Internal problem ID [10116]
Internal file name [OUTPUT/9063_Monday_June_06_2022_06_19_27_AM_33787835/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1795 (book 6.204).
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]]
\[ \boxed {a b y \left (-1+y\right ) y^{\prime \prime }-\left (\left (2 a b -a -b \right ) y+\left (-a +1\right ) b \right ) {y^{\prime }}^{2}+f y \left (-1+y\right ) y^{\prime }=0} \]
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 b \,y^{2}-y a b \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (-2 y a b p \left (y \right )+f \,y^{2}+a y p \left (y \right )+b y p \left (y \right )+a b p \left (y \right )-f y -b 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 b -a b -y a -y b +b}{\left (-1+y \right ) y b a}\\ q(y) &=\frac {-f \,y^{2}+f y}{\left (-1+y \right ) y b a} \end {align*}
Hence the ode is \begin {align*} \frac {d}{d y}p \left (y \right )-\frac {\left (2 y a b -a b -y a -y b +b \right ) p \left (y \right )}{\left (-1+y \right ) y b a} = \frac {-f \,y^{2}+f y}{\left (-1+y \right ) y b a} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int -\frac {2 y a b -a b -y a -y b +b}{\left (-1+y \right ) y b a}d y} \\ &= {\mathrm e}^{-\ln \left (-1+y \right )-\ln \left (y \right )+\frac {\ln \left (-1+y \right )}{b}+\frac {\ln \left (y \right )}{a}} \\ \end{align*} Which simplifies to \[ \mu = \left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}} \] 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}{\left (-1+y \right ) y b a}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}y}} \left (\left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}} p\right ) &= \left (\left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}}\right ) \left (\frac {-f \,y^{2}+f y}{\left (-1+y \right ) y b a}\right )\\ \mathrm {d} \left (\left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}} p\right ) &= \left (-\frac {f \,y^{\frac {-a +1}{a}} \left (-1+y \right )^{\frac {-b +1}{b}}}{a b}\right )\, \mathrm {d} y \end {align*}
Integrating gives \begin {align*} \left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}} p &= \int {-\frac {f \,y^{\frac {-a +1}{a}} \left (-1+y \right )^{\frac {-b +1}{b}}}{a b}\,\mathrm {d} y}\\ \left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}} p &= -\frac {f \left (-1\right )^{-\frac {-b +1}{b}} y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [-\frac {-b +1}{b}, \frac {-a +1}{a}+1\right ], \left [2+\frac {-a +1}{a}\right ], y\right )}{b} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =\left (-1+y \right )^{\frac {-b +1}{b}} y^{\frac {-a +1}{a}}\) results in \begin {align*} p \left (y \right ) &= -\frac {\left (-1+y \right )^{\frac {b -1}{b}} y^{\frac {a -1}{a}} f \left (-1\right )^{-\frac {-b +1}{b}} y^{\frac {1}{a}} \operatorname {hypergeom}\left (\left [-\frac {-b +1}{b}, \frac {-a +1}{a}+1\right ], \left [2+\frac {-a +1}{a}\right ], y\right )}{b}+c_{1} \left (-1+y \right )^{\frac {b -1}{b}} y^{\frac {a -1}{a}} \end {align*}
which simplifies to \begin {align*} p \left (y \right ) &= \frac {\left (-1+y \right )^{\frac {b -1}{b}} \left (y \left (-1\right )^{-\frac {1}{b}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{b}\right ], \left [1+\frac {1}{a}\right ], y\right ) f +b \,y^{\frac {a -1}{a}} c_{1} \right )}{b} \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 } = \frac {\left (-1+y\right )^{\frac {b -1}{b}} \left (y \left (-1\right )^{-\frac {1}{b}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{b}\right ], \left [1+\frac {1}{a}\right ], y\right ) f +b y^{\frac {a -1}{a}} c_{1} \right )}{b} \end {align*}
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {\left (-1+\textit {\_a} \right )^{-\frac {b -1}{b}} b}{\textit {\_a} \left (-1\right )^{-\frac {1}{b}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{b}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +b \,\textit {\_a}^{\frac {a -1}{a}} c_{1}}d \textit {\_a} = x +c_{2} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {\left (-1+\textit {\_a} \right )^{-\frac {b -1}{b}} b}{\textit {\_a} \left (-1\right )^{-\frac {1}{b}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{b}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +b \,\textit {\_a}^{\frac {a -1}{a}} c_{1}}d \textit {\_a} &= x +c_{2} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {\left (-1+\textit {\_a} \right )^{-\frac {b -1}{b}} b}{\textit {\_a} \left (-1\right )^{-\frac {1}{b}} \operatorname {hypergeom}\left (\left [\frac {1}{a}, 1-\frac {1}{b}\right ], \left [1+\frac {1}{a}\right ], \textit {\_a}\right ) f +b \,\textit {\_a}^{\frac {a -1}{a}} c_{1}}d \textit {\_a} = x +c_{2} \] 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: 58
dsolve(a*b*y(x)*(-1+y(x))*diff(diff(y(x),x),x)-((2*a*b-a-b)*y(x)+(1-a)*b)*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}{b a}}+\int _{}^{y \left (x \right )}\textit {\_a}^{\frac {-a +1}{a}} \left (\textit {\_a} -1\right )^{\frac {1-b}{b}}d \textit {\_a} -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.25 (sec). Leaf size: 69
DSolve[f[x]*(-1 + y[x])*y[x]*y'[x] - ((1 - a)*b + (-a - b + 2*a*b)*y[x])*y'[x]^2 + a*b*(-1 + y[x])*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \text {InverseFunction}\left [-a \text {$\#$1}^{\frac {1}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{a},1-\frac {1}{b},1+\frac {1}{a},\text {$\#$1}\right )\&\right ]\left [\int _1^x\exp \left (-\int _1^{K[1]}\frac {f(K[1])}{a b}dK[1]\right ) c_1dK[1]+c_2\right ] \]