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justchill64 · May 23, 2013, 06:20 AM

Can somebody tell me from this dy/dt = (1-a)y-y^2

I got y= a + ce^-yt is this right?

justchill64 · May 23, 2013, 06:45 AM

when u multiply inside the bracket

1y-ay-y2 = dy/dt
dy/dt+y2-y = ay

y^-2+y = ay

found the integrating factor
e^yt

IF*P = Q*IF
e^yt *y = ay*e^yt

y= ay + c
y = ay + ce^-yt

justchill64 · May 23, 2013, 06:46 AM

Originally Posted by justchill64

when u multiply inside the bracket

1y-ay-y2 = dy/dt
dy/dt+y2-y = ay

y^-2+y = ay

found the integrating factor
e^yt

IF*P = Q*IF
e^yt *y = ay*e^yt

y= ay + c
y = ay + ce^-yt

That's how I did it

ebaines · May 23, 2013, 09:03 AM

I think you have misapplied the integrating factor concept. That approach works for differential equations of the form:

$ \frac {dy}{dt} + y P(t) = Q(t) $

Note that the functions P and Q are functions of the variable t, not y, but your equation has a y^2 term in it.

Also, please explain how you go from dy/dt+y2-y = ay to y^-2+y = ay, or written using LaTeX from::

$ \frac {dy}{dt} + y^2 -y = ay$

to

$y^{-2}+y = ay$ .

justchill64 · May 23, 2013, 10:23 AM

Please ignore that but look at the final answer

ebaines · May 23, 2013, 11:46 AM

Here's how I would evaluate it. First let k = 1-a, since these are just constants - makes it easier to deal with.

$ \frac {dy}{dt} = ky-y^2 \\ \\ \frac {dy}{ky-y^2} = dt \\ \\ \int \frac {dy}{ky-y^2} = \int dt $

$ \frac {- \ln(k-y)+ \ln y} k = t + C \\ \\ \frac 1 k \ln (\frac y {k-y}) = t + C \\ \\ \frac y {k-y} = e^{k(t+C)} \\ \\ y = (k-y) e^{k(t+C)} \\ \\ y(1+e^{k(t+C)}) = e^{k(t+C)} \\ \\ y = \frac {e^{k(t+C)} } {1+ e^{k(t+C)} } $

Now sub (1-a) for k, and with some manipulation you get:

$ y = \frac {e^{(1-a)(t+C)} } {1+ e^{(1-a)(t+C)} } = \frac {(1-a)e^{(t+C)}} {e^{(t+C)} + e^{a(t+C)}} $