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Math Help - IVP problem

  1. #1
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    IVP problem

    find the exact solution of the IVP.
    y'=2xy, y(0)=2
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  2. #2
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    Quote Originally Posted by calculuskid1 View Post
    find the exact solution of the IVP.
    y'=2xy, y(0)=2
    separate variables ...

    \frac{dy}{y} = 2x \, dx

    integrate and use the initial condition to find the solution
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  3. #3
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    Quote Originally Posted by calculuskid1 View Post
    y'=2xy, y(0)=2
    This is a very simple separable DE

    \frac{dy}{dx} = 2xy

    \frac{dy}{y} = 2x~dx

    Now tntegrate both sides.

    \int \frac{dy}{y} = \int 2x~dx
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  4. #4
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    I got that and i got ln y=x^2+c but im not sure what to do there,
    ln 2=c
    so then ln y=x^2+ln2
    then y=e^(x^2+ln2)
    can that be simplified?
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  5. #5
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    Quote Originally Posted by calculuskid1 View Post
    I got that and i got ln y=x^2+c but im not sure what to do there,
    ln 2=c
    so then ln y=x^2+ln2
    then y=e^(x^2+ln2)
    can that be simplified?
    e^{x^2+\ln{2}} = e^{x^2} \cdot e^{\ln{2}} = ?
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