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In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dy/dt also appears. Consider the following initial value problem, defined for t≥ 0: dy dt +16 y(t-w) e 8w dw = 5, y(0) = 0. a. Use convolution and Laplace transforms to find the Laplace transform of the solution. == Y(s) C{y(t)}= = 5(s+8) s(s+4)2 b. Obtain the solution y(t). y(t) = = 5-40e-4t+80te -4t

In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dy/dt also appears. Consider the following initial value problem, defined for t≥ 0: dy dt +16 y(t-w) e 8w dw = 5, y(0) = 0. a. Use convolution and Laplace transforms to find the Laplace transform of the solution. == Y(s) C{y(t)}= = 5(s+8) s(s+4)2 b. Obtain the solution y(t). y(t) = = 5-40e-4t+80te -4t

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.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 8CR
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In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dy/dt also appears. Consider the following initial value problem, defined for t≥ 0: dy dt +16 y(t-w) e 8w dw = 5, y(0) = 0. a. Use convolution and Laplace transforms to find the Laplace transform of the solution. == Y(s) C{y(t)}= = 5(s+8) s(s+4)2 b. Obtain the solution y(t). y(t) = = 5-40e-4t+80te -4t
Transcribed Image Text:In an integro-differential equation, the unknown dependent variable y appears within an integral, and its derivative dy/dt also appears. Consider the following initial value problem, defined for t≥ 0: dy dt +16 y(t-w) e 8w dw = 5, y(0) = 0. a. Use convolution and Laplace transforms to find the Laplace transform of the solution. == Y(s) C{y(t)}= = 5(s+8) s(s+4)2 b. Obtain the solution y(t). y(t) = = 5-40e-4t+80te -4t
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