nonlinear first order differential equation 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?
$y'=x^{3}y^{3}-1\\$
 A: 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:


*

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

*$y^\prime (x) = \sum_{n=0}^\infty (n+1)\ a_{n+1}\ x^n$,


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).
A: 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 $y'=f(x)y^n+g(x)$.
Approach $1$:
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.
Approach $2$:
Let $u=xy$ ,
Then $y=\dfrac{u}{x}$
$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$
$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=u^3-1$
$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{u}{x^2}+u^3-1$
Let $v=\dfrac{1}{x^2}$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=-\dfrac{2}{x^3}\dfrac{du}{dv}$
$\therefore-\dfrac{2}{x^4}\dfrac{du}{dv}=\dfrac{u}{x^2}+u^3-1$
$(uv+u^3-1)\dfrac{dv}{du}=-2v^2$
This belongs to an Abel equation of the second kind.
