Find the general solution of the logistic differential equation:                                           dy/dt= y(2-y) Using slope field analysis (and...

Find the general solution of the logistic differential equation:

dy/dt= y(2-y)

Using slope field analysis (and possibly a computer generated slope field) explain why you are confident that your family of solutions is correct?

pramodpandey | Student

`(dy)/(dt)=y(y-1)`

variables are separable.

`(dy)/(y(y-1))=dt`

`` `1/(y(2-y))=A/y+B/(1-y)`

`=(1/2)(1/y)+(1/2)(1/(2-y))`

`(1/2){1/y+1/(2-y)}dy=dt`

Integrating

`(1/2){log(y)-log(2-y)}=t`

`log(y/(2-y))=2t`

`e^(2t)=y/(2-y)`

`y=2e^(2t)/(1+e^(2t))+c`

where c is constant of integration.

C can be determined by given condition.

oldnick | Student

`dy/dx=` `y(2-y)`        `(dy)/(y(2-y))=dx`

integrating both sides:

`=1/2(1/y +1/(2-y)) dy= dx`

`log[sqrt(y(2-y))] = x`

power :

`sqrt(cy(2-y))= e^x`

developing:

`2cy -cy^2 - e^(2x)= 0`

rewriting in a fit shape:

`cy^2-2cy+e^(2x)=0`

`cy^2-2cy+c -c+e^(2x) =0`

`c(y^2-2y+1)+e^(2x) -c=0`

`c(y-1)^2=c-e^(2x)`

`(y-1)^2=1-(e^(2x))/c`

`y-1=sqrt(1-e^(2x)/c)`

so the function we'r searching for is:

`y= 1+sqrt(1-e^(2x)/c)`

this funcion is definite for `c` `!= 0` , since c depends by a initial  condition it doesn't affetc define field of the function .

Once we have assigned c, can lok at field, as the argument inside root is to be greater equal zero.
so that:

`c>e^(2x)`

since  the function  `e^(2x)` increasing  it means :   `x<|log c|`

from arbitray choice of c,(however it does suggest us c>0)

Now we calculate slope:

`y'= (-e^(2x))/(sqrt(c(c-e^(2x))))`