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    Suppose that y (a function of x) satisfies the following:

    y+dy/dx=xe^-x

    where dy/dx is the derivative of y. It is known also that y=1 when x=0.
    Suppose that the function z is obtained from y as follows:

    z=ye^x

    Show that

    dz/dx=x

    Use this to find an expression for y in terms of x.

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    Applying the product rule to z = y * e^x gives

    dz/dx
    = dy/dx * e^x + y * e^x
    = e^x * (dy/dx + y).

    Since we know that dy/dx + y = x * e^(-x) it follows that

    dz/dx = e^x * x * e^(-x) = x.

    So z = (1/2)x^2 + constant.

    When x = 0 we have y = 1 and hence z = 1. So z = (1/2)x^2 + 1.

    So y = e^(-x) * z = e^(-x) * [(1/2)x^2 + 1].
 
 
 

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