Question

Differential equations:

y' = y(3 − ty)
Describe how solutions appear to behave as t increases and how their behavior depends on the initial value y0 ("0" is a subscript) when t = 0.

How do you solve this question? I don't think the equation is linear so I am confused as to how to solve it.. Can someone help me please?

Answer 1

i think it's more of a bernoulli

Answer 2

it is a bernoulli...make it linear by dividing the differential equation by y^2...and put 1/y = z and then it becomes linear in z..

Answer 3

I hope you got it..proceed as a linear differential therefurther...

Answer 4

Oh I see, but how do you divide y' by y^2?

Answer 5

when you put z=1/y, you get by differentiating
dz/dt = -1/y^2 x dy/dx
So, -dz/dt = (dy/dx)(1/y^2)

Answer 6

I got
(1/y^2)(dy/dt) = (1/y)[(3/y^2)-(t/y)]
After I put 1/y as z,
(z^2)(-1/z^2)(dy/dt) = z +3z -tz
-(dy/dt) = z-tz+3z^2

Answer 7

Is this right?

Answer 8

just simply write DE as
\[\Large y'=3y-ty^2\]
now multiply whole equation with \[\Large y^{-2}\]
\[\Large y^{-2}y'=3y^{-1}-t\]
let
\[\Large v=y^{-1}\]
\[\Large v'=-y^{-2}y'\]
or
\[\Large -v'=y^{-2}y'\]
just substitute in above for ^-2y' and y^-1
\[\Large -v'=2v-t\]
or
\[\Large v'+3v=t\]
this is now Linear equation just find the integrating factor and solve it .

Answer 9

does that help you ?

Answer 10

there is typo mistake from third last line it is
\[\Large -v=3v-t\]

Answer 11

Thanks a lot!