## frank0520 one year ago Use separation of variables to solve the differential equation: K dN/dt = -r(N-K)(N-A) , K,r,A are constants.

1. frank0520

here is a picture

2. freckles

$K \frac{dN}{dt}=-r(N-K) (N-A) \\ K dN=-r(N-K)(N-A) dt \text{ divide both sides by } (N-K)(N-A) \\ \frac{K}{(N-K)(N-A)} dN=-r dt \text{ now integrate both sides }$

3. frank0520

that's the part I'm stuck on. the left hand side because the right is easy

4. freckles

$\frac{K}{(N-K)(N-A)}=\frac{a}{N-K}+\frac{b}{N-A} \\ \frac{K}{(N-K)(N-A)}=\frac{a(N-A)+b(N-K)}{(N-K)(N-A)} \\ \frac{K}{(N-K)(N-A)}=\frac{(a+b)N+(-aA-bK)}{(N-K)(N-A)} \\ \text{ so we have teh system } \\ a+b=0 \\ -aA-bK=K \\ \text{ so the first equation gives } a=-b \\ \text{ and using the first \in the second we have } \\ bA-bK=K \\ \text{ solving for } b \\ b=\frac{K}{A-K} \\ \text{ and so } a=-b=\frac{-K}{A-K}$

5. freckles

what I'm trying to suggest from the above is to use partial fractions

6. freckles

are you there? still having trouble?

7. freckles

the left hand side you should be evaluating.. $\frac{-K}{A-K} \int\limits \frac{1}{N-K} dN+ \frac{K}{A-K} \int\limits \frac{1}{N-A} dN$

8. frank0520

so would it be: $\frac{ -K }{ A-K }\ln|N-K| + \frac{ K }{ A-K }\ln|N-A| = -r t + C$

9. freckles

yep

10. frank0520

Ok, Thank for the Help!

11. freckles

did you understand what I did above for the partial fraction part?

12. frank0520

Yes, when I did myself I put the -r with the partial fractions, which made it harder like: $\frac{ K }{ -r(N-K)(N-A) } dN$ but now it more clearer with your way.

13. freckles

cool stuff