How to Integrate a Simple Differential Equation with Initial Conditions?

RentonT
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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}
 
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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.
 
Dick said:
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}
 
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RentonT said:
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.
 
Dick said:
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}?
 
Nevermind. My poor algebra skills were my downfall this time. I figured out that it's \int \frac{-500}{y(y-1000)}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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