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First order ordinary differential equation help please! watch

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    This is the first time ie used integrating factors and I am close to the solution gien..heres my working:

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

     I = e^{\int Pdx}

     = e^{\int \frac{3}{x}}dx

     \int \frac{3}{x}dx = 3lnx = lnx^3

     I = e^{lnx^3} = x^3

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

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

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

     \int x^3\frac{dy}{dx}dx + \int 3x^2ydx = \int e^xdx

     \int x^3dy + y \int 3x^2dx = e^x + C

     x^3 \int{dy} + y \int 3x^2dx = e^x + C

     yx^3 + yx^3 = e^x + C

     2yx^3 = e^x + C

     y = \frac{e^x + C}{2x^3}

    They have:

     y = \frac{e^x + C}{x^3}

    Where have I gone wrong please?

    Thanks
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    It's all correct until you started integrating - you can't take out variables from the integrals like that. The idea of the integrating factor is that you can get it into a form where you can integrate it by inspection, since it is in the form v du/dx + u dv/dx, which is the product rule for differentiation. If you compare x³ dy/dx + 3x²y with that form, can you see what it must integrate to?
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    (Original post by Chemhistorian)
    This is the first time ie used integrating factors and I am close to the solution gien..heres my working:

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

     I = e^{\int Pdx}

     = e^{\int \frac{3}{x}}dx

     \int \frac{3}{x}dx = 3lnx = lnx^3

     I = e^{lnx^3} = x^3

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

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

     x^3\frac{dy}{dx} + 3x^2y = e^x
    It is right up to this.
    You can not integrate pointwise, but now you can write the equation as
     \frac{d}{dx} \left (x^3y)=e^x
    using the product rule for differentiation.
    This form is why you use integration factor, I think.
    (the equation may be solved another ways, too).
    Multiplying by dx you get differential of
    d(x^3y)=e^xdx
    and integrating this gives the solution


     \int x^3\frac{dy}{dx}dx + \int 3x^2ydx = \int e^xdx

     y = \frac{e^x + C}{2x^3}

    They have:

     y = \frac{e^x + C}{x^3}

    Where have I gone wrong please?

    Thanks
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    Just to add to that - the integrating factor method is obtained from the assumption that the left hand side can be written as a derivative of a product, and from this assumption the factor e^integral of P dx is obtained. Because of that, whenever you multiply a first order ordinary differential equation by the integrating factor, the left hand side will always be in the form of a derivative of a product.
 
 
 
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