# Differential equation problem

1. Jan 15, 2014

### fayled

I'm solving a differential equation to do with quadratic resistance and it seems to be acting very strangely - I get the opposite sign of answer than I should. If anybody could have a quick look through that would be much appreciated.

For a particle moving downward and taking positive upwards, I have
mdv/dt=-mg+bv2

The terminal velocity comes out at vl=-√(mg/b) (negative as it obviously has its terminal velocity downwards).

Substituting this in gives
dv/dt=-g(1-(v/vl)2)

Now make the substitution z=v/vl so dv/dt=vldz/dt. Then we obtain

vldz/dt=-g(1-z2)

Noting that 1/1-z2=0.5[(1/1+z)+(1/1-z)] and separating variables we get

∫[(1/1+z)+(1/1-z)]dz=-2g/vl∫dt

Integrating each side then gives (the initial conditions are v=0 at t=0 so the constant of integration is zero)

ln(1+z/1-z)=-2gt/vl.

Next let k=vl/2g so that

ln(1+z/1-z)=-t/k

e-t/k=1+z/1-z

e-t/k-ze-t/k=1+z

z(1+e-t/k)=e-t/k-1

z=(e-t/k-1)/(e-t/k+1)

Therefore

v=vl[(e-t/k-1)/(e-t/k+1)]

Now, the correct answer should be

v=vl[(1-e-t/k)/(1+e-t/k)]

i.e

v=-vl[(e-t/k-1)/(e-t/k+1)]

which must be right because as t→∞, v→vl which by the definition of vl is expected.

I've spent a lot of time trying to work out where I'm going wrong and it's driving me crazy. Thanks in advance :)

2. Jan 15, 2014

### pasmith

Probably better to define $v_0 = (mg/b)^{1/2} > 0$ so that terminal velocity is $-v_0$ and $v$ and $z$ increase in the same direction.

I believe this is correct.

I think you have confused yourself. You've defined $k = v_l/2g$ and you initially defined $v_l$ to be negative, so $k$ is negative. It follows that $-1/k$ is positive, so that $e^{-t/k} \to \infty$ as $t \to \infty$. Thus your answer
$$v=v_l \frac{e^{-t/k}-1}{e^{-t/k}+1} \to v_l < 0$$
as required.

On the other hand, if you define $v_l$ to be positive then $-1/k$ is negative, so that your answer tends to $-v_l < 0$ as required.

Last edited: Jan 15, 2014
3. Jan 15, 2014

### fayled

But I defined it as vl=-(mg/b)1/2...

4. Jan 15, 2014

### pasmith

I only noticed that after replying, since it's extremely unnatural to do that. More normal would be to define $v_l = (mg/b)^{1/2}$.

I have now edited my post accordingly.

5. Jan 15, 2014

### fayled

Ah right, thankyou very much.