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Math Help - Particular Solution - Calculus based question

  1. #1
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    Particular Solution - Calculus based question

    So I have a fairly simple looking question I just forgot how to complete it.

    2ydx = 3xdy when y(-2) = 1

    dx/(3x) = dy(2y)

    ln(3x) = ln(2y) + C

    This is where I get stuck. I need to solve for y but I've just forgotten how to do this and don't know how to describe this type of problem.
    Thanks for any help!
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  2. #2
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    Quote Originally Posted by coldize View Post
    So I have a fairly simple looking question I just forgot how to complete it.

    2ydx = 3xdy when y(-2) = 1

    dx/(3x) = dy(2y)

    ln(3x) = ln(2y) + C

    This is where I get stuck. I need to solve for y but I've just forgotten how to do this and don't know how to describe this type of problem.
    Thanks for any help!
    Exponentiate both sides and you get 3x = Ay where A = 2e^C ....
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  3. #3
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    Any chance you could walk through that with me? I'm afraid I don't know what you mean by exponentiate.
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    Quote Originally Posted by coldize View Post
    Any chance you could walk through that with me? I'm afraid I don't know what you mean by exponentiate.
    e^{ln(3x)} = e^{ln(2y) + C} and apply the usual index and log rules.
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  5. #5
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    Great. Thank you!
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