60.1.532 problem 545
Internal
problem
ID
[10546]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
545
Date
solved
:
Sunday, March 30, 2025 at 05:59:09 PM
CAS
classification
:
[_quadrature]
\begin{align*} {y^{\prime }}^{4}-\left (y-a \right )^{3} \left (y-b \right )^{2}&=0 \end{align*}
✓ Maple. Time used: 0.042 (sec). Leaf size: 131
ode:=diff(y(x),x)^4-(y(x)-a)^3*(y(x)-b)^2 = 0;
dsolve(ode,y(x), singsol=all);
\begin{align*}
y &= a \\
y &= b \\
x -\int _{}^{y}\frac {1}{\left (\left (\textit {\_a} -a \right )^{3} \left (\textit {\_a} -b \right )^{2}\right )^{{1}/{4}}}d \textit {\_a} -c_1 &= 0 \\
x -i \int _{}^{y}\frac {1}{\left (\left (\textit {\_a} -a \right )^{3} \left (\textit {\_a} -b \right )^{2}\right )^{{1}/{4}}}d \textit {\_a} -c_1 &= 0 \\
x +i \int _{}^{y}\frac {1}{\left (\left (\textit {\_a} -a \right )^{3} \left (\textit {\_a} -b \right )^{2}\right )^{{1}/{4}}}d \textit {\_a} -c_1 &= 0 \\
x +\int _{}^{y}\frac {1}{\left (\left (\textit {\_a} -a \right )^{3} \left (\textit {\_a} -b \right )^{2}\right )^{{1}/{4}}}d \textit {\_a} -c_1 &= 0 \\
\end{align*}
✓ Mathematica. Time used: 1.109 (sec). Leaf size: 333
ode=-((-a + y[x])^3*(-b + y[x])^2) + D[y[x],x]^4==0;
ic={};
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*}
y(x)\to \text {InverseFunction}\left [-\frac {4 \sqrt [4]{a-\text {$\#$1}} \sqrt {\frac {\text {$\#$1}-b}{a-b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {a-\text {$\#$1}}{a-b}\right )}{\sqrt {b-\text {$\#$1}}}\&\right ]\left [-\sqrt [4]{-1} x+c_1\right ] \\
y(x)\to \text {InverseFunction}\left [-\frac {4 \sqrt [4]{a-\text {$\#$1}} \sqrt {\frac {\text {$\#$1}-b}{a-b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {a-\text {$\#$1}}{a-b}\right )}{\sqrt {b-\text {$\#$1}}}\&\right ]\left [\sqrt [4]{-1} x+c_1\right ] \\
y(x)\to \text {InverseFunction}\left [-\frac {4 \sqrt [4]{a-\text {$\#$1}} \sqrt {\frac {\text {$\#$1}-b}{a-b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {a-\text {$\#$1}}{a-b}\right )}{\sqrt {b-\text {$\#$1}}}\&\right ]\left [-(-1)^{3/4} x+c_1\right ] \\
y(x)\to \text {InverseFunction}\left [-\frac {4 \sqrt [4]{a-\text {$\#$1}} \sqrt {\frac {\text {$\#$1}-b}{a-b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {a-\text {$\#$1}}{a-b}\right )}{\sqrt {b-\text {$\#$1}}}\&\right ]\left [(-1)^{3/4} x+c_1\right ] \\
y(x)\to a \\
y(x)\to b \\
\end{align*}
✓ Sympy. Time used: 26.635 (sec). Leaf size: 218
from sympy import *
x = symbols("x")
a = symbols("a")
b = symbols("b")
y = Function("y")
ode = Eq(-(-a + y(x))**3*(-b + y(x))**2 + Derivative(y(x), x)**4,0)
ics = {}
dsolve(ode,func=y(x),ics=ics)
\[
\left [ - \frac {2 i \sqrt {- b + y{\left (x \right )}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {\left (- b + y{\left (x \right )}\right ) e^{2 i \pi }}{\operatorname {polar\_lift}{\left (a - b \right )}}} \right )}}{\operatorname {polar\_lift}^{\frac {3}{4}}{\left (a - b \right )}} = C_{1} - \sqrt [4]{-1} x, \ - \frac {2 i \sqrt {- b + y{\left (x \right )}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {\left (- b + y{\left (x \right )}\right ) e^{2 i \pi }}{\operatorname {polar\_lift}{\left (a - b \right )}}} \right )}}{\operatorname {polar\_lift}^{\frac {3}{4}}{\left (a - b \right )}} = C_{1} + \sqrt [4]{-1} x, \ - \frac {2 i \sqrt {- b + y{\left (x \right )}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {\left (- b + y{\left (x \right )}\right ) e^{2 i \pi }}{\operatorname {polar\_lift}{\left (a - b \right )}}} \right )}}{\operatorname {polar\_lift}^{\frac {3}{4}}{\left (a - b \right )}} = C_{1} - \left (-1\right )^{\frac {3}{4}} x, \ - \frac {2 i \sqrt {- b + y{\left (x \right )}} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {\left (- b + y{\left (x \right )}\right ) e^{2 i \pi }}{\operatorname {polar\_lift}{\left (a - b \right )}}} \right )}}{\operatorname {polar\_lift}^{\frac {3}{4}}{\left (a - b \right )}} = C_{1} + \left (-1\right )^{\frac {3}{4}} x\right ]
\]