Internal problem ID [10115]
Internal file name [OUTPUT/9062_Monday_June_06_2022_06_19_20_AM_21973797/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1794 (book 6.203).
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], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {a y \left (-1+y\right ) y^{\prime \prime }-\left (a -1\right ) \left (2 y-1\right ) {y^{\prime }}^{2}+f y \left (-1+y\right ) y^{\prime }=0} \]
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 (a \,y^{2}-y a \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (f \,y^{2}-2 a y p \left (y \right )+2 y p \left (y \right )-f y +a p \left (y \right )-p \left (y \right )\right ) p \left (y \right ) = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d y}p \left (y \right ) + p(y)p \left (y \right ) &= q(y) \end {align*}
Where here \begin {align*} p(y) &=-\frac {2 y a -a -2 y +1}{a y \left (-1+y \right )}\\ q(y) &=\frac {-f \,y^{2}+f y}{a y \left (-1+y \right )} \end {align*}
Hence the ode is \begin {align*} \frac {d}{d y}p \left (y \right )-\frac {\left (2 y a -a -2 y +1\right ) p \left (y \right )}{a y \left (-1+y \right )} = \frac {-f \,y^{2}+f y}{a y \left (-1+y \right )} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 y a -a -2 y +1}{a y \left (-1+y \right )}d y} \\ &= {\mathrm e}^{-\frac {\left (a -1\right ) \ln \left (y \left (-1+y \right )\right )}{a}} \\ \end{align*} Which simplifies to \[ \mu = \left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}}\left ( \mu p\right ) &= \left (\mu \right ) \left (\frac {-f \,y^{2}+f y}{a y \left (-1+y \right )}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (\left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}} p\right ) &= \left (\left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}}\right ) \left (\frac {-f \,y^{2}+f y}{a y \left (-1+y \right )}\right )\\ \mathrm {d} \left (\left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}} p\right ) &= \left (-\frac {\left (y \left (-1+y \right )\right )^{\frac {-a +1}{a}} f}{a}\right )\, \mathrm {d} y \end {align*}
Integrating gives \begin {align*} \left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}} p &= \int {-\frac {\left (y \left (-1+y \right )\right )^{\frac {-a +1}{a}} f}{a}\,\mathrm {d} y}\\ \left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}} p &= -\left (-1\right )^{-\frac {-a +1}{a}} f \,y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [-\frac {-a +1}{a}, \frac {-a +1}{a}+1\right ], \left [2+\frac {-a +1}{a}\right ], y\right ) + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\left (y \left (-1+y \right )\right )^{-\frac {a -1}{a}}\) results in \begin {align*} p \left (y \right ) &= -\left (y \left (-1+y \right )\right )^{\frac {a -1}{a}} \left (-1\right )^{-\frac {-a +1}{a}} f \,y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [-\frac {-a +1}{a}, \frac {-a +1}{a}+1\right ], \left [2+\frac {-a +1}{a}\right ], y\right )+c_{1} \left (y \left (-1+y \right )\right )^{\frac {a -1}{a}} \end {align*}
which simplifies to \begin {align*} p \left (y \right ) &= \left (\left (-1\right )^{-\frac {1}{a}} y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{a}\right ], \left [1+\frac {1}{a}\right ], y\right ) f +c_{1} \right ) y \left (-1+y \right ) \left (y \left (-1+y \right )\right )^{-\frac {1}{a}} \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 } = \left (\left (-1\right )^{-\frac {1}{a}} y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{a}\right ], \left [1+\frac {1}{a}\right ], y\right ) f +c_{1} \right ) y \left (-1+y\right ) \left (y \left (-1+y\right )\right )^{-\frac {1}{a}} \end {align*}
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {\left (\textit {\_a} \left (-1+\textit {\_a} \right )\right )^{\frac {1}{a}}}{\left (\left (-1\right )^{-\frac {1}{a}} \textit {\_a}^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{a}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +c_{1} \right ) \textit {\_a} \left (-1+\textit {\_a} \right )}d \textit {\_a} = x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {\left (\textit {\_a} \left (-1+\textit {\_a} \right )\right )^{\frac {1}{a}}}{\left (\left (-1\right )^{-\frac {1}{a}} \textit {\_a}^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{a}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +c_{1} \right ) \textit {\_a} \left (-1+\textit {\_a} \right )}d \textit {\_a} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {\left (\textit {\_a} \left (-1+\textit {\_a} \right )\right )^{\frac {1}{a}}}{\left (\left (-1\right )^{-\frac {1}{a}} \textit {\_a}^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{a}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +c_{1} \right ) \textit {\_a} \left (-1+\textit {\_a} \right )}d \textit {\_a} = x +c_{2} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 48
dsolve(a*y(x)*(-1+y(x))*diff(diff(y(x),x),x)-(a-1)*(2*y(x)-1)*diff(y(x),x)^2+f*y(x)*(-1+y(x))*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 1 \\ y \left (x \right ) &= 0 \\ c_{1} {\mathrm e}^{-\frac {f x}{a}}-c_{2} +\int _{}^{y \left (x \right )}\frac {\left (\textit {\_a} \left (\textit {\_a} -1\right )\right )^{\frac {1}{a}}}{\textit {\_a} \left (\textit {\_a} -1\right )}d \textit {\_a} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.198 (sec). Leaf size: 83
DSolve[f[x]*(-1 + y[x])*y[x]*y'[x] - (-1 + a)*(-1 + 2*y[x])*y'[x]^2 + a*(-1 + y[x])*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \text {InverseFunction}\left [a \text {$\#$1}^{-1/a} (-((\text {$\#$1}-1) \text {$\#$1}))^{\frac {1}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{a},\frac {a-1}{a},1+\frac {1}{a},1-\text {$\#$1}\right )\&\right ]\left [\int _1^x\exp \left (-\int _1^{K[1]}\frac {f(K[1])}{a}dK[1]\right ) c_1dK[1]+c_2\right ] \]