Differential Equation...

sebbie

Question 8 of June 2005, how do you do part b?

Is the answer to 6c, y=8/(4+x^2) 0<x<infinity ?

Reply 1

spikey_aman

whats the actual question?

Reply 2

sebbie

liquid pouring into a container at 20cm^3 per second and leaking out at rate proportional to volume of liquid in container....

therefore, dv/dt = 20-kv

b) the container is initially empty, by solving differential equation show that:

V=A+Be^-kt giving A and B in terms of k

Reply 3

spikey_aman

Okay.....

Seperate the variables:

Int: dV/20-kV = Int: dt

so

1/-kln|20-kv| = t + c

It is initially empty... so substitute the zeros in and you get c as -1/kln20

1/-kln|20-kv| = t -1/kln20

times by -k

ln|20-kV| = -kt + ln20

20 - kV = e^-kt + ln20

20 - e^-kt + ln20 = kV

20 - 20e^-kt = kV

V = 20/k - 20/k.e^-kt

Reply 4

meef cheese

dv/dt = 20-kv
=> INT dv/(20-kv) = INT dt
=> t + C = (-1/k)ln(20-kv)

At t=0, v=0 => C = (-1/k)ln20

t = 1/k[ln(20/(20-kv))]
tk = ln[20/(20-kv)]
e^tk = 20/(20-kv)

etc.

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