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Math Help - Rearranging Differential Equation

  1. #1
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    Rearranging Differential Equation

    Hi, I was going over my notes and didnt understand one of the steps made in a calculation, was hoping someone could explain it to me.

    Im trying to find an analytic solution for an ODE: dY/dX + X*Y = X
    The integrating factor is I = e^((X^2)/2)

    Multiplying the equation with the integrating factor

    I*(dY/dX) + X*I*Y = X*I

    This next step I dont follow where it gets simplified to:

    (d/dX)*(Y*e^((X^2)/2)) = X*(e^((X^2)/2))

    The following line's after are:

    YI = I

    Y = 1 + C*I

    Any help would be appriciated, thank you.
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  2. #2
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    The step you're wondering about is the heart of the integrating factor procedure. Suppose you had the DE

    \dfrac{dy}{dx}+P(x)\,y=Q(x), and you multiply through by the integrating factor e^{\int P(x)\,dx}. Then you get

    e^{\int P(x)\,dx}\dfrac{dy}{dx}+e^{\int P(x)\,dx}P(x)\,y=Q(x)e^{\int P(x)\,dx}.

    The whole point of the procedure is that the LHS is now a total derivative. Indeed, if we examine

    \dfrac{d}{dx}\left[e^{\int P(x)\,dx} y\right]=\left(e^{\int P(x)\,dx}\right)\left(\dfrac{dy}{dx}\right)+y\left  (\dfrac{d}{dx}e^{\int P(x)\,dx}\right)=<br />
e^{\int P(x)\,dx}\,\dfrac{dy}{dx}+ye^{\int P(x)\,dx}\,P(x),
    which is just our new LHS after we multiplied through by the integrating factor.

    Someone was clever enough to discover that the integrating factor was indeed e^{\int P(x)\,dx}, and after that, it became a procedure.

    Does that clear things up a bit for you, perhaps?
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  3. #3
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    Quote Originally Posted by Miika89 View Post
    This next step I dont follow where it gets simplified to:

    (d/dX)*(Y*e^((X^2)/2)) = X*(e^((X^2)/2))
    If you consider the product rule before this line you had

    vu'+uv' = f(x)

    (uv)' = f(x)

    And integrating both sides

    uv = \int f(x)
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  4. #4
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    Yes, thank you.

    If I didnt know this procedure is there a way of getting from I*(dY/dX) + X*I*Y = X*I to YI = I? Or would you suggest its too complicated and I just memorise how this procedure works?
    ------------------------------------------------------------------------------

    Thank you both, I think pickslides just answered my 2nd question
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  5. #5
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    Quote Originally Posted by Miika89 View Post
    Yes, thank you.

    If I didnt know this procedure is there a way of getting from I*(dY/dX) + X*I*Y = X*I to YI = I? Or would you suggest its too complicated and I just memorise how this procedure works?
    I say memorise it. It comes in very handy and enhances your understanding of calculus.
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  6. #6
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    I'd say as well as memorising it i.e. that after you have multiplied by the Integrating Factor the LHS reduces to \displaystyle \frac{d}{dx}(I\,y), get good at recognising Product Rule expansions, i.e. that \displaystyle u\,\frac{dv}{dx} + v\,\frac{du}{dx} = \frac{d}{dx}(u\,v).
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  7. #7
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    Not having gotten much further, Im stuck on how this next stage works:

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

    I am assuming he has now integrated the RHS?
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  8. #8
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    Given \displaystyle I= e^{\frac{x^2}{2}}

    \displaystyle Iy'+Ixy = xI

    Note \displaystyle I'=Ix then

    \displaystyle (Iy)' = xI

    \displaystyle Iy = \int xI~dx

    \displaystyle e^{\frac{x^2}{2}}y = \int xe^{\frac{x^2}{2}}~dx

    Then by substitution \displaystyle u = \frac{x^2}{2} integrate the right hand side.

    Final step is to divide both sides by \displaystyle e^{\frac{x^2}{2}}
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