bernoulli equation/partial fraction

**treetheta** · October 14th 2009, 07:02 PM

so i know this question can be done both way

y' = 3y - y^2

<=>

y' = y(3-y)

alright so i dont think i can do it by bernoulli because i dont have two variables so i dont think thats gonna work but
I need to solve the initial value problem y(0) = 2

i get up to

dy
-- = 3dt
y(1-y/3)

i just don't know how to to setup and do the partial fraction

**redsoxfan325** · October 14th 2009, 08:12 PM

Originally Posted by treetheta

so i know this question can be done both way

y' = 3y - y^2

<=>

y' = y(3-y)

alright so i dont think i can do it by bernoulli because i dont have two variables so i dont think thats gonna work but
I need to solve the initial value problem y(0) = 2

i get up to

dy
-- = 3dt
y(1-y/3)

i just don't know how to to setup and do the partial fraction

$\frac{1}{y(3-y)}=\frac{A}{y}+\frac{B}{3-y}\implies A(3-y)+By=1\implies$ $(B-A)y+3A=1\implies A=\frac{1}{3},B=\frac{1}{3}$

So,

$\int\frac{y'}{y(3-y)}\,dx=\frac{1}{3}\int\frac{y'}{y}\,dx+\frac{1}{3 }\int\frac{y'}{3-y}\,dx=\int\,dx$

Letting $k=e^{3c_0}$ , You end up with

$\frac{1}{3}(\ln(y)-\ln(3-y))=x+c_0\implies\ln\left(\frac{y}{3-y}\right)=3x+3c_0\implies\frac{y}{3-y}=e^{3x}\cdot \underbrace{e^{3c_0}}_{k}\implies$ $y=3ke^{3x}-kye^{3x}\implies y(1+ke^{3x})=3ke^{3x}\implies y=\frac{3ke^{3x}}{1+ke^{3x}}$

At the point $(0,2)$ , we have $2=\frac{3k}{1+k}\implies k=2$ .

So $\boxed{y=\frac{6e^{3x}}{1+2e^{3x}}}$

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