Integrating d²x/dt² = -k²/x³ for particle motion

[Saptarshi Sarkar]

Homework Statement

A particle of unit mass moves under the action of force F=-k²/x³. What is the time required by the particle to reach the centre of force from a distance d.

Relevant Equations

d²x/dt² = -k²/x³

I tried solving the differential equation by integrating both sides wrt t but this does not work out and I am doing something wrong. How do I integrate such an equation properly?

d²x/dt² = -k²/x³
=> dx/dt = -k²t/x³
=> x = -k²t²/2x³ + d

 

[PeroK]

Homework Statement:: A particle of unit mass moves under the action of force F=-k²/x³. What is the time required by the particle to reach the centre of force from a distance d.
Homework Equations:: d²x/dt² = -k²/x³

I tried solving the differential equation by integrating both sides wrt t but this does not work out and I am doing something wrong. How do I integrate such an equation properly?

d²x/dt² = -k²/x³
=> dx/dt = -k²t/x³
=> x = -k²t²/2x³ + d

$x$ is a function of $t$, not a constant. You can't just integrate $\frac{1}{x^3}$ as though it were a constant function.

As a first step you could multiply the equation by $2\frac{dx}{dt}$, which is called an integrating factor.

 

[ehild]

d²x/dt² = -k²/x³
=> dx/dt = -k²t/x³
=> x = -k²t²/2x³ + d

As @PeroK said, you can not integrate as if x were a constant.
d²x/dt² is the acceleration, the time derivative of the velocity, v=dx/dt. In terms of v, your equation is dv/dt=-k²/x³.
You can consider v as function of x, v(x(t)), and determine dv/dt using the chain rule:
dv/dt= dv/dx * dx/dt = v dv/dx.
Now you have a first-order differential equation for v as function of x:
v dv/dx = - k²/x³. This is a separable differential equation.
Get v(x), then integrate v=dx/dt with respect to t.