MATH 202 Exam 2, Spring 2002

MATHEMATICS 202 SPRING SEMESTER 2007-08 SOLUTION OF QUIZ II Time: 70 MINUTES Date: April 19, 2008 Name:—————————————————————————- ID Number:———————————————————————– Section Number:—————————————————————— Course Instructors: Professors Abdallah Lyzzaik and Dolly Fayyad Question Grade 1 /15 2 /15 3 /15 4 /15 5 /15 6 /15 7 /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. Solve the initial-value problem xdy dx + 4y = x4y2; y(1) = 1. (15 points) Solution. By writing the DE as dy dx + 4 xy = x3y2, it becomes a Bernoulli DE of degree 2. Now write the latter equation as y−2 dy dx + 4 xy−1 = x3, and let u = y−1. Then −u′ = y−2y′ and substitution yields du dx + −4 x u = −x3, which is a linear DE of order 1 with an integrating factor µ(x) = � exp �−4 x � dx = 1 x4. Multiplying µ with the latter DE yields 1 x4 du dx + −4 x5 u = −x−1, or d dx � u x4 � = −x−1. Integration of both sides with respect to x gives � u x4 � = − ln |x| + c, or � 1 yx4 � = − ln |x| + c. Since y(1) = 1, c = 1 and the solution of the initial-value problem is y = 1 x4(1 − ln |x|). 3 2. Find the general solution of the differential equation (3x2y + y3) dx + (x3 + 3xy2) dy = 0. (15 points) Solution. Since 3x2y+y3 and 3x2y+y3 are homogenous functions of same degree 3, the DE is homogenous. If y = ux, then dy = udx+xdu and, by substitution the DE becomes x3(u2 + 3)dx + x3(3u2 + 1)(udx + xdu) = 0, or 4u(u2 + 1)dx + x(3u2 + 1)du = 0, or dx x + 1 4 �1 u + 2u u2 + 1 � du = 0. Integration of both sides then gives ln � x4|u|(u2 + 1) � = c1, or ln � x4 ���� y x ���� ��y x �2 + 1 �� = c1, or xy(x2 + y2) = c, where c ∈ R since by inspection y = 0 is a trivial solution of the DE, is the general solution of the DE. 4 3. Find the general solution of the differential equation cos x dx + � 1 + 2 y � sin x dy = 0 by finding an appropriate integrating factor. (15 points) Solution. Let M = cos x and N = (1 + 2/y) sin x. Since My = 0 ̸= � 1 + 2 y � cos x = Nx, the DE is not exact. But My − Nx N = − cot x; hence the DE has integrating factor µ(x) = exp � (− cot x)dx = 1 sin x. Multiplying this with the DE yields the equation cot x dx + � 1 + 2 y � dy = 0 which is exact and separable. Integration of both sides gives ln | sin x| + y + ln y2 = c1, or y2ey sin x = c, where c ̸= 0 since y = 0 is not a solution of the DE. 5 4. Find the general solution of the differential equation e−xy′′ − 2e−xy′ + 2e−xy = tan x. (15 points) Solution. Write the DE as y′′ − 2y′ + 2y = ex tan x. which is a linear nonhomogeneous DE of constant coefficients with auxiliary equation m2 − 2m + 2 = 0 whose roots are m = 1 ± i. Thus the complementary solution is yc = ex(c1 cos x + c2 sin x); x ∈ R, By the variation of parameters method a particular solution of the DE is yp = ex(u1 cos x + u2 sin x); x ∈ R, where u′ 1 = W1/W, u′ 2 = W2/W, W = ����� ex cos x ex sin x ex(cos x − sin x) ex(cos x + sin x) ����� = e2x, W1 = ����� 0 ex sin x ex tan x ex(cos x + sin x) ����� = −e2x sin x tan x, and W2 = ����� ex cos x 0 ex(cos x − sin x) ex tan x ����� = e2x sin x. Thus u′ 1 = − sin x tan x and u′ 2 = sin x; hence u1 = ln | sec x − tan x| + sin x and u2 = −cosx. Thus yp = ex cos x ln | sec x − tan x|; x ∈ R, and the general solution of the DE is y = ex(c1 cos x + c2 sin x) + ex cos x ln | sec x − tan x|. 6 5. Solve by two different methods the Cauchy-Euler differential equa- tion x3y′′′ + xy′ − y = 0. (15 points) Solution. Method 1. By using the substitution x = et, we find y′ = y./x, y′′ = (y.. − y.)/x2, and y′′′ = (y... − 3y.. + 2y.), where y. = dy/dt, y.. = d2y/dt2, and y..0 = d3y/dt3. and the differentia equation becomes (y... − 3y.. + 3y. − y = 0 which is a homogenous linear DE with constant coefficients whose aux- iliary equation is m3 − 3m2 + 3m − 1 = (m − 1)3 = 0 and a single root m = 1 repeated thrice. Thus the general solution of the differential equation is y = (c1 + c2t + c3t2)et = (c1 + c2 ln x + c3(ln x)2)x, x > 0. Method 2. By using the substitution y = xm the differential equation becomes x3(m3 − 3m2 + 3m − 1) = 0, or for x > 0, m3 − 3m2 + 3m − 1 = (m − 1)3 = 0, which is the auxiliary equation for the DE. Thus the DE has a funda- mental set {x, x ln x, x(ln x)2} and the general solution of the differen- tial equation is y = (c1 + c2 ln x + c3(ln x)2)x, x > 0. 7 6. Find the general solution of the differential equation y′′′ − 2y′′ − 4y′ + 8y = 6xe2x by using the method of undetermined coefficients. (Hint: m=2) (15 points) Solution. This is a homogenous linear DE with constant coefficients whose auxiliary equation is m3 − 2m2 − 4m + 8 = (m − 2)2(m + 2) = 0 and whose roots are m = 2(twice) and m = −2. Thus the complemen- tary solution is yc = (c1 + c2x)e2x + c3e−2x. By the method of undetermined coefficients the particular solution has the form yp = x2(ax + b)e2x. To find a and b we write y′ p = [2ax3 + (3a + 2b)x2 + 2bx]e2x, y′′ p = [4ax3 + 4(3a + b)x2 + (6a + 8b)x + 2b]e2x], and y′′′ p = [8ax3 + (36a + 8b)x2 + (36a + 24b)x + 6a + 12b]e2x]. Then by substitution in the differential equation we obtain [24ax + (6a + 8b)]e2x = 6xe2x, or 24ax + (6a + 8b) = 6x. Hence, 24a = 6 and 6a + 8b = 0. Thus a = 1/4, b = −3/16, and yp = x2(x/4 − 3/16)e2x. Therefore the general solution is y = [c1 + c2x − 3/16x2 + x3/4]e2x + c3e−2x. 8 7. Consider a linear homogeneous differential equation a2(x)y′′ + a1(x)y′ + a0(x)y = 0, −∞ < x < ∞, where a2, a1, and a0 are continuous functions with a2(x) ̸= 0 for all x. (a) Can a set {y1, y2}, where y1(0) = 1, y2(0) = −1, y′ 1(0) = −1, y′ 2(0) = 1, be a fundamental set of solutions for the differential equa- tion? Justify your answer. (5 points) Solution. NO since W(y1(0), y2(0)) = 0. (b) Can a set {y1, y2, y3} of three nontrivial solutions of the differen- tial equation be linearly independent? Justify your answer. (5 points) Solution. NO since a fundamental set consists only of two solutions; thus one solution must be a linear combination of the other two.