] 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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful ------------------- * Tackling next 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 <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 241
dsolve(y(x)^3*diff(y(x),x$2)^2+y(x)*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= 0 \\ -4 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -4 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \frac {-16 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (-12 \ln \left (\textit {\_a} \right )+8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )+2 i \left (-x -c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (-i \sqrt {3}-1\right )^{2}} &= 0 \\ \frac {-16 \left (\int _{}^{y \left (x \right )}\frac {1}{\left (12 \ln \left (\textit {\_a} \right )-8 c_{1} \right )^{\frac {2}{3}}}d \textit {\_a} \right )+2 i \left (x +c_{2} \right ) \sqrt {3}+2 x +2 c_{2}}{\left (1-i \sqrt {3}\right )^{2}} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.526 (sec). Leaf size: 459
DSolve[y[x]^3*y''[x]^2+y[x]*y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i (-c_1)} (-\log (\text {$\#$1})-i (-1) c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i (-1) c_1-\log (\text {$\#$1})\right )}{(-i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{-i c_1} (-\log (\text {$\#$1})-i c_1){}^{2/3} \Gamma \left (\frac {1}{3},-i c_1-\log (\text {$\#$1})\right )}{(c_1-i \log (\text {$\#$1})){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i (-c_1)} (-\log (\text {$\#$1})+i (-c_1)){}^{2/3} \Gamma \left (\frac {1}{3},i (-c_1)-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})-c_1){}^{2/3}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\left (\frac {2}{3}\right )^{2/3} e^{i c_1} (-\log (\text {$\#$1})+i c_1){}^{2/3} \Gamma \left (\frac {1}{3},i c_1-\log (\text {$\#$1})\right )}{(i \log (\text {$\#$1})+c_1){}^{2/3}}\&\right ][x+c_2] \\ \end{align*}