Psst.
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Could someone explain to me why my solution is wrong?

 \frac{dy}{dx} = x(1-x)

 dy = x(1-x) \ dx

\int \ dy = \int x(1-x) \ dx

 y = \frac{x^2}{2} - \frac{x^3}{3} + k
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Google22
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That looks right to me. Is there more to the question?
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Psst.
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(Original post by Google22)
That looks right to me. Is there more to the question?
Yes,  0<x<1 . I was told my solution is incorrect for some reason.
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Muttley79
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(Original post by Psst.)
Could someone explain to me why my solution is wrong?

 \frac{dy}{dx} = x(1-x)

 dy = x(1-x) \ dx

\int \ dy = \int x(1-x) \ dx

 y = \frac{x^2}{2} - \frac{x^3}{3} + k
Ifyou've copied the question corectly then it's fine - were you given values to calculate k?
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Psst.
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(Original post by Muttley79)
Ifyou've copied the question corectly then it's fine - were you given values to calculate k?
If you're not too busy, could you scroll down Exercise A, Question 12 and check if they did it wrong. I understand that they changed the dependent variable from y to t by mistake, but that's trivial.

http://pmt.physicsandmathstutor.com/...hapter%204.pdf
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the bear
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it all looks good to me :dontknow:
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ghostwalker
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@OP

Agree with everyone else - your working is fine

If they'd asked solve "dx/dy=...." then it would be a more interesting question, more in line with their given solution (apart from the "t").
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Muttley79
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(Original post by Psst.)
If you're not too busy, could you scroll down Exercise A, Question 12 and check if they did it wrong. I understand that they changed the dependent variable from y to t by mistake, but that's trivial.

http://pmt.physicsandmathstutor.com/...hapter%204.pdf
Their working does not fit their question ... have they miscopied the exercise?

Needs to be dx/dy or dx/dt if they meant t.
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