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Math Help - [SOLVED] Linear DE

  1. #1
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    [SOLVED] Linear DE

    Hi, the question is

    y'=\frac{y}{x}+2x^2
    and y(1)=0

    I put it to this form

    y' -y(x^{-1})=2x^2

    so f(x)=x^{-1} and h(x)=2x^2

    is that right?

    Integrating Factor I(x)=exp(\int f(x)dx)
    =exp(\int x^{-1}dx)

    which means I(x) = e

    So i mulptiply both sides by e?

    \frac{d}{dx}ey=2ex^2

    and then i'm lost and confused, any help will be marvellous
    Last edited by Rapid_W; June 4th 2009 at 06:50 AM.
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  2. #2
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    Quote Originally Posted by Rapid_W View Post
    Hi, the question is

    y'=\frac{y}{x}+2x^2
    and y(1)=0

    I put it to this form

    y' -y(x^{-1})=2x^2

    so f(x)=x^{-1} and h(x)=2x^2

    is that right?

    Integrating Factor I(x)=exp(\int f(x)dx)
    =exp(\int x^{-1}dx)

    which means I(x) = e

    So i mulptiply both sides by e?

    \frac{d}{dx}ey=2ex^2

    and then i'm lost and confused, any help will be marvellous
    I(x)=exp(\int - x^{-1}dx) = e^{- \ln x} = \frac{1}{x}

    Now continue.
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  3. #3
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    Where is the dunce hat, I need it! Omg, if I can't remember x^(-1) integrates to ln(x) how am I going to pass my tests
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  4. #4
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    Nah, i can't work it out still

    I obvioulsy now know that I(x) = 1/x so i mulitply the equation by this...

    [tex] y' \frac{y}{x} = \frac{2x^2}{x}[/maTH]

    so now i'm meant to integrate both sides, i now get this

    ylnx = x^2 + c

    So now what do i do? The initial value is y(1)=0
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  5. #5
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    Your equation was:

    y' - \frac{1}{x} y = 2 x^2 .

    Multiplying by the integrating factor gives you:

    \frac{1}{x} y' - \frac{1}{x^2} y = 2 x

    \Leftrightarrow \frac{d}{dx} \left( \frac{y}{x} \right) = 2 x

    so

    \frac{y}{x} = \int 2x \, \mathrm{d}x.

    You should be able to take it from there, right?
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  6. #6
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    I found the answer doing it this way...

    f(x)=-x^{-1}
    h(x)=2x^2
    I(x)=\frac{1}{x}
    so
    \int I(x)h(x)dx
    =\int (\frac{1}{x})(2x^2)dx
    =\int 2x
    x^2+c

    z(x)=x^2+c

    general solution = y=\frac{z(x)}{I(x)}[/tex]
    so
    y=\frac{x^2+c}{\frac{1}{x}}
    =x^3+cx

    and the initial value was y(1)=0

    so 0=1^3+c
    c=-1
    so finaly
    y=x^3-x
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