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1. 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.

Too difficult for me
2. 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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Updated: March 19, 2004
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