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

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

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

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

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.

Please ignore that but look at the final answer

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)}}