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Name: Solutions Math 1B: Quiz 5 April 8, 2015 1. Find the forms of the particular solutions for the following problems. Do not solve for the unknown coefficients. (a) 2y ′′ + 7y ′ + 6y = e4x sin(x). The corresponding homogeneous equation is 2y ′′ + 7y ′ + 6y = 0 which has the auxiliary equation 2r2 + 7r + 6 = 0 ⇒ (2r + 3)(r + 2) = 0 ⇒ r = −3 , −2. The 2 −3x −2x complementary solution therefore is yc (x) = c1 e 2 + c2 e . For the trial particular solution, try yp (x) = e4x (A cos(x) + B sin(x)). Since there is no overlap between the trial yp and the yc found above, this form of yp is good enough. (b) y ′′ − 4y ′ + 4y = xe2x . The corresponding homogeneous equation is y ′′ − 4y ′ + 4y = 0 which has the auxiliary equation r2 − 4r + 4 = 0 ⇒ (r − 2)2 = 0 ⇒ r = 2. The complementary solution therefore is yc (x) = c1 e2x + c2 xe2x . For the trial particular solution, try yp (x) = e2x (A + Bx). Since both e2x and xe2x appear in yc , we need to consider instead the form yp (x) = e2x (Ax2 + Bx3 ) to avoid any overlap between the particular and complementary solutions. 1 2. Find the general solution of y ′′ + y = csc(x). The corresponding homogeneous equation is y ′′ +y = 0 which has the auxiliary equation r2 + 1 = 0 ⇒ r = ±i. The complementary solution therefore is yc (x) = c1 cos(x) + c2 sin(x) with basic solutions y1 (x) = cos(x) and y2 (x) = sin(x). We assume the particular solution is of the form yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x) where u1 and u2 are unknown functions. By the method of variation of parameters, we know then that u′1 and u′2 satisfy u′1 (x)y1 (x) + u′2 (x)y2 (x) = 0 u′1 (x)y1′ (x) + u′2 (x)y2′ (x) = csc(x). (1) (2) (1) : cos(x)u′1 (x) + sin(x)u′2 (x) = 0 (2) : − sin(x)u′1 (x) + cos(x)u′2 (x) = csc(x). (3) (4) We have Note that sin(x)(3) + cos(x)(4) gives u′2 (x)(sin2 (x) + cos2 (x)) = csc(x) cos(x) ⇒ u′2 (x) = cot(x). ∫ Use (3) to get u′1 (x) = − tan(x) cot(x) = −1. Thus, u1 (x) = −1 dx = −x and ∫ ∫ cos(x) u2 (x) = cot(x) dx = dx = ln | sin(x)|. sin(x) We therefore have yp (x) = −x cos(x) + sin(x) ln | sin(x)| so the general solution to the problem is y(x) = yc (x) + yp (x) = c1 cos(x) + c2 sin(x) − x cos(x) + sin(x) ln | sin(x)|. 2