4.21.25 \(y'(x)^3=(y(x)-a)^2 (y(x)-b)^2\)

ODE
\[ y'(x)^3=(y(x)-a)^2 (y(x)-b)^2 \] ODE Classification

[_quadrature]

Book solution method
Missing Variables ODE, Independent variable missing, Solve for \(y'\)

Mathematica
cpu = 0.721639 (sec), leaf count = 236

\[\left \{\left \{y(x)\to \text {InverseFunction}\left [-\frac {3 \sqrt [3]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\& \right ][x+c_1]\right \},\left \{y(x)\to \text {InverseFunction}\left [-\frac {3 \sqrt [3]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\& \right ]\left [-\sqrt [3]{-1} x+c_1\right ]\right \},\left \{y(x)\to \text {InverseFunction}\left [-\frac {3 \sqrt [3]{a-\text {$\#$1}} \left (\frac {\text {$\#$1}-b}{a-b}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {a-\text {$\#$1}}{a-b}\right )}{(b-\text {$\#$1})^{2/3}}\& \right ]\left [(-1)^{2/3} x+c_1\right ]\right \}\right \}\]

Maple
cpu = 1.102 (sec), leaf count = 126

\[\left [y \left (x \right ) = a, y \left (x \right ) = b, x -\left (\int _{}^{y \left (x \right )}\frac {1}{\left (\left (\textit {\_a} -a \right )^{2} \left (\textit {\_a} -b \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} \right )-\textit {\_C1} = 0, x -\left (\int _{}^{y \left (x \right )}-\frac {2}{\left (1+i \sqrt {3}\right ) \left (\left (\textit {\_a} -a \right )^{2} \left (\textit {\_a} -b \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} \right )-\textit {\_C1} = 0, x -\left (\int _{}^{y \left (x \right )}\frac {2}{\left (-1+i \sqrt {3}\right ) \left (\left (\textit {\_a} -a \right )^{2} \left (\textit {\_a} -b \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} \right )-\textit {\_C1} = 0\right ]\] Mathematica raw input

DSolve[y'[x]^3 == (-a + y[x])^2*(-b + y[x])^2,y[x],x]

Mathematica raw output

{{y[x] -> InverseFunction[(-3*Hypergeometric2F1[1/3, 2/3, 4/3, (a - #1)/(a - b)]
*(a - #1)^(1/3)*((-b + #1)/(a - b))^(2/3))/(b - #1)^(2/3) & ][x + C[1]]}, {y[x] 
-> InverseFunction[(-3*Hypergeometric2F1[1/3, 2/3, 4/3, (a - #1)/(a - b)]*(a - #
1)^(1/3)*((-b + #1)/(a - b))^(2/3))/(b - #1)^(2/3) & ][-((-1)^(1/3)*x) + C[1]]},
 {y[x] -> InverseFunction[(-3*Hypergeometric2F1[1/3, 2/3, 4/3, (a - #1)/(a - b)]
*(a - #1)^(1/3)*((-b + #1)/(a - b))^(2/3))/(b - #1)^(2/3) & ][(-1)^(2/3)*x + C[1
]]}}

Maple raw input

dsolve(diff(y(x),x)^3 = (y(x)-a)^2*(y(x)-b)^2, y(x))

Maple raw output

[y(x) = a, y(x) = b, x-Intat(1/((_a-a)^2*(_a-b)^2)^(1/3),_a = y(x))-_C1 = 0, x-I
ntat(-2/(1+I*3^(1/2))/((_a-a)^2*(_a-b)^2)^(1/3),_a = y(x))-_C1 = 0, x-Intat(2/(-
1+I*3^(1/2))/((_a-a)^2*(_a-b)^2)^(1/3),_a = y(x))-_C1 = 0]