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dy 2. Solve = dt -y +28(t− 3), y(0) = 1.

dy 2. Solve = dt -y +28(t− 3), y(0) = 1.

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 36E
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please solve question 2 with explanation 

Note: Recall that the delta function 8(t) is the first derivative of the step function H(t). The key property of 8(t) is the way it behaves in integrals: for any number c and any function f, ·b [ = f(s) 6 (s - c) ds i.e. the integral "picks out" the value of the function f at the location c where 6(t – c) has a spike. The integral is zero if the spike lies outside the domain of integration. For integrals from 0 to t, when c > 0, we can rewrite the rule more concisely with a step function: 1. Evaluate the following integrals involving 8(t): ·3 (a) So e e¹⁹ cos(s)√(s — 2) ds. 2. Solve [ƒ(c), a<c<b, 0, otherwise, [*ƒ(s) 5 (s — c) ds = f(c)H(t – c). (8³ (b) S + s)6(s - 2) ds. (c) S s² ln(1 + s)8(s − 1) ds. (Your answer will depend on t.) dy dt = −y +26(t− 3), y(0) = 1.
Transcribed Image Text:Note: Recall that the delta function 8(t) is the first derivative of the step function H(t). The key property of 8(t) is the way it behaves in integrals: for any number c and any function f, ·b [ = f(s) 6 (s - c) ds i.e. the integral "picks out" the value of the function f at the location c where 6(t – c) has a spike. The integral is zero if the spike lies outside the domain of integration. For integrals from 0 to t, when c > 0, we can rewrite the rule more concisely with a step function: 1. Evaluate the following integrals involving 8(t): ·3 (a) So e e¹⁹ cos(s)√(s — 2) ds. 2. Solve [ƒ(c), a<c<b, 0, otherwise, [*ƒ(s) 5 (s — c) ds = f(c)H(t – c). (8³ (b) S + s)6(s - 2) ds. (c) S s² ln(1 + s)8(s − 1) ds. (Your answer will depend on t.) dy dt = −y +26(t− 3), y(0) = 1.
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