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Math Help - 1st order linear differential equation

  1. #1
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    1st order linear differential equation

    can anyone explain me the principle to solve:

    y' + 2y = x^2
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  2. #2
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    This equation is first-order linear. There's a procedure called variation of parameters, or the integrating factor method, that you can use to solve this equation. Take a look here for detailed instructions on solving these equations, along with some examples.
    Last edited by mr fantastic; November 20th 2010 at 11:54 AM.
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  3. #3
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    Consider a DE of the form

    \displaystyle \frac{dy}{dx} + P\,y = Q where \displaystyle y, P, Q are all functions of \displaystyle x and you hope to solve the DE for \displaystyle y. We call this type of DE First-Order Linear.

    What we aim to do is to find a function of \displaystyle x, which we will call \displaystyle I (I use \displaystyle I because it's called the Integrating Factor) which will enable us to write the LHS as a single derivative.

    Let's say we had multiplied both sides by \displaystyle I. That would give

    \displaystyle I\,\frac{dy}{dx} + I\,P\,y = I\,Q.


    Now let's take a closer look at the product rule.

    If you wished to find the derivative of \displaystyle f\,y, where \displaystyle f and \displaystyle y are functions of \displaystyle x. Using the product rule this would give

    \displaystyle \frac{d}{dx}(f\,y) = f\,\frac{dy}{dx} + \frac{df}{dx}\,y.

    The LHS of your DE (after multiplying by \displaystyle I) looks suspiciously similar. In fact, if \displaystyle I = f and \displaystyle I\,P = \frac{df}{dx}, then it IS in fact a product rule expansion. Note that if \displaystyle I=f then \displaystyle \frac{dI}{dx} = \frac{df}{dx}.

    Substituting gives \displaystyle I\,P = \frac{dI}{dx}, which is separable, and solving for \displaystyle I gives

    \displaystyle P = \frac{1}{I}\,\frac{dI}{dx}

    \displaystyle \int{P\,dx} = \int{\frac{1}{I}\,\frac{dI}{dx}\,dx}

    \displaystyle \int{P\,dx} = \int{\frac{1}{I}\,dI}

    \displaystyle \int{P\,dx} + C = \ln{|I|}

    \displaystyle Ae^{\int{P\,dx}} = I where \displaystyle A = \pm e^C.


    Any value of \displaystyle A will work when we multiply the DE by the integrating factor, so we choose \displaystyle A = 1 for simplicity.

    So you will need to multiply through by your integrating factor, which is of the form \displaystyle e^{\int{P\,dx}}, and that will reduce your LHS into a product rule expansion, which will then make the function integrable.



    So looking at your example...

    \displaystyle \frac{dy}{dx} + 2y = x^2.

    Notice that your \displaystyle P = 2, thus your integrating factor is \displaystyle e^{\int{2\,dx}} = e^{2x}.

    Multiplying through by your integrating factor gives

    \displaystyle e^{2x}\,\frac{dy}{dx} + 2e^{2x}y = x^2e^{2x}

    \displaystyle \frac{d}{dx}(e^{2x}y) = x^2e^{2x} (check using the product rule...)

    \displaystyle e^{2x}y = \int{x^2e^{2x}\,dx}

    \displaystyle e^{2x}y = \frac{1}{2}x^2e^{2x} - \int{x\,e^{2x}\,dx}

    \displaystyle e^{2x}y = \frac{1}{2}x^2e^{2x} - \left(\frac{1}{2}x\,e^{2x} - \int{\frac{1}{2}e^{2x}\,dx}\right) + C

    \displaystyle e^{2x}y = \frac{1}{2}x^2e^{2x} - \frac{1}{2}x\,e^{2x} + \frac{1}{4}e^{2x} + C

    \displaystyle y = \frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{4} + Ce^{-2x}.
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  4. #4
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    thanks a lot for your explanation
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