Internal problem ID [10038]
Internal file name [OUTPUT/8985_Monday_June_06_2022_06_08_04_AM_88660805/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1717 (book 6.126).
ODE order: 2.
ODE degree: 1.
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]]
\[ \boxed {y^{\prime \prime } y+a \left ({y^{\prime }}^{2}+1\right )=0} \]
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*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y +a p \left (y \right )^{2} = -a \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {a \left (p^{2}+1\right )}{p y} \end {align*}
Where \(f(y)=-\frac {a}{y}\) and \(g(p)=\frac {p^{2}+1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {p^{2}+1}{p}} \,dp &= -\frac {a}{y} \,d y \\ \int { \frac {1}{\frac {p^{2}+1}{p}} \,dp} &= \int {-\frac {a}{y} \,d y} \\ \frac {\ln \left (p^{2}+1\right )}{2}&=-a \ln \left (y \right )+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {p^{2}+1} &= {\mathrm e}^{-a \ln \left (y \right )+c_{1}} \end {align*}
Which simplifies to \begin {align*} \sqrt {p^{2}+1} &= c_{2} {\mathrm e}^{-a \ln \left (y \right )} \end {align*}
Which simplifies to \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} y^{-a} {\mathrm e}^{c_{1}} \] The solution is \[ \sqrt {p \left (y \right )^{2}+1} = c_{2} y^{-a} {\mathrm e}^{c_{1}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \sqrt {{y^{\prime }}^{2}+1} = c_{2} y^{-a} {\mathrm e}^{c_{1}} \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 }&=\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2 a}}\, y^{-a} \tag {1} \\ y^{\prime }&=-\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-y^{2 a}}\, y^{-a} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} = x +c_{3} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} = x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} &= x +c_{3} \\ \tag{2} \int _{}^{y}-\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \int _{}^{y}-\frac {\textit {\_a}^{a}}{\sqrt {c_{2}^{2} {\mathrm e}^{2 c_{1}}-\textit {\_a}^{2 a}}}d \textit {\_a} = x +c_{4} \] Verified OK.
Maple trace
`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)+a*(_b(_a)^2+1)/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.093 (sec). Leaf size: 61
dsolve(diff(diff(y(x),x),x)*y(x)+a*(diff(y(x),x)^2+1)=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {\textit {\_a}^{a}}{\sqrt {-\textit {\_a}^{2 a}+c_{1}}}d \textit {\_a} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {\textit {\_a}^{a}}{\sqrt {-\textit {\_a}^{2 a}+c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 3.471 (sec). Leaf size: 526
DSolve[a*(1 + y'[x]^2) + y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {\text {$\#$1} \sqrt {1-e^{2 c_1} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 c_1} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 c_1} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt {1-e^{2 c_1} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 c_1} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 c_1} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\text {$\#$1} \sqrt {1-e^{2 (-c_1)} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 (-c_1)} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 (-c_1)} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt {1-e^{2 (-c_1)} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 (-c_1)} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 (-c_1)} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\text {$\#$1} \sqrt {1-e^{2 c_1} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 c_1} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 c_1} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt {1-e^{2 c_1} \text {$\#$1}^{-2 a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {1}{2 a},1-\frac {1}{2 a},e^{2 c_1} \text {$\#$1}^{-2 a}\right )}{\sqrt {-1+e^{2 c_1} \text {$\#$1}^{-2 a}}}\&\right ][x+c_2] \\ \end{align*}