Internal problem ID [4678]
Internal file name [OUTPUT/4171_Sunday_June_05_2022_12_35_19_PM_64352952/index.tex]
Book: Differential Gleichungen, Kamke, 3rd ed, Abel ODEs
Section: Abel ODE’s with constant invariant
Problem number: problem 51.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "abelFirstKind"
Maple gives the following as the ode type
[_Abel]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\left (y-f \left (x \right )\right ) \left (y-g \left (x \right )\right ) \left (y-\frac {a f \left (x \right )+b g \left (x \right )}{a +b}\right ) h \left (x \right )-\frac {f^{\prime }\left (x \right ) \left (y-g \left (x \right )\right )}{f \left (x \right )-g \left (x \right )}-\frac {g^{\prime }\left (x \right ) \left (y-f \left (x \right )\right )}{g \left (x \right )-f \left (x \right )}=0} \]
This is Abel first kind ODE, it has the form \[ y^{\prime }= f_0(x)+f_1(x) y +f_2(x)y^{2}+f_3(x)y^{3} \] Comparing the above to given ODE which is \begin {align*} y^{\prime }&=-\frac {\left (-h \left (x \right ) f \left (x \right ) a -h \left (x \right ) f \left (x \right ) b +g \left (x \right ) h \left (x \right ) a +h \left (x \right ) g \left (x \right ) b \right ) y^{3}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {\left (2 h \left (x \right ) a f \left (x \right )^{2}+h \left (x \right ) f \left (x \right )^{2} b -h \left (x \right ) f \left (x \right ) g \left (x \right ) a +h \left (x \right ) f \left (x \right ) g \left (x \right ) b -g \left (x \right )^{2} h \left (x \right ) a -2 h \left (x \right ) b g \left (x \right )^{2}\right ) y^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {\left (-h \left (x \right ) f \left (x \right )^{3} a -h \left (x \right ) a f \left (x \right )^{2} g \left (x \right )-2 h \left (x \right ) f \left (x \right )^{2} g \left (x \right ) b +2 h \left (x \right ) f \left (x \right ) g \left (x \right )^{2} a +h \left (x \right ) b g \left (x \right )^{2} f \left (x \right )+h \left (x \right ) g \left (x \right )^{3} b +g^{\prime }\left (x \right ) a +g^{\prime }\left (x \right ) b -f^{\prime }\left (x \right ) a -f^{\prime }\left (x \right ) b \right ) y}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {f \left (x \right )^{3} g \left (x \right ) h \left (x \right ) a -f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) a +f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) b -f \left (x \right ) g \left (x \right )^{3} h \left (x \right ) b -g^{\prime }\left (x \right ) f \left (x \right ) a -g^{\prime }\left (x \right ) f \left (x \right ) b +f^{\prime }\left (x \right ) g \left (x \right ) a +f^{\prime }\left (x \right ) g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \frac {g^{\prime }\left (x \right ) f \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {g^{\prime }\left (x \right ) f \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {f^{\prime }\left (x \right ) g \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {f^{\prime }\left (x \right ) g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {f \left (x \right )^{3} g \left (x \right ) h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {f \left (x \right ) g \left (x \right )^{3} h \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}\\ f_1(x) &= \frac {h \left (x \right ) f \left (x \right )^{3} a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) g \left (x \right )^{3} b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {h \left (x \right ) a f \left (x \right )^{2} g \left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {2 h \left (x \right ) f \left (x \right )^{2} g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {2 h \left (x \right ) f \left (x \right ) g \left (x \right )^{2} a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) b g \left (x \right )^{2} f \left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {g^{\prime }\left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {g^{\prime }\left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {a f^{\prime }\left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {b f^{\prime }\left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}\\ f_2(x) &= -\frac {2 h \left (x \right ) a f \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right )^{2} b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {g \left (x \right )^{2} h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {2 h \left (x \right ) b g \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}\\ f_3(x) &= \frac {a f \left (x \right ) h \left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {h \left (x \right ) f \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {g \left (x \right ) h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {g \left (x \right ) h \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )} \end {align*}
Since \(f_2(x)=-\frac {2 h \left (x \right ) a f \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right )^{2} b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {g \left (x \right )^{2} h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {2 h \left (x \right ) b g \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}\) is not zero, then the first step is to apply the following transformation to remove \(f_2\). Let \(y = u(x) - \frac {f_2}{3 f_3}\) or \begin {align*} y &= u(x) - \left ( \frac {-\frac {2 h \left (x \right ) a f \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right )^{2} b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {g \left (x \right )^{2} h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {2 h \left (x \right ) b g \left (x \right )^{2}}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {h \left (x \right ) f \left (x \right ) g \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}}{\frac {3 a f \left (x \right ) h \left (x \right )}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}+\frac {3 h \left (x \right ) f \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {3 g \left (x \right ) h \left (x \right ) a}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}-\frac {3 g \left (x \right ) h \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )}} \right ) \\ &= \frac {f \left (x \right ) \left (2 a +b \right )+\left (a +2 b \right ) g \left (x \right )+3 \left (a +b \right ) u \left (x \right )}{3 a +3 b} \end {align*}
The above transformation applied to (1) gives a new ODE as \begin {align*} \text {Expression too large to display}\tag {2} \end {align*}
This is Abel first kind ODE, it has the form \[ u^{\prime }\left (x \right )= f_0(x)+f_1(x) u \left (x \right ) +f_2(x)u \left (x \right )^{2}+f_3(x)u \left (x \right )^{3} \] Comparing the above to given ODE which is \begin {align*} u^{\prime }\left (x \right )&=\text {Expression too large to display}\tag {1} \end {align*}
Therefore \begin {align*} f_0(x) &= \text {Expression too large to display}\\ f_1(x) &= \text {Expression too large to display}\\ f_2(x) &= 0\\ f_3(x) &= \frac {a^{3} f \left (x \right ) h \left (x \right )}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}+\frac {f \left (x \right ) h \left (x \right ) b^{3}}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}-\frac {h \left (x \right ) g \left (x \right ) a^{3}}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}-\frac {h \left (x \right ) g \left (x \right ) b^{3}}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}+\frac {3 f \left (x \right ) h \left (x \right ) a^{2} b}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}+\frac {3 f \left (x \right ) h \left (x \right ) a \,b^{2}}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}-\frac {3 h \left (x \right ) g \left (x \right ) a^{2} b}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}}-\frac {3 h \left (x \right ) g \left (x \right ) a \,b^{2}}{f \left (x \right ) a^{3}+3 f \left (x \right ) a^{2} b +3 f \left (x \right ) a \,b^{2}+f \left (x \right ) b^{3}-g \left (x \right ) a^{3}-3 g \left (x \right ) a^{2} b -3 g \left (x \right ) a \,b^{2}-g \left (x \right ) b^{3}} \end {align*}
Since \(f_2(x)=0\) then we check the Abel invariant to see if it depends on \(x\) or not. The Abel invariant is given by \begin {align*} -\frac {f_{1}^{3}}{f_{0}^{2} f_{3}} \end {align*}
Which when evaluating gives \begin {align*} \text {Expression too large to display} \end {align*}
Since the Abel invariant depends on \(x\) then unable to solve this ode at this time.
Unable to complete the solution now.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } f \left (x \right ) a +y^{\prime } f \left (x \right ) b -y^{\prime } g \left (x \right ) a -y^{\prime } g \left (x \right ) b -g^{\prime }\left (x \right ) f \left (x \right ) a -g^{\prime }\left (x \right ) f \left (x \right ) b +f^{\prime }\left (x \right ) g \left (x \right ) a +f^{\prime }\left (x \right ) g \left (x \right ) b -f^{\prime }\left (x \right ) y a -f^{\prime }\left (x \right ) y b +g^{\prime }\left (x \right ) y a +g^{\prime }\left (x \right ) y b -y^{3} f \left (x \right ) h \left (x \right ) a -y^{3} f \left (x \right ) h \left (x \right ) b +y^{3} g \left (x \right ) h \left (x \right ) a +y^{3} g \left (x \right ) h \left (x \right ) b +2 y^{2} f \left (x \right )^{2} h \left (x \right ) a +y^{2} f \left (x \right )^{2} h \left (x \right ) b -y^{2} g \left (x \right )^{2} h \left (x \right ) a -2 y^{2} g \left (x \right )^{2} h \left (x \right ) b -y f \left (x \right )^{3} h \left (x \right ) a +y g \left (x \right )^{3} h \left (x \right ) b +f \left (x \right )^{3} g \left (x \right ) h \left (x \right ) a -f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) a +f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) b -f \left (x \right ) g \left (x \right )^{3} h \left (x \right ) b +y f \left (x \right ) g \left (x \right )^{2} h \left (x \right ) b +2 y f \left (x \right ) g \left (x \right )^{2} h \left (x \right ) a -y^{2} f \left (x \right ) g \left (x \right ) h \left (x \right ) a +y^{2} f \left (x \right ) g \left (x \right ) h \left (x \right ) b -y f \left (x \right )^{2} g \left (x \right ) h \left (x \right ) a -2 y f \left (x \right )^{2} g \left (x \right ) h \left (x \right ) b =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {g^{\prime }\left (x \right ) f \left (x \right ) a +g^{\prime }\left (x \right ) f \left (x \right ) b -f^{\prime }\left (x \right ) g \left (x \right ) a -f^{\prime }\left (x \right ) g \left (x \right ) b +f^{\prime }\left (x \right ) y a +f^{\prime }\left (x \right ) y b -g^{\prime }\left (x \right ) y a -g^{\prime }\left (x \right ) y b +y^{3} f \left (x \right ) h \left (x \right ) a +y^{3} f \left (x \right ) h \left (x \right ) b -y^{3} g \left (x \right ) h \left (x \right ) a -y^{3} g \left (x \right ) h \left (x \right ) b -2 y^{2} f \left (x \right )^{2} h \left (x \right ) a -y^{2} f \left (x \right )^{2} h \left (x \right ) b +y^{2} g \left (x \right )^{2} h \left (x \right ) a +2 y^{2} g \left (x \right )^{2} h \left (x \right ) b +y f \left (x \right )^{3} h \left (x \right ) a -y g \left (x \right )^{3} h \left (x \right ) b -f \left (x \right )^{3} g \left (x \right ) h \left (x \right ) a +f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) a -f \left (x \right )^{2} g \left (x \right )^{2} h \left (x \right ) b +f \left (x \right ) g \left (x \right )^{3} h \left (x \right ) b -y f \left (x \right ) g \left (x \right )^{2} h \left (x \right ) b -2 y f \left (x \right ) g \left (x \right )^{2} h \left (x \right ) a +y^{2} f \left (x \right ) g \left (x \right ) h \left (x \right ) a -y^{2} f \left (x \right ) g \left (x \right ) h \left (x \right ) b +y f \left (x \right )^{2} g \left (x \right ) h \left (x \right ) a +2 y f \left (x \right )^{2} g \left (x \right ) h \left (x \right ) b}{a f \left (x \right )-g \left (x \right ) a +b f \left (x \right )-b g \left (x \right )} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact trying Abel <- Abel successful`
✓ Solution by Maple
Time used: 0.156 (sec). Leaf size: 648
dsolve(diff(y(x),x)-(y(x)-f(x))*(y(x)-g(x))*(y(x)-(a*f(x)+b*g(x))/(a+b))*h(x)-diff(f(x),x)*(y(x)-g(x))/(f(x)-g(x))-diff(g(x),x)*(y(x)-f(x))/(g(x)-f(x)) = 0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (f \left (x \right )-g \left (x \right )\right ) \left (a +2 b \right ) {\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} \,a^{4}-b^{4} \ln \left (\frac {9 a^{3}+18 a^{2} b +18 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{2 a +b}\right )+\ln \left (\frac {9 a^{2} b +9 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{a -b}\right ) a^{4}+\ln \left (\frac {9 a^{2} b +9 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{a -b}\right ) b^{4}+3 c_{1} a^{3} b +6 c_{1} a^{2} b^{2}+3 c_{1} a \,b^{3}-2 a \,b^{3} \left (\int f \left (x \right ) g \left (x \right ) h \left (x \right )d x \right )-2 a^{3} b \left (\int f \left (x \right ) g \left (x \right ) h \left (x \right )d x \right )-2 a^{2} b^{2} \left (\int f \left (x \right ) g \left (x \right ) h \left (x \right )d x \right )+a^{2} b^{2} \left (\int f \left (x \right )^{2} h \left (x \right )d x \right )+a \,b^{3} \left (\int f \left (x \right )^{2} h \left (x \right )d x \right )+a^{3} b \left (\int f \left (x \right )^{2} h \left (x \right )d x \right )+a^{2} b^{2} \left (\int g \left (x \right )^{2} h \left (x \right )d x \right )+a \,b^{3} \left (\int g \left (x \right )^{2} h \left (x \right )d x \right )+a^{3} b \left (\int g \left (x \right )^{2} h \left (x \right )d x \right )-2 \textit {\_Z} \,a^{3} b -2 \textit {\_Z} \,a^{2} b^{2}-\textit {\_Z} a \,b^{3}+3 \ln \left (\frac {9 a^{2} b +9 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{a -b}\right ) a^{3} b +4 \ln \left (\frac {9 a^{2} b +9 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{a -b}\right ) a^{2} b^{2}+3 \ln \left (\frac {9 a^{2} b +9 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{a -b}\right ) a \,b^{3}-a^{3} b \ln \left (\frac {9 a^{3}+18 a^{2} b +18 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{2 a +b}\right )-2 a^{2} b^{2} \ln \left (\frac {9 a^{3}+18 a^{2} b +18 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{2 a +b}\right )-2 a \,b^{3} \ln \left (\frac {9 a^{3}+18 a^{2} b +18 a \,b^{2}+9 b^{3}+a \,{\mathrm e}^{\textit {\_Z}}+2 b \,{\mathrm e}^{\textit {\_Z}}}{2 a +b}\right )\right )}+9 f \left (x \right ) \left (a +b \right ) \left (a^{2}+b a +b^{2}\right )}{9 a^{3}+18 a^{2} b +18 a \,b^{2}+9 b^{3}} \]
✓ Solution by Mathematica
Time used: 1.124 (sec). Leaf size: 355
DSolve[y'[x]-(y[x]-f[x])*(y[x]-g[x])*(y[x]-(a*f[x]+b*g[x])/(a+b))*h[x]-f'[x]*(y[x]-g[x])/(f[x]-g[x])-g'[x]*(y[x]-f[x])/(g[x]-f[x]) == 0,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [-\frac {1}{3} (a-b)^{2/3} (2 a+b)^{2/3} (a+2 b)^{2/3} \text {RootSum}\left [\text {$\#$1}^3 (a-b)^{2/3} (2 a+b)^{2/3} (a+2 b)^{2/3}-3 \text {$\#$1} a^2-3 \text {$\#$1} a b-3 \text {$\#$1} b^2+(a-b)^{2/3} (2 a+b)^{2/3} (a+2 b)^{2/3}\&,\frac {\log \left (\frac {\frac {-2 a f(x) h(x)-a g(x) h(x)-b f(x) h(x)-2 b g(x) h(x)}{a+b}+3 h(x) y(x)}{\sqrt [3]{\frac {(f(x)-g(x))^3 \left (2 a^3 h(x)^3+3 a^2 b h(x)^3-3 a b^2 h(x)^3-2 b^3 h(x)^3\right )}{(a+b)^3}}}-\text {$\#$1}\right )}{-\text {$\#$1}^2 (a-b)^{2/3} (2 a+b)^{2/3} (a+2 b)^{2/3}+a^2+a b+b^2}\&\right ]=\int _1^x\frac {\left (\frac {(f(K[1])-g(K[1]))^3 \left (2 a^3 h(K[1])^3-2 b^3 h(K[1])^3-3 a b^2 h(K[1])^3+3 a^2 b h(K[1])^3\right )}{(a+b)^3}\right )^{2/3}}{9 h(K[1])}dK[1]+c_1,y(x)\right ] \]