Need help with a simple Differential Equation

[Frogger888]

I am stumped on a simple diff eq. Can anyone Help?

y'=y(y-2) y(0)=1

What method do I use and how do I go about doing it without an x in the equation

Thank you :shock:

 

[Unco]

I'm sure you're familiar with separation of variables:

\(\displaystyle \mbox{ \frac{1}{y(y - 2)} \frac{dy}{dx} = 1}\)

 

[soroban]

Hello, Frogger888!

\(\displaystyle y'\:=\:y(y\,-\,2),\;\;y(0)\,=\,1\)

What method do I use and how do I go about doing it without an x in the equation?

There's always an x in the equation!

We have: \(\displaystyle \:\frac{dy}{dx}\;=\;y(y\,-\,2)\)

Separate varaibles: \(\displaystyle \:\frac{dy}{y(y\,-\,2)}\;=\;dx\)

Can you finish it now?

 

[Frogger888]

After that then what?

I get the integral

1/2(log(y-2)-(log(y))=x

What steps do you do to solve for y(0)=1
is it some form of e^x? Since that is the only thing that can give you 1 when y= 0
If so how do you get there

It has been a while since I have taken a math coarse so it is tough for me to get back into it. Thank you for the help

 

[Unco]

Re: After that then what?

There should be a 1/2 in front of log|y|, too, and you're missing a constant of integration.

Remember the rule for subtracting logs; solve for y and then substitute your initial condition.

 

[Frogger888]

I got it! Thank you guys for the help!

 

[Frogger888]

Is the answer c=-1???
Thanks

 

[Unco]

Yep, so it comes out as y = 2/(1 + e^(2x)).