MATH 202 Exam 2, Spring 2002

MATHEMATICS 202 SPRING SEMESTER 2006-07 QUIZ II Time: 70 MINUTES. Date: April 28, 2007. Name:—————————————————————————- ID Number:———————————————————————– Section Number:—————————————————————— Course Instructors: Professors Abdallah Lyzzaik and Hassan Yamani Question Grade 1 /20 2 /20 3 /20 4 /20 5 /10 6 /10 TOTAL /100 Answer The Following Six Questions On The Page Allocated For Each Question (You May Use The Back Of The Pages If Needed). 1 2 1. Find the general solution of the differential equation y(4) + 2y′′′ + 11y′′ + 2y′ + 10y = 0 knowing that one of its solutions is y = cos x. (20 points) Solution. Since cos x is a solution, two characteristic root are ±i. Thus, by Synthetic Division by ±i, the CE is found to be m4 + 2m3 + 11m2 + 2m + 10 = (m − i)(m + i)(m2 + 2m + 10) = 0 and the characteristic root are ±i. and −1 ± 3i. Hence, the GS for the differential equation is y − c1 cos x + c2 sin x + e−x(c3 cos 3x + c4 sin 3x), x ∈ R. 3 2. Find the general solution of the differential equation x2y′′ − xy′ + y = x2. (20 points) Solution. This is a nonhomogeneous Cauchy-Euler differential equa- tion with CE m2 − 2m + 1 = (m − 1)2 = 0. Hence the complementary solution is yc = c1x + c2x ln x. By the method of VP, a particular solution is yp = u1x + u2x ln x, where u′ 1 = W1/W and u′ 2 = W2/W, with W = ����� x x ln x 1 1 + ln x ����� = x, W1 = ����� 0 x ln x 1 1 + ln x ����� = −x ln x, and W2 = ����� x 0 1 1 ����� = x. Thus, u′ 1 = − ln x and u′ 2 = 1, and consequently, u1 = − � ln x dx = x − x ln x and u2 = � dx = x. Hence, yp = (x − x ln x)x + x2 ln x = x2 and the GS is y = c1x + c2x ln x + x2. 4 3. Use the method of undetermined coefficients to find the general solution of the differential equation y′′ + y = ex + sin x. (20 points) Solution. This is a nonhomogeneous linear differential equation with CE m2 + 1 = 0. Hence the characteristic roots are m = ±i, and the complementary solution is yc = c1 cos x + c2 sin x, x ∈ R. A particular solution of the differential equation is yp = yp1 + yp2, where yp1 and yp2 are particular solutions of the differential equation y′′ + y = ex and y′′ + y = sin x, respectively. A particular solution of the first differential equation is given by yp1 = x(a cos x + b sin x). Then y′ p1 = (a cos x + bsinx) + x(−a sin x + b cos x) and y′′ p1 = −2a sin x + 2b cos x + x(−a cos x − b sin x). Hence, by substitution, in the first differential equation, we obtain −2a sin x + 2b cos x = sin x, which gives a = −1/2 and b = 0. Thus yp1 = −(x cos x)/2. A particular solution of the second differential equation is given by yp2 = aex. Then y′ p2 = y′′p2 = aex. Hence, by substitution, in the second differential equation, we obtain 2aex = ex, or a = 1/2. Thus yp2 = ex/2. Therefore, yp = −(x cos x)/2 + ex/2, and the GS is c1 cos x + c2 sin x − (x cos x)/2 + ex/2, x ∈ R. 5 4. Use the substitution x = et to transform the Cauchy-Euler differ- ential equation x2y′′ + 10xy′ + 8y = x2 to a differential equation of constant coefficients, then solve the differ- ential equation. Show all the details of your work. (20 points) Solution. By substitution, we find y′ = 1 xy., y′′ = 1 x2(y.. − y.), and the differential equation becomes y.. + 9y. + 8y = e2t. Hence, the CE is m2 + 9m + 8 = (m + 1)(m + 8) = 0 and the CRs are −1, −8. Thus complementary solution of the differential equation is yc = c1e−t + c2e−8t. Let yp = Ae2t. By substitution in the latter differential equation yields, (4A + 18A + 8A)e2t = e2t. Then A = 1/30, and yp = e2t/30. therefore, the GS in t is y = c1e−t + c2e−8t + e2t/30, and in terms in it is y = c1x−1 + c2x−8 + x2/30, 6 5. Find the general solution of the differential equation xy′′ − (x + 1)y′ + y = 0 knowing that y1 = ex is a solution. (10 points) Solution. Write the DE as y′′ − x + 1 x y′ + 1 xy = 0. By the Reduction of Order method, y2 = ex � exp � ((x + 1)/x) dx e2x dx = ex � xe−x dx = −x − 1. Hence, the GS is y = c1ex + c2(x + 1), x ∈ R. 7 6. Can the set {x2, x3} be a fundamental set over (−∞, ∞) for a linear homogeneous differential equation a2(x)y′′ + a1(x)y′ + a0(x)y = 0, where a2, a1, and a0 are continuous functions of (−∞, ∞) with a2(x) ̸= 0 for all x? Justify your answer. Solution. The set {x2, x3} is not a fundamental set over (−∞, ∞) for any of the suggested differential equations since the Wronskian W(x2, x3) = ����� x2 x3 2x 3x2 ����� = x4 is zero when x = 0 and nonzero otherwise. (10 points)