Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation?

share|cite|improve this question
could you integrate with respect to y? –  Ben Mar 22 '12 at 9:13
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. –  doraemonpaul Sep 10 '12 at 1:47

2 Answers 2

Honestly, I don't think the solutions of your ODE can be written in elementary terms.

Actually, any substitution of the type $u=y^\alpha$ won't simplify the ODE, because of that evil constant term $-1$.

Neverthless, you could look for a power series solution of your ODE using the Frobenius method, that is:

  • assume that a solution of your ODE can be expanded in a power series $\sum_{n=0}^\infty a_n\ x^n$,
  • evaluate the power series expansion of $y^3(x)$ and $y^\prime (x)$ and plug them into the ODE,
  • deduce from the ODE a recurrence relation for the coefficients $a_n$,
  • try to prove that the radius of convergence of $\sum_{n=0}^\infty a_n\ x^n$ is $>0$;

then the sum $y(x):=\sum_{n=0}^\infty a_n\ x^n$ will be an analytic solution of your ODE.

It is easy to prove that if $y(x)=\sum_{n=0}^\infty a_n\ x^n$ then:

  1. $y^3(x) = \sum_{n=0}^\infty b_n\ x^n$, where $b_n:=\sum_{k=0}^n \sum_{h=0}^{n-k} a_k\ a_h\ a_{n-k-h}$ satisfies: $$\begin{cases} b_0=a_0^3\\ b_n = \frac{1}{n\ a_0}\ \sum_{k=1}^{n} (4k-n)a_k\ b_{n-k} \end{cases}$$
  2. $y^\prime (x) = \sum_{n=0}^\infty (n+1)\ a_{n+1}\ x^n$,

therefore plugging 1 and 2 into your ODE gives: $$a_1+2a_2\ x+3a_3\ x^2 + \sum_{n=3}^\infty (n+1)\ a_{n+1}\ x^n = -1 + \sum_{n=3}^\infty b_{n-3} x^n\; .$$ Equating the coefficients of like powers of $x$, you obtain: $$\begin{cases} a_1=-1\\ a_2=0\\ a_3=0\\ (n+1)\ a_{n+1} = b_{n-3} &\text{, for } n\geq 3 \end{cases}$$ i.e.: $$\tag{1} \begin{cases} a_1=-1\\ a_2=0\\ a_3=0\\ a_{n+1} = \frac{1}{n+1}\ \sum_{k=0}^{n-3} \sum_{h=0}^{n-3-k} a_k\ a_h\ a_{n-3-k-h} &\text{, for } n\geq 3. \end{cases}$$ Note that $a_0$ cannot be determined using (1).

Now there remains to be solved the problem of finding the radius of convergence of the power series whose coefficients are given by (1).

share|cite|improve this answer

You should highly notice that Bernoulli differental equation is of the form $y'=f(x)y^n+g(x)y$ rather than of the form $y'=f(x)y^n+g(x)$.

In fact $y'=x^3y^3-1$ belongs to an Abel equation of the first kind. To find its exact solution, please refer to

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.