Question

answered question.

Answer 1

$$\frac{dy}{dx}=(ay)(1-by)\\\frac1{ay(1-by)}dy=dx$$?

Answer 2

Is that what you're saying?

Answer 3

no master i need to prove that the answer is y= 1over b+C e^-ax

Answer 4

Okay I get that but I mean the first line -- is that the equation?$$\frac{dy}{dx}=(ay)(1-by)$$

Answer 5

yes

Answer 6

Okay!

Answer 7

$$\frac{dy}{dx}=ay(1-by)\\\frac1{ay(1-by)}dy=dx$$Note we can break our first using partial fractions:$$\frac1{ay(1-by)}=\frac{A}{ay}+\frac{B}{1-by}=\frac{A(1-by)+B(ay)}{ay(1-by)}$$

Answer 8

So we have:$$A(1-by)+B(ay)=1$$If we let \(y=0\) we observe \(A=1\). If we allow \(y=1/b\), we have:$$B(a/b)=1$$therefore \(B=b/a\). Therefore we can rewrite:$$\frac1{ay(1-by)}=\frac1{ay}+\frac{b/a}{1-by}$$

Answer 9

$$\left(\frac1{ay}+\frac{b/a}{1-by}\right)dy=dx\\\left(\frac1{ay}-\frac{b/a}{by-1}\right)dy=dx\\\frac1a\left(\frac{b}{by-1}-\frac1y\right)dy=-dx\\\int\frac1a\left(\frac{b}{by-1}-\frac1y\right)dy=\int -dx\\\frac1a(\log(by-1)-\log y)=-x+C\\\log(by-1)-\log y=-ax+C\\\log\frac{by-1}y=-ax+C\\\frac{by-1}y=Ce^{-ax}\\b-\frac1y=Ce^{-ax}\\-\frac1y=-b+Ce^{-ax}\\\frac1y=b-Ce^{-ax}=b+Ce^{-ax}\\y=\frac1{b+Ce^{-ax}}$$

Answer 10

Note we had to play with it a little bit so that it'd come out like they wanted...

Answer 11

thanks :D

Answer 12

hey master can i use the ln there

Answer 13

Well I used \(\log\) to mean natural logarithm so yes

Answer 14

I meant \(\log x=\ln x\) -- it's already using the natural logarithm.