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    Need to solve xy' = y(3-y), where y=2 x=2, giving y as a function of x.

    (We showed the integral of 1/(y(3-y)) = 1/3ln(y/3-y) in previous part)

    I keep doing it and getting a different answer to online solvers/ the textbook. Any chance someone can do the question and show me their working? I need to be able to see a perfect solution Thanks!
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    (Original post by ComputerMaths97)
    Need to solve xy' = y(3-y), where y=2 x=2, giving y as a function of x.

    (We showed the integral of 1/(y(3-y)) = 1/3ln(y/3-y) in previous part)

    I keep doing it and getting a different answer to online solvers/ the textbook. Any chance someone can do the question and show me their working? I need to be able to see a perfect solution Thanks!
    Which is the way you've attempted it? Try splitting the variables and then integrating, using your previous result, and go from there. From that, you should be able to sub in y=2 and x=2 to find the value of C.
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    separate the variables then integrate the LHS and RHS.
 
 
 

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