# Simple Differential Equation

1. Mar 13, 2010

### RentonT

1. The problem statement, all variables and given/known data
"Solve this differential equation algebraically, subject to the initial condition that $$y=10$$ at $$t=0$$

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

3. 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?
$$\int \frac {dy}{\frac{-y^2}{500}+2y}$$

Last edited: Mar 13, 2010
2. Mar 13, 2010

### 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.

3. Mar 13, 2010

### RentonT

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 $$\int \frac {Linear}{Quadratic}$$

Last edited: Mar 13, 2010
4. Mar 13, 2010

### Dick

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

5. Mar 13, 2010

### RentonT

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 $$\frac{dy}{dt}$$?

6. Mar 13, 2010

### RentonT

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

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