MATH 4B Midterm 2 Solutions

# MATH 4B Midterm 2 Solutions - MATH 4BDifferential Equations...

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MATH 4B–Differential Equations, Spring 2016 Midterm Exam 2 Solutions Show all your work. Correct answers without the corresponding procedure get only half credit. No technology is permitted. Name: Permit #: Peter Merkx: Monday 5:00pm Tuesday 5:00pm Tuesday 6:00pm Tuesday 7:00pm Harris Enniss: Tuesday 8:00am Tuesday 4:00pm Question Points Score 1 20 2 20 3 20 4 20 5 20 Total: 100

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f ( t ) = L - 1 { F ( s ) } F ( s ) = L{ f ( t ) } 1 1 s , s > 0 e at 1 s - a , s > a t n , n positive integer n ! s n +1 , s > 0 cos at s s 2 + a 2 , s > 0 sin at a s 2 + a 2 , s > 0 cosh at s s 2 - a 2 , s > | a | sinh at a s 2 - a 2 , s > | a | e at cos bt s - a ( s - a ) 2 + b 2 , s > a e at sin bt b ( s - a ) 2 + b 2 , s > a t n e at , n positive integer n ! ( s - a ) n +1 , s > a u c ( t ) e - cs s , s > 0 u c ( t ) f ( t - c ) e - cs F ( s ) e ct f ( t ) F ( s - c ) Page 2
1. 20 points (a) Verify that y 1 ( t ) = t and y 2 ( t ) = te 2 t form a fundamental set of solutions for t 2 y 00 - 2 t ( t + 1) y 0 + 2( t + 1) y = 0 , t > 0 . NOTE: You DO NOT need to verify that y 1 ( t ) and y 2 ( t ) are solutions, they are! Solution: We only need to verify that W ( y 1 , y 1 )( t ) 6 = 0 for t > 0: W ( y 1 , y 2 )( t ) = y 1 ( t ) y 2 ( t ) y 0 1 ( t ) y 0 2 ( t ) = t te 2 t 1 e 2 t + 2 te 2 t = te 2 t + 2 t 2 e 2 t - te 2 t = 2 t 2 e 2 t 6 = 0 .

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