29.34.16 problem 1018
Internal
problem
ID
[5587]
Book
:
Ordinary
differential
equations
and
their
solutions.
By
George
Moseley
Murphy.
1960
Section
:
Various
34
Problem
number
:
1018
Date
solved
:
Sunday, March 30, 2025 at 09:05:06 AM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\begin{align*} {y^{\prime }}^{3}&=\left (a +b y+c y^{2}\right ) f \left (x \right ) \end{align*}
✓ Maple. Time used: 0.132 (sec). Leaf size: 189
ode:=diff(y(x),x)^3 = (a+b*y(x)+c*y(x)^2)*f(x);
dsolve(ode,y(x), singsol=all);
\begin{align*}
\int _{}^{y}\frac {1}{\left (\textit {\_a}^{2} c +\textit {\_a} b +a \right )^{{1}/{3}}}d \textit {\_a} -\frac {\int _{}^{x}{\left (\left (a +b y+c y^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a}}{\left (a +b y+c y^{2}\right )^{{1}/{3}}}+c_1 &= 0 \\
\int _{}^{y}\frac {1}{\left (\textit {\_a}^{2} c +\textit {\_a} b +a \right )^{{1}/{3}}}d \textit {\_a} +\frac {\left (1+i \sqrt {3}\right ) \int _{}^{x}{\left (\left (a +b y+c y^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a}}{2 \left (a +b y+c y^{2}\right )^{{1}/{3}}}+c_1 &= 0 \\
\int _{}^{y}\frac {1}{\left (\textit {\_a}^{2} c +\textit {\_a} b +a \right )^{{1}/{3}}}d \textit {\_a} -\frac {\left (i \sqrt {3}-1\right ) \int _{}^{x}{\left (\left (a +b y+c y^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a}}{2 \left (a +b y+c y^{2}\right )^{{1}/{3}}}+c_1 &= 0 \\
\end{align*}
✓ Mathematica. Time used: 14.676 (sec). Leaf size: 405
ode=(D[y[x],x])^3 ==(a+b y[x]+c y[x]^2) f[x];
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\&\right ]\left [\int _1^x\sqrt [3]{f(K[1])}dK[1]+c_1\right ] \\
y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\&\right ]\left [\int _1^x-\sqrt [3]{-1} \sqrt [3]{f(K[2])}dK[2]+c_1\right ] \\
y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {3}{2},\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\&\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 c} \\
y(x)\to \frac {\sqrt {b^2-4 a c}-b}{2 c} \\
\end{align*}
✓ Sympy. Time used: 54.407 (sec). Leaf size: 372
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
c = symbols("c")
y = Function("y")
f = Function("f")
ode = Eq((-a - b*y(x) - c*y(x)**2)*f(x) + Derivative(y(x), x)**3,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
\text {Solution too large to show}
\]