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y" (t) – y(t) = 4cos(t); y (0)=0 , y'(0)=1 Answer: y(t) = 2cosh(t)+ sinh (t)–2cos(t) y"(x) – 4y'(x) +4y(x)=4e2x ; y (0) = -1 , y' (0) = -4 Answer: y(x) = e2x(2x²–2x-1)

y" (t) – y(t) = 4cos(t); y (0)=0 , y'(0)=1 Answer: y(t) = 2cosh(t)+ sinh (t)–2cos(t) y"(x) – 4y'(x) +4y(x)=4e2x ; y (0) = -1 , y' (0) = -4 Answer: y(x) = e2x(2x²–2x-1)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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SOLVE D.E. USING LAPLACE TRANSFORMS. y" (t) – y(t) = 4cos(t); y (0)=0, y'(0)=1 Answer: y(t) = 2cosh(t)+ sinh (t)–2cos(t) y"(x) – 4y'(x) +4y(x)=4e2x ; y (0) = -1, y' (0) = -4 Answer: y(x) = e2x(2x²-2x-1) x"'(t) – 3x"(t) – 4x'(t) + 12x(t) = 12e-t ; x(0) = 4 , x'(0) = 2 , x"(0) = 18 Answer: x(t) =e-t+ e3t + 2cosh(2t)
Transcribed Image Text:SOLVE D.E. USING LAPLACE TRANSFORMS. y" (t) – y(t) = 4cos(t); y (0)=0, y'(0)=1 Answer: y(t) = 2cosh(t)+ sinh (t)–2cos(t) y"(x) – 4y'(x) +4y(x)=4e2x ; y (0) = -1, y' (0) = -4 Answer: y(x) = e2x(2x²-2x-1) x"'(t) – 3x"(t) – 4x'(t) + 12x(t) = 12e-t ; x(0) = 4 , x'(0) = 2 , x"(0) = 18 Answer: x(t) =e-t+ e3t + 2cosh(2t)
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