Given y'' = y^2, where y'' = (d^2y / dx^2) then, integrate once, get y' = y^3 / 3 ...

[suman_b]

Is this correct? Given y'' = y^2, where y'' = (d^2y / dx^2) then, integrate once, get y' = y^3 / 3 + C1 and y = y^4 / 12 + C2

 

[Ishuda]

Is this correct? Given y'' = y^2, where y'' = (d^2y / dx^2) then, integrate once, get y' = y^3 / 3 + C1 and y = y^4 / 12 + C2

That would be the correct process for y'' = x³. Lets look at that first integration of yours:
y' = (1/3) y³
if we had this then
y'' = dy' / dx = d[(1/3) y³ ] = y² (dy/dx) = y² y'
which is not what you started with.

EDIT: Of course, I meant it would be correct as corrected by Subhotosh Khan

 

[Deleted member 4993]

Is this correct? Given y'' = y^2, where y'' = (d^2y / dx^2) then, integrate once, get y' = y^3 / 3 + C1 and y = y^4 / 12 + C2

This is a horribly non-linear ODE. The solution involves elliptic functions.

http://www.wolframalpha.com/input/?i=y"+-+y^2+=+0