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Example 5

EXAMPLE 5

Solve the differential equation

(x2 + sin y) dx + (x + 1) cos y dy = 0.

First we test for exactness.

13_vector_calculus-144.gif

Next we find a function with the given total differential. That is, we find a potential function for the vector field

(x2 + sin y)i + (x + 1) cos yj.

Step 1

Take (0,0) for the initial point.

Step 2

Let C be the rectangular curve from (0,0) to (0, y0) to (x0,y0), shown in Figure 13.3.6.

13_vector_calculus-145.gif

Figure 13.3.6

Step 3

13_vector_calculus-146.gif

f(x, y) = sin y + ⅓x3 + x sin y.

Step 4

f(x, y) is a constant k because df = 0. The general solution is

13_vector_calculus-147.gif

Home Vector Calculus Independence of Path Examples Example 5

Last Update: 2006-11-22