# Homework Help: Differential Equation

1. Jan 14, 2010

### PennyGirl

1. The problem statement, all variables and given/known data
Solve the differential equation...
X'' = .5*x^-2
taken with respect to t
at t=0, x'=0 and x=10

2. Relevant equations
N/A

3. The attempt at a solution
I tried to split the variables (ie d^2 X * X^2 = .5 dt^2) but didn't get the right answer with this (i plugged it bak in and it didn't work?)
How should I start this problem?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 14, 2010

### rochfor1

3. Jan 14, 2010

### PennyGirl

d^2 X * X^2 = .5 dt^2
integrate both sides...
dx * x^3/3=(.5*t + C) *dt
at t=0, dx/dt=0 and x=10...
0*10^3/3 = .5*0 + C
C=0

dx * x^3/3 = .5*t*dt
integrate both sides again...
x^4 / 12 = .25*t^2 + C
same refs....
10^4/12 = .25*0 +C
C = 833.3

x^4/12 = .25*t^2 + 833.3

algebra...

x= (3*t^2 + 10000)^(1/4)

but that doesn't work...

4. Jan 14, 2010

### rock.freak667

Do you know the variation of parameters method? That might work.

5. Jan 14, 2010

### PennyGirl

I didn't think that would work because I thought the exponent on the x term would have to be 1, while in this case its -2...

6. Jan 14, 2010

### Dick

Just integrate (1/2)x^(-2) twice. Don't forget to keep the constants of integration around.

7. Jan 14, 2010

### PennyGirl

so....
integrating once...
x'=.0345*x^-2*t

then again...
x=.0345/2*x^-2*t^2+C\
???

Last edited: Jan 14, 2010
8. Jan 14, 2010

### Dick

No, you aren't missing anything. I am. I didn't notice t was the independent variable.

9. Jan 15, 2010

### HallsofIvy

NO. You cannot separate a second derivative like this. A second derivative cannot be treated like a fraction.

Your original equation can be written as [itex]x^2d^2x/dt^2= 1/2[/math]. Let y= dx/dt. Then, by the chain rule, [itex]d^2x/dt= dy/dt= (dy/dx)(dx/dt)= y dy/dx. Your equation becomes y dy/dx= (1/2)x-2. Since that is now a first derivative, it can be treated like a fraction and separated: ydy= (1/2)x-2dx. Integrate that to find y, as a function of x, and then integrate dx/dt= y to find x.` Since you know that x(0)= 10 and x'(0)= 0, you know that y= 0 when x= 10 so can find the first constant of integration before the second integral.