MATH 2280 Practice Final Solutions: Laplace Transforms of Specific Functions, Exams of Differential Equations

Download MATH 2280 Practice Final Solutions: Laplace Transforms of Specific Functions and more Exams Differential Equations in PDF only on Docsity! MATH 2280 PRACTICE FINAL SOLUTIONS 1. Problem 1 Let f : [0,∞) → R be Riemann integrable on [0,M ] for every M ≥ 0. Define the Laplace transform of f . Definition 1.1. The Laplace transform of f is given by (1) L(f) = ∫ ∞ 0 e−stf(t)dt for every s such that the improper integral exists. The above definition is all that is required, but further comments are in order. Remark 1.2. Since f is Riemann integrable on [0,M ], (2) ∫ M 0 e−stf(t)dt exists and hence the improper integral has a chance of existing, at least for some s. Typically, Re (s) > a for some a ∈ R although there is no a priori reason why this has to be the case. 2. Problem 2 Show that the Laplace transform of t2 is 2 s3 . Do not use table 6.2.1. 1 2 MATH 2280 PRACTICE FINAL SOLUTIONS Solution 2.1. We compute: L(t2) = ∫ ∞ 0 t2e−stdt = lim M→∞ ∫ M 0 t2e−stdt = lim M→∞ (2 s ∫ M 0 te−stdt− t 2e−st s ∣∣∣ M 0 ) ( u = t2, u̇ = 2t, v̇ = e−st, v = −e −st s ) = lim M→∞ (2 s ∫ M 0 te−stdt− M 2e−sM s ) = lim M→∞ ( 2 s2 ∫ M 0 e−stdt− 2te −st s2 ∣∣∣ M 0 − M 2e−sM s ) ( u = t, u̇ = 1, v̇ = e−st, v = −e −st s ) = lim M→∞ ( 2 s2 ∫ M 0 e−stdt− 2Me −sM s2 − M 2e−sM s ) = lim M→∞ ( − 2e −st s3 ∣∣∣ M 0 − 2Me −sM s2 − M 2e−sM s ) = lim M→∞ (2− 2e−sM s3 − 2Me −sM s2 − M 2e−sM s ) = lim M→∞ ( 2 s3 − 2 s3esM − 2M s2esM − M 2 sesM ) (3) Now by l’Hôpital’s rule, (4) lim M→∞ M2 sesM = lim M→∞ 2M s2esM = lim M→∞ 2 s3esM = 0. Hence, the limit in the preceding calculation is 2 s3 since the other terms all go to 0 as M →∞. Thus, (5) L(t2) = 2 s3 as required. 3. Problem 3 Let uc be the Heaviside function at c. Let Y be the Laplace transform of y. Solve the following initial value problem for Y :