1. ## differential equation

How do I solve this differential equation:
$\displaystyle m\frac{d^2x(t)}{dt^2}=\frac{-c}{x(t)^2}$, c and m are constants larger than 0. Initial conditions are x(0)=a, x'(0)=0.

2. I don't understand what did I do wrong with latex syntax.

If someone doesn't understand what I wrote:
m*x''(t)=-c/x(t)^2

3. FYI LaTeX is down at the moment. Please stay tuned for futher details. News at 11.

-Dan

4. Originally Posted by Ene Dene
How do I solve this differential equation:
$\displaystyle m\frac{d^2x(t)}{dt^2}=\frac{-c}{x(t)^2}$, c and m are constants larger than 0. Initial conditions are x(0)=a, x'(0)=0.
Solve initial value problem from the general solution.

5. I don't understand something, if you have a harmonic oscilator differential equation:
x''(t)+a^2*x(t)=0,
the solution is:
x(t)=Acos(a*t)+Bsin(a*t).

So why don't you get solution x(t)? Or do you suggest that solution is:
x(t)=c1+c2*t+c/m*ln(t) ?
If so, I don't think that's a solution, try puting it in initial equation.

6. Originally Posted by Ene Dene
I don't understand something, if you have a harmonic oscilator differential equation:
x''(t)+a^2*x(t)=0,
the solution is:
x(t)=Acos(a*t)+Bsin(a*t).

So why don't you get solution x(t)? Or do you suggest that solution is:
x(t)=c1+c2*t+c/m*ln(t) ?
If so, I don't think that's a solution, try puting it in initial equation.
Well, if c = 0 it works. Then the solution is x(t) = c1 + c2*t

-Dan

7. Originally Posted by Ene Dene
I don't understand something, if you have a harmonic oscilator differential equation:
x''(t)+a^2*x(t)=0,
the solution is:
x(t)=Acos(a*t)+Bsin(a*t).

So why don't you get solution x(t)? Or do you suggest that solution is:
x(t)=c1+c2*t+c/m*ln(t) ?
If so, I don't think that's a solution, try puting it in initial equation.
No sure what you are asking.

x^2x''=.....

But you wrote,
x''+a^2x=.....

They are different equations.

I solved the upper one, and I think that is the one you wanted to be solved.

8. I can't understand how is x(t)=C1+C2*t+c/m*ln(t) solution of equation x''(t)+c/(m*(x(t))^2)=0.

x''(t)=-(c/m)/t^2.

And then we have:

-(c/m)/t^2+c/(m*(C1+C2*t+c/m*ln(t))^2)=!0

9. Originally Posted by Ene Dene
I can't understand how is x(t)=C1+C2*t+c/m*ln(t) solution of equation x''(t)+c/(m*(x(t))^2)=0.

x''(t)=-(c/m)/t^2.

And then we have:

-(c/m)/t^2+c/(m*(C1+C2*t+c/m*ln(t))^2)=!0
Originally Posted by topsquark
Well, if c = 0 it works. Then the solution is x(t) = c1 + c2*t

-Dan
Did you miss this post? You have a solution that works! (Whether or not the ln term is extraneous. I assume there's a reason it comes into the solution.)

-Dan

10. Originally Posted by topsquark
Did you miss this post? You have a solution that works! (Whether or not the ln term is extraneous. I assume there's a reason it comes into the solution.)
In my first post I said that c>0. x(t)=c1+c2*t+c/m*ln(t) is just not a solution.

Anyway it doesn't matter anymore, I've maneged to solve the problem by avoiding this differential equation (it was a physics problem).