Help me Diff Equation population model

[benz31345]

Homework Statement

The population x of a certain city satisfies the logistic law
dx/dt = x/100 - x^2/10^8

where time t is measured in years.given that the population of this city is 100000 in 1980,determine the population as a function of this for t>1980 .In particular,answer the following question
1. What will be the population in 2020

Homework Equations

I don't know how to do.

The Attempt at a Solution

 

[clamtrox]

That differential equation is separable, so you can solve it right away,
[tex] t-t0 = \int^x_{x_0} \frac{dx}{x/100-x^2/10^8} [/tex]
Then just calculate the integral, solve x(t) and use the initial values given to fix the constant x₀

 

[benz31345]

So when integral finish already it just have two x
I don't know how to do next step.

 

[HallsofIvy]

I don't know what you are talking about. If by "two x" you mean you have x in two different places and want x= f(t), solve the equation for x!

Please don't just say "I don't know how to do it" and then say "I have already done that". Show us what you got when you integrated and how you have tried to solve for x.

 

[benz31345]

I get. lnx + 10^6/x = t/10^+c

When take e

X + e^10^6/x = ce^t/10^2

X(t) = ce^t/10^2 - e^10^6/x

Have two x right? X is answer
How to solve?

 

[clamtrox]

I get. lnx + 10^6/x = t/10^+c

When take e

X + e^10^6/x = ce^t/10^2

X(t) = ce^t/10^2 - e^10^6/x

Have two x right? X is answer
How to solve?

That's not how you integrate that! Split it into fractions!

[tex] \frac{1}{x - bx^2} = \frac{1}{x} + \frac{b}{1-bx} [/tex]
or something close to that