How to Integrate a Simple Differential Equation with Initial Conditions?

[RentonT]

Homework Statement

"Solve this differential equation algebraically, subject to the initial condition that $$y=10$$ at $$t=0$$

Homework Equations

$$\frac{dy}{dt} = 2y*\frac{1000-y}{1000}$$

The Attempt at a Solution

I first reduced the right side to $$\frac{-y^2}{500} + 2y$$
After that I separated the variables, but I don't know how to integrate that function.
Can anyone point me in the right step?
$$\int \frac {dy}{\frac{-y^2}{500}+2y}$$

 

[Dick]

Have you done any partial fractions recently? You can write the integrand as A/(y)+B/(1000-y). You just need to find A and B.

 

[RentonT]

Have you done any partial fractions recently? You can write the integrand as A/(y)+B/(1000-y). You just need to find A and B.

I am in fact working on the section over partial fractions. Thank you very much Dick. I see how partial fractions plays into this. Might I ask how you got $$\frac{A}{y} + \frac{B}{1000-y}$$

I'm having trouble seeing the partial fractions. Most of them I have worked with I use the shortcut to Heaviside's Method. Many have been in the form of $$\int \frac {Linear}{Quadratic}$$

 

[Dick]

I am in fact working on the section over partial fractions. Thank you very much Dick. I see how partial fractions plays into this. Might I ask how you got $$\frac{A}{y} + \frac{B}{1000-y}$$

y and (1000-y) were the factors of your original expression before you multiplied it out. You didn't really want to do that.

 

[RentonT]

y and (1000-y) were the factors of your original expression before you multiplied it out. You didn't really want to do that.

OK. So what would be on the left-hand side of the equal sign?
$$x=\frac{A}{y}+\frac{B}{1000-y}$$

I understand you multiply the left side by one of the denominators on the right side and then plug in the x value that would have zeroed the denominator. I just don't know what goes on the left side. Is it the $$\frac{dy}{dt}$$?

 

[RentonT]

Nevermind. My poor algebra skills were my downfall this time. I figured out that it's $$\int \frac{-500}{y(y-1000)}$$