MATH 202 Exam 2 Solution, Spring 2009

American University of Beirut MATH 202 Differential Equations Spring 2009 quiz # 2 - solution Exercise 1 Find the general solution of the given differential equation (do not find the con- stants) a) y′′′ − y′′ = 6 y = c0 + c1x + c2ex � �� � + c3x2 ���� yc yp b) y′′ − 4y′ + 5y = e−x + 2 cos(2x) y = e2x(c0 cos x + c1 sin x) � �� � + c2e−x � �� � + c3 cos(2x) + c4 sin(2x) � �� � yc yp1 yp2 c) y(4) − 2y′′ + y = 1 + x − xex + sin x y = c1ex + c2xex + c3e−x + c4xe−x � �� � + c5 + c6x � �� � + (c7x2 + c8x3)ex � �� � + c9 sin x + c10 cos x � �� � yc yp1 yp2 yp3 Exercise 2 Find the general solution of y′′ − y = 2ex ex + e−x yc = c1ex + c2e−x; W = −2, c′ 1 = 1 ex + e−x = ex 1 + e2x , and c1 = � ex 1 + e2x dx = tan−1(ex) c′ 2 = − e2x ex + e−x , and c2 = − � e2x ex + e−x dx = − � u2 u + 1 u du u = − � u2 1 + u2 du = − � � 1 − 1 1 + u2 � du = tan−1(u) − u = tan−1(ex) − ex hence yp = (ex + e−x) tan−1(ex) − 1 and the general solution is: y = yc + yp Exercise 3 Consider the differential equation (E) : x2y′′ − (x2 + 2x)y′ + (x + 2)y = x3 a) check that y1 = x is solution of (E0): obvious by direct substitution. b) let y2 = xu(x). Show that u(x) satisfies a first order linear differential equation; find u(x) then find the general solution of (E) on (0, ∞). y′ 2 = u(x) + xu′(x), y′′ 2 = 2u′(x) + xu′′(x), and substituting in (E) yields: u′′(x) − u′(x) = 1. Taking v = u′, then v′ − v = 1, and then v(x) = ex − 1, and u(x) = ex − x. The general solution of (E) is then: y = c1x + c2xex � �� � + −x2 ���� yc yp Exercise 4 Use the substitution x = et to solve the Cauchy-Euler differential equation x2y′′ − xy′ + y = x(ln x)2 on (0, ∞). dy dx = dy dt × dt dx = 1 x dy dt and d2y dx2 = − 1 x2 dy dt + 1 x2 d2y dt2 Substituting in the equation yields: d2y dt2 − 2dy dt + y = t2et, and the solution is y = c1et + c2tet � �� � + (c3t2 + c4t3 + c5t4)et � �� � yc yp hence y = c1x + c2x ln x + c3x(ln x)2 + c4x(ln x)3 + c5x(ln x)4 Exercise 5 Find two power series solutions of the differential equation y′′ − xy′ + y = 0 about the ordinary point x = 0. Give the radius of convergence. Let y = ∞ � n=0 cnxn be a series solution of the differential equation. Deriving and substituting in the equation yields the following: � c0 + 2c2 = 0 (1) (n + 2)(n + 1)cn+2 − (n − 1)cn = 0 n ≥ 1 (2) equation (1) implies c2 = −1 2 c0 c1 ∈ R, and from equation (2), we find that c3 = c5 = c7 = . . . = c2n+1 = . . . = 0, then y1 = x is a solution. form equation (2), c2n = 2n − 3 2n(2n − 1) c2n−2; solving the recurrence yields c2n = −2n − 3 2n n! c0, and y2 = ∞ � n=0 c2nx2n is the other solution of the differential equation. The radius of convergence is R = ∞ (by ratio test). The general solution is then: y = c0 ∞ � n=0 2n − 3 2n n! x2n + c1x