Differential equation word problem

[timm3r]

Homework Statement

the population of a town grows at a rate proportional to the population at any time. its initial population of 500 increases by 15% in 10 years. what will be the population in 30 years?

Homework Equations

my teacher sucks at teaching and i have no idea what on Earth he's trying to say. all i got from him was this

dp/dt = Kt
the derivative of the population with respect to time is equal to a constant, which is 15% i think and time.

The Attempt at a Solution

 

[HallsofIvy]

You mean that incompetent teacher expects you to be able to do basic calculus? I assume, of course, that you have never taken calculus- other wise that equation would be all you need. (Odd, this is posted in the "Calculus and Beyond" section!)

Sarcasm aside (which you triggered with "my teacher sucks at teaching". Blaming others for your problems guarentees that you can't solve them.):

No, K is not 15% (0.15). Integrate that equation to find P as a function of both t and K. The answer will, of course, involve a constant of integration. Use the fact that p(0)= 500 and p(10)= 500+ (0.15)(500)= 575 to determine both K and the constant of integration.

 

[kukumaluboy]

Just solving this if others nid help

Let x be the number of people at any time t (years);

dx/dt = kx , Where K>0 (Since is growth);

Use seperable Equation method;

(1/x) dx = k dt;

Integrate Both side w.r.t dx and dt respectively;

ln|X| = kt + C;

Initial Population is 500 --> at time 0, x is 500;
ln|500| = k(0) + C;
C = ln|500|;

Putting back at the original equation,
ln|X| = kt + ln|500|
x = e^(kt + ln|500|)
x = e^(kt) * e^(ln|500|)
x= 500e^(kt);

In 10 years, population will increase by 15% -> 115% of 500 = 575
x= 575 when t = 10,

575 = 500e^(10k);
k=0.0140(3sf)

hence x=500e^(0.014t);

At 30 years, x=500e^(0.014 * 30);
x=761