Advanced Engineering Mathematics, 6th Edition by Peter V. O'Neil

By Peter V. O'Neil

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We saw this previously by showing that this differential equation can have no potential function. 18 Consider x2 + 3xy + 4xy + 2x y = 0 Here M x y = x2 + 3xy and N x y = 4xy + 2x. Now N = 4y + 2 x and M = 3x y and 3x = 4y + 2 is satisfied by all x y on a straight line. However, N/ x = M/ y cannot hold for all x y in an entire rectangle in the plane. Hence this differential equation is not exact on any rectangle. 19 Consider ex sin y − 2x + ex cos y + 1 y = 0 With M x y = ex sin y − 2x and N x y = ex cos y + 1, we have M N = ex cos y = x y for all x y .

Next, compute x = x x M x0 = M x y0 + = M x y0 + y0 d + y y0 y y0 y Nx x d y0 N x x d M x y d = M x y0 + M x y − M x y0 = M x y and the proof is complete. For example, consider again y + y = 0. Here M x y = y and N x y = 1, so N =0 x and M =1 y throughout the entire plane. Thus, y + y = 0 cannot be exact on any rectangle in the plane. We saw this previously by showing that this differential equation can have no potential function. 18 Consider x2 + 3xy + 4xy + 2x y = 0 Here M x y = x2 + 3xy and N x y = 4xy + 2x.

This is a simple enough problem for, say, a soda can, but not quite so easy for a large oil storage tank or chemical facility. We need two principles from physics. The first is that the rate of discharge of a fluid flowing through an opening at the bottom of a container is given by dV = −kAv dt in which V t is the volume of fluid in the container at time t, v t is the discharge velocity of fluid through the opening, A is the cross sectional area of the opening (assumed constant), and k is a constant determined by the viscosity of the fluid, the shape of the opening, and the fact that the cross-sectional area of fluid pouring out of the opening is slightly less than that of the opening itself.

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