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    ,1 \frac{dy}{dx} = y + y^2 through (0,1). I can't do this one. I tried taking the reciprocal, splitting into partial fractions and then anti differentiating to get x = \ln y - \ln (1+y) + c and I can't get it in the form y = f(x). I got it to  e^{x-c} = \frac{y}{1+y} but I couldn't do anything else. How is it done?
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    What does the question actually ask you to do? Do you need to get it in the form y = f(x)? If you really must then carry out the division in y/(1+y) long division style or otherwise.
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    by anti differentiating do you mean integrating
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    (Original post by gangsta316)
     e^{x-c} = \frac{y}{1+y}
    Assuming that's correct, it quite easy to rearrange.
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    It asks to find the solution curve. The answer at the back of the book is given in the form y = f(x).
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    (Original post by rnd)
    Assuming that's correct, it quite easy to rearrange.
    How?
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    a = b/(1 + b)

    Rearrange for b. What's the most intuitive way?
 
 
 
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