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Hello science companions ;)

I was wondering for a while how to solve this ODE (y''=f(y)) and I finally found one way to integrate it. I don't know actually if it is correct but as I see them, all steps are mathematically consistent. Here I show how I came to the general solution (for all steps consider y=y(x):

-1st multiply each side of the ODE by 2y'

2·y'·y''=2·f(y)·y'

-2nd using (y'^{2})'=2·y'·y'' write

(y'^{2})'=2·f(y)·y'

-3rd first both-side-with-respect-to-x integration

[tex]\int[/tex](y'^{2})'dx=2[tex]\int[/tex]f(y)y'dx+A (1st integration costant)

-4th cancelling derivative with integral and using dy=y'dx

y'^{2}=2[tex]\int[/tex]f(y)dy+A

-5th the integral is the primitive of f(y) so we call it F(y), and aplying squareroots to both sides we finally have:

y'=[tex]\pm[/tex][tex]\sqrt{2F(y)+A}[/tex]

These are two ecuations that only differ in sign and represent two "families of families" of solutions. In any of them the equation is separable as we have the first derivative equal to an only function of y, so reorganizing we may write:

[tex]\frac{1}{\sqrt{2F(y)+A}}·y'=\pm1[/tex]

Then integration as usual which gives, using again dy=y'dx:

[tex]\int\frac{1}{\sqrt{2F(y)+A}}dy=\pm x+B[/tex] (2nd integration constant)

What represents the general set of solutions if I wasn't mistaken. Please any comments, corrections, advise... will be more than welcome. Hope I made myself clear enough ;)

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# Opinion about y =f(y)

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