memarf1
Im trying to turn this equation into 2 seperate equations in order to place it in a runge kutta problem. This is the proposed problem and conditions:

$$\frac{d^2f}{dx^2} + f = 0$$

allowing

$$f (x) = A\cos x + B\sin x$$
$$f ' (x) = -A\sin x + B\cos x$$
$$f '' (x) = -A\cos x - B\sin x$$

and
$$g = \frac{df}{dx}$$

meaning
$$\frac{df}{dx} - g = 0$$ which is identical to $$\frac{d^2f}{dx^2} + f = 0$$

so
$$\frac{dg}{dx} + f = 0$$

the initial conditions for equation 1 are:
$$f (0) = 1$$
$$f ' (0) = 0$$

and for equation 2 are:
$$f (0) = 0$$
$$g (0) = 1$$

I hope this formatting is more easy to read.
any suggestions??

Last edited:

Homework Helper
memarf1 said:
Im trying to turn this equation into 2 seperate equations in order to place it in a runge kutta problem. This is the proposed problem and conditions:

d''f______________________f (x) = Acosx + Bsinx
--- + f = 0______________f ' (x) = -Asinx + Bcosx
dx''______________________f '' (x) = -Acosx - Bsinx
____df_________df
g = ----_______---- - g = 0 is identical to d''f
____dx_________dx____________________---- + f = 0
_____________________________________dx''
so
___dg
__---- + f = 0______________if f(0) = 1____and if____f(0) = 0
___dx_______________________f '(0) = 0____________g(0) = 1

My organization may be confusing, ignore long underscore lines, and some of the stuff is organized up and down instead of left and right.
any suggestions??

Right, you really need to learn Latex. So your post I tjink would go like this:

$$\frac{d''f}{dx''} + f = 0$$

Therefore:

$$f (x) = A\cos x + B\sin x$$
$$f ' (x) = -A\sin x + B\cos x$$
$$f '' (x) = -A\cos x - B\sin x$$

However, before I try to translate the rest, I feel it worth noting that this is very confusing:

$$\frac{d''f}{dx''}$$

Please stick to something like this:

$$\frac{d^{2}y}{dx^{2}} \quad \text{or} \quad y''$$

memarf1
yes, that is correct.

Off Subject, but what is latex?

Homework Helper
memarf1 said:
yes, that is correct.

Off Subject, but what is latex?