Can this equation be solved analytically?

[sccv]

Dear all,
I have this differential equation:

dy
---- = K*(A * y^3 + B * y^2 + C * y + 1)
dx

where A, B, C, K are non-zero constants

I tried to solve it analytically but achieve no result so far. Could anyone help me to solve this equation ?

 

[ReyChiquito]

you can separate your equation

\frac{dy}{Ay^3+By^2+Cy+1}=dx

all you have to do is integrate both sides

\int_{y(x_0)}^{y(x)} \frac{d\xi}{A\xi^3+B\xi^2+C\xi+1}=K(x-x_0)

 

[Tide]

For arbitrary values of A, B and C you will not get a nice analytic solution. Even if you could factor the polynomial, you would get three different logarithmic terms (if the factors are distinct) and you would not be able to solve for y = y(x).

 

[Tide]

P.S. What Chiquito has shown you is sometimes referred to as "reducing to quadratures" and is useful when all else fails!

 

[sccv]

You are right, Tide. That is why I have to find another way to solve for y = y(x)

 

[ReyChiquito]

Yup, in principle, you can find the primitive of the integral above, and that will give you an implicit solution for your ode, but to find a explicit solution? i doubt youll be able to, at least with separation of variables.

 

[saltydog]

The cusp catastrophe

Nice! It's the cubic differential equation, the canonical form of the cusp catastrophe. Are you aware of this? Know what "shocks" are in regards to dynamics?

SD

 

[Hurkyl]

For arbitrary values of A, B and C you will not get a nice analytic solution. Even if you could factor the polynomial, you would get three different logarithmic terms (if the factors are distinct) and you would not be able to solve for y = y(x).

You sure? You can always factor the polynomial (over C, of course), and while you get three logarithmic terms, once you exponentiate both sides, you're left with a polynomial in y.

The only problem is when the polynomial has a double root. (A triple root is fine)