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
:
Monday, January 27, 2025 at 12:12:52 PM
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*}
✓ Solution by Maple
Time used: 0.151 (sec). Leaf size: 189
dsolve(diff(y(x),x)^3 = (a+b*y(x)+c*y(x)^2)*f(x),y(x), singsol=all)
\begin{align*}
\int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{{1}/{3}}}d \textit {\_a} -\frac {\int _{}^{x}{\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a}}{\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{{1}/{3}}}+c_{1} &= 0 \\
\int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{{1}/{3}}}d \textit {\_a} +\frac {\left (1+i \sqrt {3}\right ) \left (\int _{}^{x}{\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a} \right )}{2 \left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{{1}/{3}}}+c_{1} &= 0 \\
\int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{{1}/{3}}}d \textit {\_a} -\frac {\left (i \sqrt {3}-1\right ) \left (\int _{}^{x}{\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )}^{{1}/{3}}d \textit {\_a} \right )}{2 \left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{{1}/{3}}}+c_{1} &= 0 \\
\end{align*}
✓ Solution by Mathematica
Time used: 14.676 (sec). Leaf size: 405
DSolve[(D[y[x],x])^3 ==(a+b y[x]+c y[x]^2) f[x],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*}