9 11 Solve the initial-value problem Evaluate the integral. 1. y sin 3 x cos 2 x dx. calculation. 1 + i i23

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1 Mock Exam Solve the differential equation. 7. d dt te t s1 Mock Exam 9 11 Solve the initial-value problem. 9. x ln x, Match the differential equation with its direction field (labeled I IV). Give reasons for our answer I 1 x 4. x 6. 3 x 3 II 9 10 Sketch a direction field for the differential equation. Then use it to sketch three solution curves Evaluate the integral. 1. sin 3 x cos x dx _ x _ x Find the general solution to the differential equation = 0. III IV Find the particular solution to the initial value problem - + = 0, (0) = 1, (0) = -1. _ x _ x Find a particular solution of - = sin x. 7. Use the direction field labeled I (for Exercises 3 6) to sketch the graphs of the solutions that satisf the given initial conditions. (a) 0 1 (b) 0 0 (c) 0 1 Express the complex numbers in Exercises in the form re iu, with r Ú 0 and -p 6 u p. Draw an Argand diagram for each calculation. 1 + i i Evaluate the integral t 4 t 1 dt 18. Find the two square roots of i. Solve the boundar value problem +4 = 0, (0) = 0, a p 1 b = 1. Draw the phase plane diagram of the sstem ẋ = 4 ẏ = 1 3x Solve the nonhomogeneous equation - -3 = 1 - x. Express the complex numbers in Exercises in the form re iu, with r Ú 0 and -p 6 u p. Draw an Argand diagram for each calculation. 1 + i 11. A1 + -3B i

2 Mock Exam 1 solutions Solution The auxiliar equation is r + 4 = 0, which has the complex roots r = ;i. The general solution to the differential equation is = c 1 cos x + c sin x. The boundar conditions are satisfied if (0) = c 1 # 1 + c # 0 = 0 a p. 1 b = c 1 cos a p 6 b + c sin a p 6 b = 1 It follows that c 1 = 0 and c =. The solution to the boundar value problem is = sin x.

3 Solution The auxiliar equation for the complementar equation - -3 = 0 is It has the roots r = -1 and r = 3 giving the complementar solution c = c 1 e - x + c e 3x. Now G(x) = 1 - x is a polnomial of degree. It would be reasonable to assume that a particular solution to the given nonhomogeneous equation is also a polnomial of degree because if is a polnomial of degree, then - -3 is also a polnomial of degree. So we seek a particular solution of the form We need to determine the unknown coefficients A, B, and C. When we substitute the polnomial and its derivatives into the given nonhomogeneous equation, we obtain p or, collecting terms with like powers of x, r - r - 3 = (r + 1)(r - 3) = 0. p = Ax + Bx + C. A - (Ax + B) - 3(Ax + Bx + C) = 1 - x -3Ax + (-4A - 3B)x + (A - B - 3C) = 1 - x. This last equation holds for all values of x if its two sides are identical polnomials of degree. Thus, we equate corresponding powers of x to get -3A = -1, -4A - 3B = 0, and A - B - 3C = 1. These equations impl in turn that A 1>3, B -4>9, and C 5>7. Substituting these values into the quadratic expression for our particular solution gives p = 1 3 x x B Theorem 7, the general solution to the nonhomogeneous equation is = c + p = c 1e - x + c e 3x x x A1 + -3B = = i 1 - i =

4 Mock Exam solutions Solution The auxiliar equation is r - 4r + 5 = 0. The roots are the complex pair r = (4 ; 16-0)> or r 1 = + i and r = - i. Thus, a = and b = 1 give the general solution = e x (c 1 cos x + c sin x). Solution The auxiliar equation is r - r + 1 = (r - 1) = 0. The repeated real root is r = 1, giving the general solution = c 1 e x + c xe x. Then, =c 1 e x + c (x + 1)e x. From the initial conditions we have 1 = c 1 + c # 0 and -1 = c1 + c # 1. Thus, c 1 = 1 and c = -. The unique solution satisfing the initial conditions is = e x - xe x.

5 Solution If we tr to find a particular solution of the form and substitute the derivatives of equation in the given equation, we find that A must satisf the for all values of x. Since this requires A to equal both - and 0 at the same time, we conclude that the nonhomogeneous differential equation has no solution of the form A sin x. It turns out that the required form is the sum p = A sin x + B cos x. The result of substituting the derivatives of this new trial solution into the differential equation is or p -A sin x + A cos x = sin x -A sin x - B cos x - (A cos x - B sin x) = sin x (B - A) sin x - (A + B) cos x = sin x. This last equation must be an identit. Equating the coefficients for like terms on each side then gives B - A = p = A sin x and A + B = 0. Simultaneous solution of these two equations gives A = -1 and B = 1. Our particular solution is p = cos x - sin x i3 1 - i3 = 18. ± = ± = ± Phase diagram: Note that the equilibrium points are found b solving 4 0, 1 3x 0 x, and hence (-, ) and (, ) are the equilibrium points. (-, ) is a center and (, ) is a saddle point.

PRE-LEAVING CERTIFICATE EXAMINATION, 2010

L.7 PRE-LEAVING CERTIFICATE EXAMINATION, 00 MATHEMATICS HIGHER LEVEL PAPER (300 marks) TIME : ½ HOURS Attempt SIX QUESTIONS (50 marks each). WARNING: Marks will be lost if all necessary work is not clearly