 By Peter V. O'Neil

ISBN-10: 0495082376

ISBN-13: 9780495082378

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Sample text

If we cannot find an integrating factor that is a function of just x or just y, then we must try something else. 13) and be observant. 22 Consider 2y2 − 9xy + 3xy − 6x2 y = 0. 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.

15 Solve the initial value problem y = 3x2 − y x y 1 =5 First recognize that the differential equation can be written in linear form: 1 y + y = 3x2 x An integrating factor is e to get 1/x dx = eln x = x, for x > 0. Multiply the differential equation by x xy + y = 3x3 or xy = 3x3 Integrate to get 3 xy = x4 + C 4 Then C 3 y x = x3 + 4 x for x > 0. For the initial condition, we need y 1 =5= 3 +C 4 so C = 17/4 and the solution of the initial value problem is 17 3 y x = x3 + 4 4x for x > 0. 6 in closed form (as a finite algebraic combination of elementary functions).

2y 1 + x2 + xy = 0 y 2 = 3 (Hint: try 8. y2 + y − xy = 0 19. sin x − y + cos x − y − cos x − y y = 0 10. 2y2 − 9xy + 3xy − 6x2 y = 0 (Hint: try xa y b ) 11. y + y = y (Hint: try x y = x y =e y ) 12. x2 y + xy = −y−3/2 (Hint: try ax b x y = xa y b ) In each of Problems 13 through 20, find an integrating factor, use it to find the general solution of the differential equation, and then obtain the solution of the initial value problem. 6 2 = xa ebx ) y 0 = 7 /6 9. 2xy2 + 2xy + x2 y + x2 y = 0 4 2 = ya ebx ) 20.