Internal problem ID [7402]
Internal file name [OUTPUT/6369_Sunday_June_05_2022_04_41_52_PM_62970962/index.tex]
Book: Second order enumerated odes
Section: section 1
Problem number: 13.
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_x]]
\[ \boxed {{y^{\prime \prime }}^{2}+y^{\prime }=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} {p^{\prime }\left (x \right )}^{2}+p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Solving the given ode for \(p^{\prime }\left (x \right )\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} p^{\prime }\left (x \right )&=\sqrt {-p \left (x \right )} \tag {1} \\ p^{\prime }\left (x \right )&=-\sqrt {-p \left (x \right )} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {-p}}d p &= \int d x \\ -2 \sqrt {-p \left (x \right )}&=x +c_{1} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {-p}}d p &= \int d x \\ 2 \sqrt {-p \left (x \right )}&=c_{2} +x \\ \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*} -2 \sqrt {-y^{\prime }} = x +c_{1} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{4} x^{2}-\frac {1}{2} c_{1} x -\frac {1}{4} c_{1}^{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (x +c_{1} \right )^{3}}{12}+c_{3} \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*} 2 \sqrt {-y^{\prime }} = c_{2} +x \end {align*}
Integrating both sides gives \begin {align*} y &= \int { -\frac {1}{4} c_{2}^{2}-\frac {1}{2} c_{2} x -\frac {1}{4} x^{2}\,\mathop {\mathrm {d}x}}\\ &= -\frac {\left (c_{2} +x \right )^{3}}{12}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (x +c_{1} \right )^{3}}{12}+c_{3} \\ \tag{2} y &= -\frac {\left (c_{2} +x \right )^{3}}{12}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (x +c_{1} \right )^{3}}{12}+c_{3} \] Verified OK.
\[ y = -\frac {\left (c_{2} +x \right )^{3}}{12}+c_{4} \] Verified OK.
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*} 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*} 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}+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}+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 )}} \tag {1} \\ \frac {d}{d y}p \left (y \right )&=\frac {1}{\sqrt {-p \left (y \right )}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int -\sqrt {-p}d p &= \int d y \\ \frac {2 \left (-p \left (y \right )\right )^{\frac {3}{2}}}{3}&=y +c_{1} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int \sqrt {-p}d p &= \int d y \\ -\frac {2 \left (-p \left (y \right )\right )^{\frac {3}{2}}}{3}&=y +c_{2} \\ \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*} \frac {2 \left (-y^{\prime }\right )^{\frac {3}{2}}}{3} = y+c_{1} \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 y+12 c_{1} \right )^{\frac {2}{3}}}{4} \tag {1} \\ y^{\prime }&=-\left (-\frac {\left (12 y+12 c_{1} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (12 y+12 c_{1} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {2} \\ y^{\prime }&=-\left (-\frac {\left (12 y+12 c_{1} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (12 y+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 -\frac {4}{\left (12 y +12 c_{1} \right )^{\frac {2}{3}}}d y &= \int d x \\ -\left (12 y+12 c_{1} \right )^{\frac {1}{3}}&=x +c_{3} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {16}{\left (12 y +12 c_{1} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= \int d x \\ \frac {2 \left (12 y+12 c_{1} \right )^{\frac {1}{3}}}{1+i \sqrt {3}}&=x +c_{4} \\ \end{align*} Solving equation (3)
Integrating both sides gives \begin{align*} \int -\frac {16}{\left (12 y +12 c_{1} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= \int d x \\ -\frac {2 \left (12 y+12 c_{1} \right )^{\frac {1}{3}}}{i \sqrt {3}-1}&=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} = y+c_{2} \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 y-12 c_{2} \right )^{\frac {2}{3}}}{4} \tag {1} \\ y^{\prime }&=-\left (-\frac {\left (-12 y-12 c_{2} \right )^{\frac {1}{3}}}{4}-\frac {i \sqrt {3}\, \left (-12 y-12 c_{2} \right )^{\frac {1}{3}}}{4}\right )^{2} \tag {2} \\ y^{\prime }&=-\left (-\frac {\left (-12 y-12 c_{2} \right )^{\frac {1}{3}}}{4}+\frac {i \sqrt {3}\, \left (-12 y-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 -\frac {4}{\left (-12 y -12 c_{2} \right )^{\frac {2}{3}}}d y &= \int d x \\ \left (-12 y-12 c_{2} \right )^{\frac {1}{3}}&=x +c_{6} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {16}{\left (-12 y -12 c_{2} \right )^{\frac {2}{3}} \left (1+i \sqrt {3}\right )^{2}}d y &= \int d x \\ \frac {2 \left (-12 y-12 c_{2} \right )^{\frac {1}{3}}}{i \sqrt {3}-1}&=x +c_{7} \\ \end{align*} Solving equation (3)
Integrating both sides gives \begin{align*} \int -\frac {16}{\left (-12 y -12 c_{2} \right )^{\frac {2}{3}} \left (i \sqrt {3}-1\right )^{2}}d y &= \int d x \\ -\frac {2 \left (-12 y-12 c_{2} \right )^{\frac {1}{3}}}{1+i \sqrt {3}}&=x +c_{8} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{12} c_{3}^{3}-\frac {1}{4} c_{3}^{2} x -\frac {1}{4} c_{3} x^{2}-\frac {1}{12} x^{3}-c_{1} \\ \tag{2} y &= -\frac {1}{12} c_{4}^{3}-\frac {1}{4} c_{4}^{2} x -\frac {1}{4} c_{4} x^{2}-\frac {1}{12} x^{3}-c_{1} \\ \tag{3} y &= -\frac {1}{12} c_{5}^{3}-\frac {1}{4} c_{5}^{2} x -\frac {1}{4} c_{5} x^{2}-\frac {1}{12} x^{3}-c_{1} \\ \tag{4} y &= -\frac {1}{12} c_{6}^{3}-\frac {1}{4} c_{6}^{2} x -\frac {1}{4} c_{6} x^{2}-\frac {1}{12} x^{3}-c_{2} \\ \tag{5} y &= -\frac {1}{12} c_{7}^{3}-\frac {1}{4} c_{7}^{2} x -\frac {1}{4} c_{7} x^{2}-\frac {1}{12} x^{3}-c_{2} \\ \tag{6} y &= -\frac {1}{12} c_{8}^{3}-\frac {1}{4} c_{8}^{2} x -\frac {1}{4} c_{8} x^{2}-\frac {1}{12} x^{3}-c_{2} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{12} c_{3}^{3}-\frac {1}{4} c_{3}^{2} x -\frac {1}{4} c_{3} x^{2}-\frac {1}{12} x^{3}-c_{1} \] Verified OK.
\[ y = -\frac {1}{12} c_{4}^{3}-\frac {1}{4} c_{4}^{2} x -\frac {1}{4} c_{4} x^{2}-\frac {1}{12} x^{3}-c_{1} \] Verified OK.
\[ y = -\frac {1}{12} c_{5}^{3}-\frac {1}{4} c_{5}^{2} x -\frac {1}{4} c_{5} x^{2}-\frac {1}{12} x^{3}-c_{1} \] Verified OK.
\[ y = -\frac {1}{12} c_{6}^{3}-\frac {1}{4} c_{6}^{2} x -\frac {1}{4} c_{6} x^{2}-\frac {1}{12} x^{3}-c_{2} \] Verified OK.
\[ y = -\frac {1}{12} c_{7}^{3}-\frac {1}{4} c_{7}^{2} x -\frac {1}{4} c_{7} x^{2}-\frac {1}{12} x^{3}-c_{2} \] Verified OK.
\[ y = -\frac {1}{12} c_{8}^{3}-\frac {1}{4} c_{8}^{2} x -\frac {1}{4} c_{8} x^{2}-\frac {1}{12} x^{3}-c_{2} \] 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 -> Calling odsolve with the ODE`, diff(diff(diff(y(x), x), x), x)+1/2, y(x)` *** Sublevel 4 *** Methods for third order ODEs: --- Trying classification methods --- trying a quadrature <- quadrature successful <- 2nd order ODE linearizable_by_differentiation 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 <- 2nd order ODE linearizable_by_differentiation successful -> Calling odsolve with the ODE`, diff(y(x), x) = 0, y(x), singsol = none` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful`
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 27
dsolve(diff(y(x),x$2)^2+diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= c_{1} \\ y \left (x \right ) &= -\frac {1}{12} x^{3}+\frac {1}{2} c_{1} x^{2}-x \,c_{1}^{2}+c_{2} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.022 (sec). Leaf size: 69
DSolve[(y''[x])^2+y'[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x^3}{12}-\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\ y(x)\to -\frac {x^3}{12}+\frac {1}{4} i c_1 x^2+\frac {c_1{}^2 x}{4}+c_2 \\ \end{align*}