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Math Help - Getting Y in terms of x

  1. #1
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    Getting Y in terms of x

    <br /> <br />
dx/dt=1/(t+1) dy/dt=2t <br />

    Write an equation expressing y in terms of x.

    So far, I have x=ln(t+1)+C and y=t^2+C

    So,

    t=e^(x-C)-1

    and y would equal:

    y=(e^(x-C)-1)^2+D

    I don't know how to get rid of the 'Cs' or if the 'D' should be a C. I thought that there should be a D because the D represnts the constant from the y equation.
    Not really sure if I'm doing this problem properly.
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  2. #2
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    e^(i*pi)'s Avatar
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    I would use the chain rule to get \dfrac{dy}{dx} and find y from there
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  3. #3
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    Is this what you mean?:

    dy/dx=2t/(t+1)

    Integrating gives,

    y=2(t-ln(t+1))+C

    so

    y=2(e^(x-C)-1-x)+C

    That still leaves me with C.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JewelsofHearts View Post
    t=e^(x-C)-1

    and y would equal:

    y=(e^(x-C)-1)^2+D

    I don't know how to get rid of the 'Cs' or if the 'D' should be a C. I thought that there should be a D because the D represnts the constant from the y equation.
    You are correct. And C and D come from different integrations, so yet they would be different.

    @ e^(i*pi)
    \displaystyle \frac{dy}{dx} = 2t^2 + 2t

    How do you get rid of the t?

    -Dan
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