We know that $$g'(x)=g(x)$$ $g(0)=1$ Then $g(x)=e^x$
If we want to solve $y'=ay$,
We can write general solution $y'=ay$ (or if derivative both sides $y''=a^2y$) $$y=k.g(a x)=k.e^{ax}$$ where $k$ is a constant.
We can express particular solution of many differential equation via using $e^{\beta x}$ but some we cannot. For example, we can find a particular solution for
$y'=1+y^2$ (if we apply derivative both sides then we get $y''=2y^3+2y$)
and particular solution for $y''=2y^3+2y$ can be written: $$y=\tan(x)=-i\frac{g(ix)-g(-ix)}{g(ix)+g(-ix)}=-i\frac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$$
but as I have known that we cannot solve the general differential equation ,$y''(x)=ay^3(x)+by^2(x)+cy(x)+d$ (where a,b,c are constants) via only using $e^{\beta x}$ . We need to define some non-elementery functions (elliptic functions) to find the solution. I thought to select an elliptic function to find a particular solution of general differential equation ,$$y''(x)=ay^3(x)+by^2(x)+cy(x)+d$$ by using the selected elliptic function and elementary function $e^{\beta x}$ (my aim to find a closed form with limited terms in combination of both functions or only with the selected elliptic function).
Let's select an elliptic function and try to use it a base function for this aim.
$$f''(x)=-2f^3(x)$$ $f(0)=0$
$f'(0)=1$
$$(f'(x))^2=1-f^4(x)$$ The defined base function can be expressed $$x= \int_{0}^{f(x)} \frac{du}{\sqrt{1-u^4}}$$
Euler found an addition formula for this selected $f(x)$. The formula at page 6 in article that was written by Jose Barrios
$$f(x+y)=\frac{f(x)\sqrt{1-f^4(y)}+f(y)\sqrt{1-f^4(x)}}{1+f^2(x)f^2(y)}$$
or we can rewrite it as:
$$f(x+y)=\frac{f(x)f'(y)+f(y)f'(x)}{1+f^2(x)f^2(y)}$$
Is it possible to find a particular solution of the general differential equation $y''=ay^3+by^2+cy+d$ by using the combination of the selected elliptic function $f(x)$ and $e^{x}$ in a closed form ?
Note:Maybe we need to select another base $f(x)$ that would be easier for this aim. I selected randomly.
I need references and hints about this idea. Thanks a lot for your helps
EDIT: 15th August ,2016
I would like to share my results:
Let's assume that we have the equation:
$$ y'=\frac{dy}{dx} = p (y^2 + my + n)$$
If we solve the differential equation:
General solution of the differential equation can be written as
$y=-\frac{m}{2}-\frac{\sqrt{m^2-4n}}{2}+\cfrac{\sqrt{m^2-4n}}{1+ke^{xp\sqrt{m^2-4n}}}$
where $k$ is a constant.
If we derivative both side $$ y'=\frac{dy}{dx} = p (y^2 + my + n)$$ $$ y''=\frac{d^2y}{dx^2} = p (2y + m)y'$$ $$ y''= p^2 (2y + m)(y^2 + my + n)$$
$$ y''= 2p^2y^3+3p^2my^2+p^2(m^2+2n)y+p^2mn$$
If we compare with our general equation , We can find a condition that it has solution via $e^{\beta x}$.
$$y''= ay^3+by^2+cy+d$$
The condition: $$2b^3+27a^2d=9abc$$
If the general equation satisfy this condition , the general equation has a solution with $e^{\beta x}$ but if the condition is not satisfy , we should express the solution with elliptic functions. It is a known fact that elliptic functions cannot be expressed by $e^{\beta x}$ in closed form.
$p=\sqrt{a/2}$ ; $m=\frac{2b}{3a}$;$n=\frac{3d}{b}$
$y_p(x)=-\frac{m}{2}-\frac{\sqrt{m^2-4n}}{2}+\cfrac{\sqrt{m^2-4n}}{1+ke^{xp\sqrt{m^2-4n}}}$ where $k$ is a constant.
I have noticed that we can express the solvable condition as:
$y_p=A\cfrac{e^{\alpha x} +B}{e^{\alpha x} +C}$
Let's transform the general diff equation
$y=A\cfrac{u +B}{u +C}$ and to check a condition for u
$$y'=A\cfrac{C-B}{(u +C)^2}u'$$
$$y''= ay^3+by^2+cy+d$$ If we integrate both side, we get: $$ \frac12 \left( \frac{dy}{dx} \right)^2 = \frac{a}{4} y^4+\frac{b}{3} y^3+\frac{c}{2} y^2+d y + e$$ where $e$ is a constant.
$$ \frac{A^2(C-B)^2}{2(u +C)^4} u'^2 = \frac{a}{4}\frac{(u +B)^4}{(u +C)^4}+\frac{b}{3} \frac{(u +B)^3}{(u +C)^3}+\frac{c}{2} \frac{(u +B)^2}{(u +C)^2}+d \frac{(u +B)}{(u +C)} + e$$
$$ \frac{A^2(C-B)^2}{2} u'^2= \frac{a}{4}(u +B)^4+\frac{b}{3} (u +B)^3(u +C)+\frac{c}{2} (u +B)^2(u +C)^2+d (u +B)(u +C)^3 + e(u +C)^4$$
I have been looking for a condition if we can transform it to one equation such as
$$ u'^2= a_4u^4+a_3u^3+a_2u^2+a_1u+a_0$$
where $a_n$ is any selected constant.
In my question , I selected to convert it to
$$ u'^2= s(1-u^4)$$
but I am not sure if we can transform the general equation into this type.
Can anybody prove that it is possible or not to convert it to a selected form such as $ u'^2= a_4u^4+a_3u^3+a_2u^2+a_1u+a_0$? (where $a_n$ are selected constants.)
EDIT:7th Sep 2016
$$ \frac{A^2(C-B)^2}{2} u'^2= \frac{a}{4}(u +B)^4+\frac{b}{3} (u +B)^3(u +C)+\frac{c}{2} (u +B)^2(u +C)^2+d (u +B)(u +C)^3 + e(u +C)^4$$
If We select $e=-(\frac{a}{4}+\frac{b}{3}+\frac{c}{2}+d)$ then $u^4$ term is canceled so we can get 3 degree polynomial form.
$$ \frac{A^2(C-B)^2}{2} u'^2= b_3 u^3+b_2u^2+b_1u+b_0$$
We can do $x=\alpha t$ then we can get the form,
$$ u'^2= u^3+k_2(A,B,C)u^2+k_1(A,B,C)u+k_0(A,B,C)$$
We still have 3 parameters ($A,B,C$) to select . We can transform into a selected form $ u'^2= u^3+a_2u^2+a_1u+a_0$. (where $a_n$ are selected constants.)
Can anybody help me to show if the transform to a selected form $ u'^2= u^3+a_2u^2+a_1u+a_0$ is possible or not? Thanks a lot for helps
In general you need to go into elliptic integrals. Set $v=\frac{dy}{dx}$ so that (I add a constant as well to the ode): $$ y'' = \frac{dv}{dx} = \frac{dv}{dy} \; \frac{dy}{dx} = v \frac{dv}{dy} = ay^3 + by^2 + cy+d$$ In other words, $$ v dv = (a y^3 + b y^2 + cy) dy$$ Thus, $$ \frac12 \left( \frac{dy}{dx} \right)^2 = \frac12 v^2 = \frac{a}{4} y^4+\frac{b}{4} y^3+\frac{c}{2} y^2+d y + e$$ from which taking square root etc you derive the usual elliptic integrals for $x$: $$ x(y) = \int \frac{dy}{\sqrt {a/2\; y^4 + 2b/3\; y^3 + c \; y^2 + 2dy +2e}}$$ (perhaps use another normalization of constants?). Does this correspond to what you had in mind?