Find f such that f'(x) = f(x)(1 - f(x))

[Idealistic]

f'(x) = f(x)(1 - f(x)) -- Im pretty sure I have to do the separation of variables, but this is as far as I can get:

f(x) = Integral(f(x)(1 - f(x))dx - and im not even sure if its correct. It looks like f(x) = e[sup:4jca0i0v]x[/sup:4jca0i0v], but I dont know how to show it. Do I have to use substitution?

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[stapel]

If you let f'(x) = dy/dx and f(x) = y, you get:

. . . . .dy/dx = y(1 - y)

. . . . .[1/(y(1 - y))] dy = 1 dx

And 1/(y(1 - y)) = 1/y + 1/(1 - y).

Does that help?

Eliz.

 

[Idealistic]

It does but heres where im stuck..

ln|y^2 + y] = ln|x| + c

how do I arrange this monstrosity of y in terms of x?

 

[Deleted member 4993]

It does but heres where im stuck..

ln|y^2 + y] = ln|x| + c

how do I arrange this monstrosity of y in terms of x?

Your solution above is incorrect. However, if it were correct - you would have simplified it with simple algebra....

ln|y^2 + y] = ln|x| + c

ln[(y^2 + y)/x] = c

[(y^2 + y)/x] = e[sup:1lbii4l2]c[/sup:1lbii4l2] = c[sub:1lbii4l2]1[/sub:1lbii4l2]

y[sup:1lbii4l2]2[/sup:1lbii4l2] + y - x * c[sub:1lbii4l2]1[/sub:1lbii4l2] = 0

Quadratic equation - you know how to handle that .....

Correct solution for your ODE was
ln[|y/(1-y)|] = x + c

Again algebra....