## Non linear ODE: y'' = a y + b y^3

I would like to solve the non linear ODE
$\frac{d}{dx}f(x)=a f(x)+ b f^3 (x)$
with the boundary
$f(0)=0\quad f(+\infty)=f_0$

How to find analitical solution?
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 If it is f ' = a f+b f^3 then integrate dx = df/(a f+b f^3) If it is y'' = a y+b y^3 then the first step is the integration of : y'' y' = (a y+b y^3) y' y'² = a y² + b (y^4)/2 +c dx = dy/sqrt(a y² + b (y^4)/2 +c) The integration will lead to x(y) on the form of elliptic functions. Very complicated to invert in order to obtain y(x)
 Pardon me, I write the uncorrect differential equation: the problem is at the second order $\frac{d^2}{dx^2}f(x)=a f(x) + b f^3 (x)$ with the boundary $f(0)=0,\ f(+\infty)=f_0$

## Non linear ODE: y'' = a y + b y^3

In my opinion, $x(y)=g(y,c_1,c_2)$ is an analytical expression for the solution but I think first, just scrap the a and b and look at:

$$y''=y+y^3$$

then do what Jacq said and get the expression in terms of:

$$x(y)=g(y,c_1,c_2)$$

then try and solve simultaneously the expressions:

$$0=g(0,c_1,c_2)$$

$$g(f_0,c_1,c_2)\to+\infty$$

for the constants $c_1,c_2$ and if need be, do so numerically for them just for starters.
 Hello ! It is a difficult problem on the analytical viewpoint. A difficulty is that elliptic functions are involved. But the major difficulty is due to the condition f(x=+infinity)=finite constant. I am not quite sure that the solution given in attachment is a correct answer to the problem, so it should be carefully verified. Attached Thumbnails
 In case of some complementary conditions specified in attachment, y(x) can be fully explicited thanks to a Jacobi elliptic function. Attached Thumbnails
 Nicely done Jacquelin and beautiful too. Thanks.
 Thank you so much!