Taking partial fractions for integration? I'm having some trouble with integrals involving partial fractions it seems. Been stuck on this forever. The equation is given below and I have to use partial fractions to solve.
dN/dt = N(1-0.0005N)

Can anyone give me step by step details on how you would do this? Much appreciate, thank you!
 A: $$\begin{align}\int \dfrac{dN}{N(1-0.0005N)} &= \int dt \end{align}\tag{1}$$
Left hand side integrand can be decomposed into partial fractions :
$$\dfrac{1}{N(1-0.0005N)} = \dfrac{A}{N} + \dfrac{B}{1-0.0005N}\tag{2}$$

You may use coverup method to find the constants $A$ and $B$  
Multiply $(2)$ by $N$ through out and plugin $N=0$, you get :
$$\begin{align}\dfrac{1}{(1-0.0005N)} &= A + N.\dfrac{B}{1-0.0005N}\\~\\\dfrac{1}{(1-0.0005(0))} &= A + 0.\dfrac{B}{1-0.0005N}\\~\\\implies 1&= A\end{align}$$
See if you can find the other constant $B$ similarly.
Once you have both the constants replace the integrand on left side in $(1)$  by $(2)$ and evaluate the integral
A: Let us take it in a more general form and suppose that the differential equation is $$\frac{dN}{dt}=(N-a)(N-b)$$ where $a\neq b$. So, we can rewrite $$\frac{dt}{dN}=\frac{dN}{(N-a)(N-b)}$$ Now, using partial fraction decomposition, write $$\frac{1}{(N-a)(N-b)}=\frac{A}{N-a}+\frac{B}{N-b}$$ Reduce everything to same denominator to get $$1=A(N-b)+B(N-a)$$ Setting $N=a$ in the expression leads to $$A=\frac{1}{a-b}$$ Setting $N=b$ in the expression leads to $$B=\frac{1}{b-a}$$ and finally $$t+C=\frac{\log (N-a)-\log (N-b)}{a-b}$$ from which $$N=a+(a-b)\frac{ e^{a \left(C+t\right)}}{e^{b \left(C+t\right)}-e^{a
   \left(C+t\right)}}$$
A: “Multiply” by dt, and divide by $N(1-0.0005N)$. Then write $\dfrac1{N(1-aN)}$ as $\dfrac xN+\dfrac y{1-aN}$, and 
integrate both sides. To determine the values of x and y, bring the sum to a common denominator, and then identify the coefficients.
A: We rewrite the DE as 
$$\frac{dN}{N(1-0.0005N)}=dt$$
and integrate. To integrate the left-hand side, express $\frac{1}{N(1-0.005N)}$
as 
$$\frac{A}{N}+\frac{B}{1-0.0005N},$$
where $A$ and $B$ are constants. 
If we bring this to a common denominator, we get
$$\frac{A(1-0.0005N)+BN}{N(1-0.0005N)}.$$
So we want $A(1-0.0005N)+BN$ to be identically equal to $1$. Thus $A=1$ and $B=0.0005$. Finally, integrate both sides. We get
$$\ln(N)-\ln(1-0.0005N)=t+C.$$
Taking the exponential of both sides, we get
$$\frac{N}{1-0.0005N}=De^t,$$
where $D$ is a constant. Now if we wish we can solve explicitly for $N$ by using a little algebra. 
