Introduction to Differential Equations - Self Quiz Solutions 36 | MATH 220, Study notes of Differential Equations

Download Introduction to Differential Equations - Self Quiz Solutions 36 | MATH 220 and more Study notes Differential Equations in PDF only on Docsity! MATH 220 Self-quiz 36 1. Find the inverse Laplace transform of the function F (s) = s s2 + s− 2 . Solution: We begin by decomposing the fraction: s (s + 2)(s− 1) = a s + 2 + b s− 1 If we clear denominators and compare numerators, then we get: as− a + bs + 2b = s =⇒ (a + b)s + (−a + 2b) = s This means that a + b = 1 and −a + 2b = 0 from which we can easily find: a = 2 3 b = 1 3 Now, F (s) becomes: 2 3 · 1 s + 2 + 1 3 · 1 s− 1 From this, we see that the inverse Laplace of this function is: 2 3 e−2t + 1 3 et 2. Use the Laplace transform in order to solve the initial value problem: y′′ − y = et y(0) = 1 y′(0) = 0 Solution: Suppose that y(t) is the sought-for solution and let F (s) be its Laplace transform. Then the Laplace transforms of y′ and y′′ are given by: sF (s)− 1 s2F (s)− s respectively. This means that if we take the Laplace transform of both sides of the equation then we get: s2F − s− F = 1 s− 1 This is an algebraic equation that we can solve for F in order to find: F (s) = s2 − s + 1 (s− 1)2(s + 1) We just need to find the inverse Laplace transform of s2 − s + 1 (s− 1)2(s + 1) . To this end, we decompose: s2 − s + 1 (s− 1)2(s + 1) = a s + 1 + b s− 1 + c (s− 1)2 Upon clearing denominators, we get: s2 − s + 1 (s− 1)2(s + 1) = a(s− 1)2 + b(s2 − 1) + c(s + 1) (s− 1)2(s + 1) This means that: a(s− 1)2 + b(s2 − 1) + c(s + 1) = s2 − s + 1 By setting s = 1, we get 2c = 1, i.e. c = 1/2. By setting s = −1, we get 4a = 3, i.e. a = 3/4. Finally by comparing the constant terms of the two polynomials we get: 2