(d^2y/dx^2) = 9y. How do i solve for y(x) ???

Answer 1

Unfortunately, the first two answerers don't know what they're doing. This can be seen by examining their answers:
(a) y=(3/2)y^3 <=> y = 0 or y^2 = 2/3. Well, y = 0 works but if y is constant nonzero then d^2y/dx^2 = 0 but 9y ≠ 0, so that doesn't work.
(b) y=9yx^2/2 <=> y = 0 or x^2 = 2/9. So apart from y=0 this doesn't even give us y as a function of x.

The correct procedure is as follows:

y'' = 9y
=> y'' - 9y = 0: a second-order, linear, constant coefficient, homogeneous ODE.
The auxiliary equation is r^2 - 9 = 0, with solutions r = ± 3. So the general solution to the homogeneous ODE is
y = Ae^(3x) + Be^(-3x)
where A and B are arbitrary constants.

Answer 2

Integrate twice.
First time:
dy/dx=(9/2)y^2
Second time:
y=(9/2)(1/3)y^3=(3/2)y^3

Answer 3

dy / dx = 9 y ² / 2 + k
y ( x ) = 9 y ³ / 6 + k x + c
y ( x ) = 3 y ² / 2 + k x + c

Answer 4

dy/dx = 9 (y^^2/2) = (9/2) y^^2

where y^^2 is y squared

y = (9/2) (y^^3/3) = (9/6) y^^3

Answer 5

integrate the equation twice
on first integration
we get
dy/dx=9yx

and again on integration
we get
y(x)=9yx^2/2