Find the function f such that `f'(x) = f(x)(1-f(x))` and `f(0) = 1/2`
1 Answer
embizze | High School Teacher | (Level 2) Educator Emeritus
Given `f'(x)=f(x)(1-f(x))`
and `f(0)=1/2:`
Let y=f(x) and rewrite as:
`(dy)/(dx)=y(1-y)`
Use separation of variables to get:
`(dy)/(y(1-y))=dx`
Rewrite the left hand side using partial fractions:
`(dy)/y - (dy)/(1-y) = dx`
So
`int (dy)/y - int(dy)/(1-y)=int dx`
lny-ln(1-y)=x+c
`ln(y)/(1-y)=x+c`
`y/(1-y)=ce^x`
`(1-y)/y=ce^(-x)`
`1/y-1=ce^(-x)`
`1/y=ce^(-x)+1 `
`y=1/(ce^(-x)+1) `
`y=(e^x)/(c+e^x) `
At 0, `y=1/2 so`
`1/2=1/(c+1) ==> c=1`
so
`y=e^x/(1+e^x)`
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embizze | High School Teacher | (Level 2) Educator Emeritus
The answer is correct. The partial fraction expansion is not written correctly.
`1/(y(1-y))=1/y+1/(1-y)`
so we would have `dy/y + dy/(1-y)`
but the derivative of 1-y is -dy so the integrals should be written as:
`int dy/y - int (-dy)/(1-y)=int dx`
and the rest will follow.
You can check the answer:
`y'=e^x/(1+e^x)^2`
`f(x)(1-f(x))=(e^x/(1+e^x))(1-e^x/(1+e^x))`
`=e^x/(1+e^x)^2`