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Math Help - [SOLVED] separable equation

  1. #1
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    [SOLVED] separable equation

    hey there.. i'm stuck in the calculation below..

    dy/dx = ye^x y(0) = 2e
    ∫1/y dy = ∫e^x dx
    ln y + C = (e^x) + C
    how to proceed so that my answer is y = 2e^(e^x)

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  2. #2
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    Quote Originally Posted by nameck View Post
    hey there.. i'm stuck in the calculation below..

    dy/dx = yex
    Is that y*e*x or y*e^x?
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  3. #3
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    Since you have ln y = (e^x)+c, you need to free the y from the ln. How you do that raise everything to the power of e.

    From that, you obtain e^(ln y) = e^[(e^x)+c].

    y = e^[(e^x)+c]

    The last step before substitution is understanding that e^(a+b) can be written as (e^a)*(e^b) and that when you have e^c is just another constant and can be written just as C
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  4. #4
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    Ok… so I proceed…
    y = e^ [(e^x) + C]
    when x = 0; y =2e

    hence, 2e = e^ [(e^0) + C]
    2e = (e^1) + C

    Apply ln to both sides
    2 = 1 + ln C
    C = e^1
    C = 0

    Hence, y = e^(e^x)

    Still didn’t get the answer provided..
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  5. #5
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    Quote Originally Posted by nameck View Post
    Ok… so I proceed…
    y = e^ [(e^x) + C]
    I made sure to mention how exponents work in my post but here is it again.

    For example if we have 2^(2+x), then we could write (2^2)*(2^x). This works because 2*2=(2^1)*(2^1)=2^(1+1)=2^2=2*2.

    Therefore, e^[e^x + c]= e^(e^x)*e^c and e^c is just another constant so e^c=C. Now we have (e^(e^x))*c=C*(e^(e^x).

    y= C*(e^(e^x))

    y(0)=C*(e^(e^0))=2*e

    e^0=1

    I think you should be able to obtain the correct solution from there.
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  6. #6
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    got it noW!!
    thanks dwsmith!!
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