Internal problem ID [6850]
Internal file name [OUTPUT/6097_Thursday_July_28_2022_04_30_22_AM_28511657/index.tex]
Book: Elementary differential equations. By Earl D. Rainville, Phillip E. Bedient. Macmilliam
Publishing Co. NY. 6th edition. 1981.
Section: CHAPTER 16. Nonlinear equations. Section 101. Independent variable missing.
EXERCISES Page 324
Problem number: 33.
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]]
\[ \boxed {\left (y y^{\prime \prime }+1+{y^{\prime }}^{2}\right )^{2}-\left (1+{y^{\prime }}^{2}\right )^{3}=0} \]
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*} \left (2 y p \left (y \right )^{2}+y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+2 y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )^{5}-2 p \left (y \right )^{3}-p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). The ode \begin {align*} \left (2 y p \left (y \right )^{2}+y^{2} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+2 y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-p \left (y \right )^{5}-2 p \left (y \right )^{3}-p \left (y \right )\right ) 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}-p \left (y \right )^{5}+2 p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right ) y -2 p \left (y \right )^{3}+2 \left (\frac {d}{d y}p \left (y \right )\right ) y -p \left (y \right )\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}-p \left (y \right )^{5}+2 p \left (y \right )^{2} \left (\frac {d}{d y}p \left (y \right )\right ) y -2 p \left (y \right )^{3}+2 \left (\frac {d}{d y}p \left (y \right )\right ) y -p \left (y \right ) = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Since \(p \left (y \right ) = 0\), 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 {\left (-1+\sqrt {p \left (y \right )^{2}+1}\right ) \left (p \left (y \right )^{2}+1\right )}{p \left (y \right ) y} \tag {1} \\ \frac {d}{d y}p \left (y \right )&=-\frac {\left (1+\sqrt {p \left (y \right )^{2}+1}\right ) \left (p \left (y \right )^{2}+1\right )}{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 {\left (-1+\sqrt {p^{2}+1}\right ) \left (p^{2}+1\right )}{p y} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=\frac {\left (p^{2}+1\right ) \left (-1+\sqrt {p^{2}+1}\right )}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (p^{2}+1\right ) \left (-1+\sqrt {p^{2}+1}\right )}{p}} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {\left (p^{2}+1\right ) \left (-1+\sqrt {p^{2}+1}\right )}{p}} \,dp} &= \int {\frac {1}{y} \,d y} \\ -\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (-1+\sqrt {p^{2}+1}\right )&=\ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (-1+\sqrt {p^{2}+1}\right )} &= {\mathrm e}^{\ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} -\frac {1}{\sqrt {p^{2}+1}}+1 &= c_{2} y \end {align*}
The solution is \[ -\frac {1}{\sqrt {p \left (y \right )^{2}+1}}+1 = c_{2} y \] Solving equation (2)
In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {\left (\sqrt {p^{2}+1}+1\right ) \left (p^{2}+1\right )}{p y} \end {align*}
Where \(f(y)=-\frac {1}{y}\) and \(g(p)=\frac {\left (\sqrt {p^{2}+1}+1\right ) \left (p^{2}+1\right )}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (\sqrt {p^{2}+1}+1\right ) \left (p^{2}+1\right )}{p}} \,dp &= -\frac {1}{y} \,d y \\ \int { \frac {1}{\frac {\left (\sqrt {p^{2}+1}+1\right ) \left (p^{2}+1\right )}{p}} \,dp} &= \int {-\frac {1}{y} \,d y} \\ -\ln \left (\sqrt {p^{2}+1}+1\right )+\frac {\ln \left (p^{2}+1\right )}{2}&=-\ln \left (y \right )+c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\ln \left (\sqrt {p^{2}+1}+1\right )+\frac {\ln \left (p^{2}+1\right )}{2}} &= {\mathrm e}^{-\ln \left (y \right )+c_{3}} \end {align*}
Which simplifies to \begin {align*} \frac {\sqrt {p^{2}+1}}{\sqrt {p^{2}+1}+1} &= \frac {c_{4}}{y} \end {align*}
The solution is \[ \frac {\sqrt {p \left (y \right )^{2}+1}}{1+\sqrt {p \left (y \right )^{2}+1}} = \frac {c_{4}}{y} \] 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 {1}{\sqrt {1+{y^{\prime }}^{2}}}+1 = c_{2} y \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {-y^{2} c_{2}^{2}+2 c_{2} y}}{-1+c_{2} y} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-y^{2} c_{2}^{2}+2 c_{2} y}}{-1+c_{2} y} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {c_{2} y -1}{\sqrt {-c_{2}^{2} y^{2}+2 c_{2} y}}d y &= \int d x \\ \frac {y \left (c_{2} y-2\right )}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}}&=x +c_{5} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {c_{2} y -1}{\sqrt {-c_{2}^{2} y^{2}+2 c_{2} y}}d y &= \int d x \\ \frac {\left (-c_{2} y+2\right ) y}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}}&=x +c_{6} \\ \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 {\sqrt {1+{y^{\prime }}^{2}}}{1+\sqrt {1+{y^{\prime }}^{2}}} = \frac {c_{4}}{y} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {\sqrt {-y^{2}+2 y c_{4}}}{y-c_{4}} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {-y^{2}+2 y c_{4}}}{y-c_{4}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin{align*} \int \frac {-c_{4} +y}{\sqrt {2 c_{4} y -y^{2}}}d y &= \int d x \\ -\sqrt {-y \left (y-2 c_{4} \right )}&=x +c_{7} \\ \end{align*} Solving equation (2)
Integrating both sides gives \begin{align*} \int -\frac {-c_{4} +y}{\sqrt {2 c_{4} y -y^{2}}}d y &= \int d x \\ \sqrt {y \left (-y+2 c_{4} \right )}&=x +c_{8} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} \frac {y \left (c_{2} y-2\right )}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}} &= x +c_{5} \\ \tag{2} \frac {\left (-c_{2} y+2\right ) y}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}} &= x +c_{6} \\ \tag{3} -\sqrt {-y \left (y-2 c_{4} \right )} &= x +c_{7} \\ \tag{4} \sqrt {y \left (-y+2 c_{4} \right )} &= x +c_{8} \\ \end{align*}
Verification of solutions
\[ \frac {y \left (c_{2} y-2\right )}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}} = x +c_{5} \] Verified OK.
\[ \frac {\left (-c_{2} y+2\right ) y}{\sqrt {-c_{2} y \left (c_{2} y-2\right )}} = x +c_{6} \] Verified OK.
\[ -\sqrt {-y \left (y-2 c_{4} \right )} = x +c_{7} \] Verified OK.
\[ \sqrt {y \left (-y+2 c_{4} \right )} = x +c_{8} \] 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 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+(1+(1+_b(_a)^2)^(1/2))*(1+_b(_a)^2)/_a = 0, _b(_a), HINT = [[_a, 0 symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.141 (sec). Leaf size: 107
dsolve((y(x)*diff(y(x),x$2)+1+diff(y(x),x)^2)^2=(1+diff(y(x),x)^2)^3,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -i x +c_{1} \\ y \left (x \right ) &= i x +c_{1} \\ y \left (x \right ) &= 0 \\ y \left (x \right ) &= -c_{1} -\sqrt {-\left (x +c_{1} +c_{2} \right ) \left (x -c_{1} +c_{2} \right )} \\ y \left (x \right ) &= -c_{1} +\sqrt {-\left (x +c_{1} +c_{2} \right ) \left (x -c_{1} +c_{2} \right )} \\ y \left (x \right ) &= c_{1} -\sqrt {-\left (x +c_{1} +c_{2} \right ) \left (x -c_{1} +c_{2} \right )} \\ y \left (x \right ) &= c_{1} +\sqrt {-\left (x +c_{1} +c_{2} \right ) \left (x -c_{1} +c_{2} \right )} \\ \end{align*}
✓ Solution by Mathematica
Time used: 45.659 (sec). Leaf size: 155
DSolve[(y[x]*y''[x]+1+(y'[x])^2)^2==(1+(y'[x])^2)^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\sqrt {e^{2 c_1}-(x+c_2){}^2}-e^{c_1} \\ y(x)\to e^{c_1}-\sqrt {e^{2 c_1}-(x+c_2){}^2} \\ y(x)\to \sqrt {e^{2 c_1}-(x+c_2){}^2}-e^{c_1} \\ y(x)\to \sqrt {e^{2 c_1}-(x+c_2){}^2}+e^{c_1} \\ y(x)\to -\sqrt {-(x+c_2){}^2} \\ y(x)\to \sqrt {-(x+c_2){}^2} \\ \end{align*}