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(x - 2xy + e^y) dx + (y - x^2 + xe^y) dy = 0

Okay so it is the standard convenient exact equation for newbies. Now here is the part that confuses me.

(a) Let P(x,y) = (x - 2xy + e^y) dx & Q(x,y) = (y - x^2 + xe^y) dy

The function is defined for all real Numbers on an x,y plane. So we start with the initial integration of x_i = 0 & y_i = 0

Now the author reasons that the integration of Q(x,y) is simply equal to the integration of y

(from 0 to y) ∫ (y - x^2 + xe^y) dy = (from o to y) ∫ y dy

Okay so it makes sense, since the initial value of x is zero, and since we are keeping x constant, it therefore follows that x will be 0 for all values and we can simply leave it at that.

Now the part that confuses me is that he does not do the same for the other function P(x,y) dx regardless of the fact that y_i = 0.

(a) gives x^2/2 - x^2y + xe^y + y^2/2 = c

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# Exact Equations

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