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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?
$y'=x^{3}y^{3}-1\\$

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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
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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).

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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 http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.

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