# Fall16 M 408D Homework 5 - leonard(kml3374 Homework 05...

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leonard (kml3374) – Homework 05 – staron – (53595) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find all values of k that don’t result in a zero function for which the function y = sin kt satisfies the differential equation y ′′ + 9 y = 0 1. k = 3 2. k = 3 , 3 correct 3. k = 9 , 9 4. k = 9 5. k = 9 6. k = 3 Explanation: We will begin by solving for y ′′ . y = sin kt y = k cos kt y ′′ = k 2 sin kt. We compute y ′′ + 9 y = k 2 sin kt + 9 sin kt = (9 k 2 ) sin kt. Hence y ′′ + 9 y = 0 if and only if k 2 = 9, i.e., if and only if k = ± 3. Thus, k = 3 , 3 . 002 10.0points Find all nonzero values of k for which the function y = A sin kt + B cos kt satisfies the differential equation y ′′ + 25 y = 0 for all values of A and B . 1. k = 5 2. k = 5 , 5 correct 3. k = 25 4. k = 25 5. k = 25 , 25 6. k = 5 Explanation: We will begin by solving for y ′′ . y = A sin kt + B cos kt y = Ak cos kt Bk sin kt y ′′ = Ak 2 sin kt Bk 2 cos kt = k 2 ( A sin kt + B cos kt ) = k 2 y. We compute y ′′ + 25 y = k 2 y + 25 y = (25 k 2 ) y. Hence y ′′ +25 y = 0 if and only if k 2 = 25, i.e., if and only if k = ± 5. Hence, k = 5 , 5 . 003 10.0points Find all values of r for which the function y = e rt satisfies the differential equation y ′′ 14 y + 48 y = 0 . 1. r = 14 2. r = 6
leonard (kml3374) – Homework 05 – staron – (53595) 2 3. r = 48 , 14 4. r = 6 , 8 correct 5. r = 8 , 6 6. r = 14 , 48 Explanation: We will begin by solving for y and y ′′ . y = e rt y = re rt y ′′ = r 2 e rt . We compute y ′′ 14 y + 48 y = r 2 e rt 14 re rt + 48 e rt = ( r 2 14 r + 48) e rt = ( r 6)( r 8) e rt . Hence y ′′ 14 y + 48 y = 0 if and only if k = 6 , 8. Thus, k = 6 , 8 . 004 10.0points Which of the following functions satisfy the differential equation y ′′ + 2 y + y = 0 ? 1. y = e 1 t , te 2 t 2. y = e 1 t , te 1 t 3. y = e 1 t , te 2 t 4. y = e 2 t , te 1 t 5. y = e 1 t , te 1 t 6. y = e 1 t , te 1 t correct Explanation: All answer choices above are of the form e rt and te rt , so let us solve for potential values of r in each of those cases. We will begin by assuming y = e rt and finding y and y ′′ . y = e rt y = re rt y ′′ = r 2 e rt . We compute y ′′ + 2 y + y = r 2 e rt + 2 re rt + e rt = ( r 2 + 2 r + 1) e rt = ( r + 1) 2 e rt Hence y ′′ + 2 y + y = 0 if and only if r = 1. This means e 1 t will satisfy this equation. Now we will assume y = te rt and find y and y ′′ . y = te rt y = rte rt + e rt = ( rt + 1) e rt y ′′ = re rt + r 2 te rt + re rt = r 2 te rt + 2 re rt = ( r 2 t + 2 r ) e rt . Now we compute y ′′ + 2 y + y = ( r 2 t + 2 r ) e rt + 2( rt + 1) e rt + te rt = ( r 2 t + 2 r + 2 rt + 2 + t ) e rt = bracketleftbig ( r 2 + 2 r + 1) t + 2 r + 2 bracketrightbig e rt = bracketleftbig ( r + 1) 2 t + 2( r + 1) bracketrightbig e rt = ( r + 1) [( r + 1) t + 2] e rt .