Question

dy/dx = x^2(1-y) and f(0)=2, find the particular solution

Answer 1

A similar example:
\(\displaystyle dy/dx = 5x^4(2-y) ,~~~~~~~~~f(1)=1 \)

Solution to this example:

Separating variables and integrating:
\(\displaystyle dy/dx = 5x^4(2-y) \)
\(\displaystyle [1/(2-y)]~ dy = 5x^4~dx \)
\(\displaystyle \int [1/(2-y)]~ dy = \int 5x^4~dx \)
\(\displaystyle \ln|2-y| = x^5+C \)

Determining C, by plugging initial value(s):
\(\displaystyle \ln|2-1| = 1^5+C \)
\(\displaystyle \ln(1) =1+ C \)
\(\displaystyle C=-1\)

Plugging the determined C into the solution:
\(\displaystyle \ln|2-y| = x^5-1 \)

Answer 2

wouldn't the integral of 1/(2-y) result in a negative?

Answer 3

@idku

Answer 4

yes you are right @gabyr5. @idku made a mistake and it should be
\[-ln|2-y|\]