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    Q. Find the general solution of these differential equations, iving the answer in explicit form where possible

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    Instead of using substitution, write 1/e^y as e^-y, and then follow from there. Hope this helps
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    (Original post by Cxletteee)
    Instead of using substitution, write 1/e^y as e^-y, and then follow from there. Hope this helps
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    (Original post by ckfeister)
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    The integration of e^-y isn't ln(e^y). In this case, y isn't a constant; treat it like an x (rather than like a 1). So the integration of e^-y gives -e^-y (the y part never changes, but the minus is brought down). I'm not very good at explaining things but I hope that makes sense. Knowing that you should be able to finish the question.
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    (Original post by ckfeister)
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    That should ring alarm bells since your answer for \displaystyle \int e^{-y} .dy should differentiate to e^{-y}, but obviously \log(e^y) = y which differentiates to just 1.
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    (Original post by RDKGames)
    That should ring alarm bells since your answer for \displaystyle \int e^{-y} .dy should differentiate to e^{-y}, but obviously \log(e^y) = y which differentiates to just 1.
    oh oops thx
 
 
 
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