so it should be y=ce^(kt)
y(o) = c e^0 = c
then y(2) = c e^(2k) = 2 * y (0)
=> 2k = ln(2) => k = ln(2) / 2
now y(10) = 1000 = c e^(10k)
=> c = 1000 / (e^(5ln2))
correct?
If all I have given is that
1. Bacteria grows at a rate proportional to it's size.
2. It doubles in 2 days.
3. At 10 days, population is 1000.
I'm not given the initial bacteria count, I need help setting up the equation.
I did:
dy/dt = ky => dy/y = kdt => lny= kt + c =>...
HallsofIvy, I understood the y1= e^t and y2=te^t
But you said "Your equation reduces completely to d^2/dt^2- 2dy/dt+ y= t"
shouldn't it reduce to:
e^2t y'' - e^t y' + y = t ?
I got this book from WILEY by Erwin Kreyszig. It tells how to solved homogenous cauchy equations. It also covers simple nonhomogenous equations.
But it doesn't cover when we have nonhomogenous Cauchy equations like this one.
x2y''-xy'+y=lnx
How do I go about solving that equation?
I substituted...
If for example, I have.
Y1=Cos(2Ln(x)) and Y2=Sin(2Ln(x)) and I have to reach the general solution.
I know how to get to the general solution is the cauchy equation:
y'' X^n + y' x + 4 y = 0
According to the answer, n=2 => y'' X^2 + y' x + 4 y = 0
How am I to know that that it is X^2 ?
Or is...