By Peter V. O'Neil

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**Extra info for Advanced Engineering Mathematics, International Student Edition **

**Sample text**

Say we begin with the first. Then integrate with respect to x: x y = x dx = 2xy3 + 2 dx = x2 y3 + 2x + g y In this integration with respect to x we held y fixed, hence we must allow that y appears in the “constant” of integration. If we calculate / x, we get 2xy2 + 2 for any function g y . Now we know to within this function g. Use the fact that we know / y to write y = 3x2 y2 + 8e4y = y x2 y3 + 2x + g y = 3x2 y2 + g y This equation holds if g y = 8e4y , hence we may choose g y = 2e4y . This gives the potential function x y = x2 y3 + 2x + 2e4y If we had chosen to integrate / y first, we would have gotten x y = 3x2 y2 + 8e4y dy = x2 y3 + 2e4y + h x Here h can be any function of one variable, because no matter how h x is chosen, y x2 y3 + 2e4y + h x as required.

However, is an integrating factor. 5 e p x dx =p x e p x dx = y p x y−q x e p x dx PROBLEMS 1. Determine a test involving M and N to tell when M + Ny = 0 has an integrating factor that is a function of y only. (b) Find an integrating factor of x alone. x that is a function 2. Determine a test to determine when M + Ny = 0 has an integrating factor of the form x y = xa yb for some constants a and b. (c) Find an integrating factor of y alone. y that is a function 3. Consider y − xy = 0. (a) Show that this equation is not exact on any rectangle.

This is not exact. 14) / y = 0 and we obtain 3xy − 6x2 x + 3y − 12x = 4y − 9x which cannot be solved for as just a function of x. Similarly, if we try = y , so / x = 0, we obtain an equation we cannot solve. We must try something else. 14) involves only integer powers of x and y. This suggests that we try x y = xa yb . 14) and attempt to choose a and b. The substitution gives us 3axa yb+1 − 6axa+1 yb + 3xa yb+1 − 12xa+1 yb = 2bxa yb+1 − 9bxa+1 yb + 4xa yb+1 − 9xa+1 yb Assume that x = 0 and y = 0.