# A tricky differential equation

1. Mar 9, 2004

### amix

Hello

I have a differential equations, which I don't know how to solve. Maybe somebody has a tip or two

The equation is:
d^2 m / d x^2 = -3x

2. Mar 9, 2004

Staff Emeritus
So m is a function of x and its second derivative with respect to x is -3x? What function gives ax when differentiated? And then what function gives THAT function when differentiated? Do you have any initial conditions?

3. Mar 9, 2004

### amix

Well, I need to find the complete solution. All I know is that I have a function m(x) and that the interval of x is from 0 to 4.

4. Mar 9, 2004

### matt grime

what selfadjoint was getting at was that this is a very easy separable solution, though you've not met those words yet perhaps.

suppose y' = 2x, ( ' means diff wrt x)
then

y= x^2+c

suppose z'' = y' in the above,
then z' = x^2+c,

so what is z?

5. Mar 13, 2004

### qmerino

I haven't taken diff. eq. in a while, so I might be wrong, but isn't that like saying:

m" = -3x ?

then just integrate: (I'm using | as the symbol for integral)

|(m" dm) = |(-3x dx)
m' = -3/2 x^2 + C

integrate again, gives:
|(m' dm) = | (-3/2 x^2 + C dx)
m = -1/2 x^3 + Cx + D

That's it!

I don't know what the interval from 0 to 4 falls-in though....

6. Mar 24, 2004

### Laplace_Wy

Well, wouldn't the interval be just a definite integral from 0 to 4? I may be wrong, if I misread your question. [b(]