I have an ODE $$ \frac{d y(x)}{dx}=\frac{a*y(x)^3+b}{c* y(x)^3+ d},$$ where $a,b,c,d$ are constants, and I would love to solve it analytically.
I tried Maple 15 and 17 and got $$ y \left( x \right) =1/2\,{ \left( {\frac {a}{d}} \right) }^{1/3}+1/2\, {3}^{1/2}{ \left( {\frac {a}{d}} \right) }^{1/3}{\it tan} \left( {\it RootOf} \left( 6\,{3}^{1/2}{ \left( {\frac {a}{d}} \right) }^{2/3}{ \it \_C1}\,{d}^{2}+6\,{3}^{1/2}{ \left( {\frac {a}{d}} \right) }^{2/3} {d}^{2}x+2\,{3}^{1/2}\ln \left( 3/2\,{ \left( {\frac {a}{d}} \right) }^{1/3}+1/2\,{3}^{1/2}{ \left( {\frac {a}{d}} \right) }^{1/3}{\it tan} \left( {\it \_Z} \right) \right) ab-2\,{3}^{1/2}\ln \left( 3/2\,{ \left( {\frac {a}{d}} \right) }^{1/3}+1/2\,{3}^{1/2}{ \left( {\frac { a}{d}} \right) }^{1/3}{\it tan} \left( {\it \_Z} \right) \right) cd-{ 3}^{1/2}\ln \left( 3/4\,{ \left( {\frac {a}{d}} \right) }^{2/3}+3/4\, { \left( {\frac {a}{d}} \right) }^{2/3}{{\it tan} \left( {\it \_Z} \right) }^{2} \right) ab+{3}^{1/2}\ln \left( 3/4\,{ \left( {\frac {a }{d}} \right) }^{2/3}+3/4\,{ \left( {\frac {a}{d}} \right) }^{2/3}{{ \it tan} \left( {\it \_Z} \right) }^{2} \right) cd-3\,{3}^{1/2}ab-9\,{ \it tan} \left( {\it \_Z} \right) ab+6\,{\it \_Z}\,ab-6\,{\it \_Z}\,cd \right) \right) $$
I tried Wolfram|Alpha ando got this: for DSolve[y'[x]=(a*y[x]^3 + b)/(d*y[x]^3+c), y, x]
and this for DSolve[y'[x]=(a*y[x]^3 + b)/(d*y[x]^3+c)]:
And so I wonder if there is any way to solve it analytically and get $y(x)$in general form not dependent on $y(x)$? Any special function? Or any package that could solve such ODEs analytically?
As mathematica already discovered for you, you will always have the equation of $$x = \int \frac{c + dy^3}{b + ay^3} dy + C.$$ Solving this integral is possible given $a,b,c,d$, but the results differ wildly depending on the constants, so no overall solution is possible. The function $y$ you are looking for is then the inverse of the integral you calculate.