Solve differential a=kxv for x(t) and x(v)

[paalfis]

Homework Statement

If the acceleration of a particle is given by a=k*x*v , where x is position, v is velocity and k is constant and positive, and in t=0: x(t=0)=0 and v(t=0)=vo; find x(t) and x(v).

The Attempt at a Solution

I arrived to dx/dt= k/2*(x²-xo)+vo , (xo=x(t=0)=0). Then I tried to solve the (definite) integral (from 0 to x) ∫1/((k/2)*(x²)+vo)dx that is equal to ∫dt = t. In this integral (which I just couldn't solve) I took vo as a constant. My idea was to solve (result of the integral)=t for x(t).
Is this the right way to do it? If so, how can I solve the integral?

Thanks!

 

[pasmith]

Homework Statement

If the acceleration of a particle is given by a=k*x*v , where x is position, v is velocity and k is constant and positive, and in t=0: x(t=0)=0 and v(t=0)=vo; find x(t) and x(v).

The Attempt at a Solution

I arrived to dx/dt= k/2*(x²-xo)+vo , (xo=x(t=0)=0). Then I tried to solve the (definite) integral (from 0 to x) ∫1/((k/2)*(x²)+vo)dx that is equal to ∫dt = t. In this integral (which I just couldn't solve) I took vo as a constant. My idea was to solve (result of the integral)=t for x(t).
Is this the right way to do it? If so, how can I solve the integral?

Thanks!

Having obtained $$\frac{d\dot x}{dx} = kx$$ it's easier to set $$\dot x = \tfrac12 k (x^2 + C)$$ where $C = 2v_0/k$ so that $$\int \frac{1}{x^2 + C}\,dx = \tfrac12 kt.$$

To do the integral, you will want to make a suitable substitution so that you can use
$$\int \frac{1}{1 + u^2}\,du = \arctan(u) + D$$

 

[paalfis]

Then for x(v) can we take equation v=k/2*x²+vo and so for x : x(v)=2(v-vo)/k ?

 

[HallsofIvy]

No, x is squared. You should have $$x= \sqrt{2(v- v_0)/k}$$

 

[paalfis]

Oh, of course, silly mistake.. Thanks!