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Math Help - [SOLVED] Seperable DE

  1. #1
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    [SOLVED] Seperable DE

    I suppose this is probably pretty easy for people in the know

    Recently I've been using Wolfram Alpha, which is an amazing tool, I mainly like it for showing the steps in calculus, however it doesn't want to give the steps for differential equations

    Anyway, the question is as follows...


    y'=5e^{4x}\sqrt{1-y^2}

    Now I know it's separable so I've put the square root bit under the y'

    \frac{y'}{\sqrt{1-y^2}}=5e^{4x}

    then I integrated both sides separately

    arcsin(y)=\frac{5}{4}e^{4x}+c

    leaving me with

    y=sin({\frac{5}{4}e^{4x}+c})

    However wolfram alpha is giving a rather more complicated answer of

    y=1+2sinh^2(\frac{1}{8}(4c+5ie^{4x}))

    So is WA wrong, or if it isn't, what steps have been taken to get to that answer?
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  2. #2
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    Quote Originally Posted by Rapid_W View Post
    I suppose this is probably pretty easy for people in the know

    Recently I've been using Wolfram Alpha, which is an amazing tool, I mainly like it for showing the steps in calculus, however it doesn't want to give the steps for differential equations

    Anyway, the question is as follows...


    y'=5e^{4x}\sqrt{1-y^2}

    Now I know it's separable so I've put the square root bit under the y'

    \frac{y'}{\sqrt{1-y^2}}=5e^{4x}

    then I integrated both sides separately

    arcsin(y)=\frac{5}{4}e^{4x}+c

    leaving me with

    y=sin({\frac{5}{4}e^{4x}+c})

    However wolfram alpha is giving a rather more complicated answer of

    y=1+2sinh^2(\frac{1}{8}(4c+5ie^{4x}))

    So is WA wrong, or if it isn't, what steps have been taken to get to that answer?
    I can't see anything wrong with your working out.

    I suspect you must have entered the original DE into Wolfram wrongly...
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  3. #3
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    Thanks for clarifying my work, but check for yourself

    http://www93.wolframalpha.com/input/...))*(5*exp(4x))

    I'm almost certian i've entered it corectly, and i'm confident WA is reading the input correctly.
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  4. #4
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    Your answer and the one by WA are both correct and are simply two different expressions for the same thing but yours is the simpler and hence better one!

    Using the identity

    1 + 2 \sinh^2 z \equiv \cosh 2z

    we have that

    y=1+2 \sinh^2(\frac{1}{8}(4c+5ie^{4x}))
    = \cosh \left( \frac{5}{4} i e^{4x} + c \right)
    = \cos \left( \frac{5}{4} e^{4x} + d \right) since \cosh iz \equiv \cos z and where c = id
    = \sin \left( \frac{\pi}{2} + \frac{5}{4} e^{4x} + d \right)
    = \sin \left(\frac{5}{4} e^{4x} + C_0 \right)

    where C_0 is the constant, c , in your expression.

    Welcome to the world of automated computer algebra!
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  5. #5
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    lol, well thanks for clarifying
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