differential equation question!?

Question

when a raindrop falls it increases in size, so its mass at time t is a function of t, m(t). The rate of growth of the mass is km(t) for some positive constant k. from law of motion. (mv)' = gm, where v is the velocity of the raindrop (directed downward) and g is the acceleration due to gravity.
Find the differential equation for the velocity of the raindrop by using law of motion!
Solve the DE and use the appropriate initial condition to evaluate the constant of integration

Answer 1

mdv/dt + vdm/dt = gm
dm/dt = km
mdv/dt + kmv = gm
dv/dt = g - kv
dv/(g - kv) = dt
-Ln(g - kv) = t + C
Ln(g - kv) = -(t + C)
g - kv = e^-(t + C)
g - kv = Ce^-t
kv = g - Ce^-t
v = g/k - Ce^-t
assumming v = 0 @ t = 0,
C = g/k, and
v = g/k(1 - e^-t)

Answer 2

I think that this a trick question because the acceleration (dv/dt) due to gravity is just g, independent of m. Thus the equation of motion is just (dv/dt) = g.

I notice that you wrote (mv)' = mg. If that means d(mv)/dt = mg and you were told to use that as the equation of motion, then the answer above (Helmut's) is correct.

Of course, this ignores air drag (which is the real situation). When you throw that in, it gets a lot more complicated, but you need more information than you've given here to solve that.

Answer 3

dy/dx = (x^2)(8 + y) First, we separate the variables via multiplying via dx and dividing via (8 + y): dy/(8 + y) = (x^2)dx combine each and each part, remembering the left will use a organic log and the right would have a consistent: ln(8 + y) = (a million/3)x^3 + C improve e to the skill of both part: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final variety: y = Ae^((a million/3)x^3) - 8 Now, we've the point (0, 3), so plug those in to hit upon a and the certain answer: 3 = Ae^((a million/3)(0)^3) - 8 sparkling up for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the certain answer is: y = 11e^((a million/3)x^3) - 8

Answer 4

(mv)'=gm

m'v+mv'=gm

(b/c rate of growth rate, km=m')

kmv+mv'=gm

m(kv+v')=gm

kv+v'=g

v'=dv/dt=g-kv

dv/(g-kv)=dt

Integrate left from v_0 to v where v_0 is your initial velocity.

Integrate right from 0 to t.

(-1/k)ln[(g-kv)/(g-kv_0)]=t

ln[(g-kv)/(g-kv_0)]=-kt

(exp is the same as e^)

[(g-kv)/(g-kv_0)]=exp(-kt)

solve for v

v=
(g/k)[1-exp(-kt)]+v_0exp(-kt).

@v_0=0

v=(g/k)[1-exp(-kt)]

therefore, as t--> infinity, v-->(g/k)
........ haha, I think.