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Thread: initial value problem

  1. #1
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    initial value problem

    solve

    $\displaystyle y'=\frac{y}{x} +2x^{2} $, y(1)=0

    really helpful if someone could show me full work through i know they're not hard but need gd understanding thanks
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  2. #2
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    Krizalid's Avatar
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    $\displaystyle \begin{aligned}
    y'&=\frac{y}{x}+2x^{2} \\
    y'-\frac{y}{x}&=2x^{2} \\
    \frac{y'}{x}-\frac{y}{x^{2}}&=2x \\
    \left\{ y\cdot \frac{1}{x} \right\}'&=2x.
    \end{aligned}$

    In the third step I found the "integrating factor." Do you know what the I.F. is?
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  3. #3
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    for integrating factor would it be $\displaystyle e^{\int\frac{1}{x}} $ when is the integrating factor negative also,thanks.
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  4. #4
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    Krizalid's Avatar
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    It's actually $\displaystyle \exp \left\{ -\int{\frac{1}{x}\,dx} \right\}.$
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  5. #5
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    hmmm why negative.
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    When you put your equation in standard form

    $\displaystyle y'+a(x)y=f(x)$

    $\displaystyle y'=\frac{y}{x} +2x^{2}$

    you need to move the $\displaystyle \frac{y}{x}$ term to the left side of the equation

    $\displaystyle y'-\frac{y}{x}=2x^2$

    Good luck
    Last edited by TheEmptySet; May 24th 2008 at 07:28 AM. Reason: don't know my left from my right :)
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