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Math Help - Differential Equation

  1. #1
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    Differential Equation

    xy' + 3y = sin x/x^2 ive tried using the "general" equation for a linear first order equation but to no avail. any help would be appreciated. im stuck at the bit where i have to integrate sin(x)/x^3 btw
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    Why? You only need to integrate \sin{x}. First, switch the DE to standard form. Then multiply the whole DE by x^3, which is the integrating factor. You've probably forgotten to multiply x^3 in the right-hand side.
    Last edited by nocturnal; October 4th 2008 at 01:54 AM. Reason: std. form
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    can u explain more in detail, im afraid i dont understand what u mean by "normal" form
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  4. #4
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    Ack, I meant standard form. That is, the coefficient of y' should be 1. Afterwards, find the integrating factor. Are you familiar with solving first-order linear ODE's?
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  5. #5
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    yes i am familiar with that and i have tried doing it, it doesnt work, can u just give it a go. im pretty sure i have done nothing wrong
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  6. #6
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    x\frac{dy}{dx} + 3y = \frac{\sin{x}}{x^2}

    \frac{dy}{dx} + \frac{3}{x}y = \frac{\sin{x}}{x^3}


    The Integrating Factor is e^{\int \frac{3}{x}\, dx} = e^{3\ln x} = e^{\ln x^3} = x^3.

    Multiply both sides by x^3.


    x^3\frac{dy}{dx} + 3x^2y = \sin{x}

    The left hand side is a product rule expansion of \frac{d}{dx}(x^3y).


    \frac{d}{dx}(x^3y) = \sin{x}

    x^3y = \int \sin{x} \, dx

     x^3y = -\cos{x} + C

     y = -x^{-3}\cos{x} + Cx^{-3}.
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  7. #7
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    Quote Originally Posted by ah-bee View Post
    can u explain more in detail, im afraid i dont understand what u mean by "normal" form
    Normal form is for second-order equations; that is, y^{\prime\prime}(t)+p(t)y(t)=0.
    It is not related for this equation.
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