Internal problem ID [8099]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 519.
ODE order: 1.
ODE degree: 3.
CAS Maple gives this as type [[_1st_order, _with_symmetry_[F(x),G(x)*y+H(x)]]]
Solve \begin {gather*} \boxed {\left (y^{\prime }\right )^{3}-f \relax (x ) \left (a y^{2}+b y+c \right )^{2}=0} \end {gather*}
✓ Solution by Maple
Time used: 0.167 (sec). Leaf size: 197
dsolve(diff(y(x),x)^3-f(x)*(a*y(x)^2+b*y(x)+c)^2=0,y(x), singsol=all)
\begin{align*} \int _{}^{y \relax (x )}\frac {1}{\left (a \,\textit {\_a}^{2}+\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} +\int _{}^{x}-\frac {\left (f \left (\textit {\_a} \right ) \left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{2}\right )^{\frac {1}{3}}}{\left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{\frac {2}{3}}}d \textit {\_a} +c_{1} = 0 \\ \int _{}^{y \relax (x )}\frac {1}{\left (a \,\textit {\_a}^{2}+\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} +\int _{}^{x}\frac {\left (f \left (\textit {\_a} \right ) \left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{2}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{2 \left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{\frac {2}{3}}}d \textit {\_a} +c_{1} = 0 \\ \int _{}^{y \relax (x )}\frac {1}{\left (a \,\textit {\_a}^{2}+\textit {\_a} b +c \right )^{\frac {2}{3}}}d \textit {\_a} +\int _{}^{x}-\frac {\left (f \left (\textit {\_a} \right ) \left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{2}\right )^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}{2 \left (a y \relax (x )^{2}+b y \relax (x )+c \right )^{\frac {2}{3}}}d \textit {\_a} +c_{1} = 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 62.21 (sec). Leaf size: 405
DSolve[-(f[x]*(c + b*y[x] + a*y[x]^2)^2) + y'[x]^3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(2 \text {$\#$1} a+b)^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x\sqrt [3]{f(K[1])}dK[1]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(2 \text {$\#$1} a+b)^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x-\sqrt [3]{-1} \sqrt [3]{f(K[2])}dK[2]+c_1\right ] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt [3]{2} (2 \text {$\#$1} a+b) \left (\frac {a (\text {$\#$1} (\text {$\#$1} a+b)+c)}{4 a c-b^2}\right )^{2/3} \text {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},\frac {(2 \text {$\#$1} a+b)^2}{b^2-4 a c}\right )}{a (\text {$\#$1} (\text {$\#$1} a+b)+c)^{2/3}}\&\right ]\left [\int _1^x(-1)^{2/3} \sqrt [3]{f(K[3])}dK[3]+c_1\right ] \\ y(x)\to -\frac {\sqrt {b^2-4 a c}+b}{2 a} \\ y(x)\to \frac {\sqrt {b^2-4 a c}-b}{2 a} \\ \end{align*}