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Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: [sin(nt), 10, Y(s) Next take the inverse transform of Y(s) to get y(t) Use step(t-c) for uc(t) Note: = y" + 9y = π π 1 1 (6²+7²) (8² +9) = 7²-9 (5²+9=2+²) 0≤t<1 1≤t y(0) = 0, y'(0) = 0

Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: [sin(nt), 10, Y(s) Next take the inverse transform of Y(s) to get y(t) Use step(t-c) for uc(t) Note: = y" + 9y = π π 1 1 (6²+7²) (8² +9) = 7²-9 (5²+9=2+²) 0≤t<1 1≤t y(0) = 0, y'(0) = 0

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
Section6.3: Implicit Differentiation
Problem 26E
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Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: (sin(7t), 0<t<1 10, 1<t Y(s) Next take the inverse transform of Y(s) to get y(t) Use step(t-c) for uc(t) Note: π (s² + π²) (s² +9) -7²) (5² y" +9y= π 1 1 7²-9 (3² +9-3²+7²) = y(0) = 0, y'(0) = 0
Transcribed Image Text:Take the Laplace transform of the following initial value and solve for Y(s) = L{y(t)}: (sin(7t), 0<t<1 10, 1<t Y(s) Next take the inverse transform of Y(s) to get y(t) Use step(t-c) for uc(t) Note: π (s² + π²) (s² +9) -7²) (5² y" +9y= π 1 1 7²-9 (3² +9-3²+7²) = y(0) = 0, y'(0) = 0
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