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The inverse Laplace transform of H(s) = + (t) ==²e²+e-²t 8 1 " (3s+2)(S-2) b) f(t) ==e+e-2t -3 d)f (t) == +1 +²e²t is:

The inverse Laplace transform of H(s) = + (t) ==²e²+e-²t 8 1 " (3s+2)(S-2) b) f(t) ==e+e-2t -3 d)f (t) == +1 +²e²t is:

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 Q8 plece
Q1) One of the following is correct for functions the f(x) = 1, g(x) = x³ and h(x) = ln(x): a) f(x), g(x) and h(x) are linearly dependent c) W (f(x), g(x), h(x)) = 0 b) f(x), g(x) and h(x) are linearly independent d) W (f(x), g(x), h(x)) = 9 Q2) L-1 -2s S²+7²) = a) -2coshπt b) -2sinhnt c)-2сost d) -2sinnt Q3) The Integrating factor which make (3x²y + 2xy + y³) dx + (x² + y²)dy = 0 exact, is: a) e-3x b) ex c)ex d) e 3x Q4) The linear form of nonlinear ODE y' - 2y = 2y, is: a) u' + 6u = -6 b) u' - 6u = -6 c) u' - 6u = 6 d) u' + 6u = 6 Q5) The general solution of 2x2y" + 3xy' - 15y = 0, is: a) y(x) = C₁x² + ₂x³ b) y(x) = ₂x + ₂x-³ 5 c) y(x) = C₁x² + ₂x−3 d) y(x) = C₁xz+C₂x³ Q6) Evaluate (2 e-2t sin4t -0.5 cos3t): 8 S b) 8 (S-2)²+16 2s²+18 4 (s+2)² +16 2s²+18 c) (8+2)²+16 28² +18 (S-2)²+16 2s²+18 Q7) The general solution of y"-4y' +9y = 0, is: a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t) c)y(t)=c₁e²t cos(√5 t) + c₂e²tsin (√5 t) b)y(t)=c₁et cos(√5 t) + c₂etsin (√5t) d) y(t) = c₁et cos(5 t) + c₂esin (5 t) Q8) The inverse Laplace transform of H(s): = is: I (3s+2)(S-2) e2t |a) ƒ (t) = ¹ + ²t | c) ƒ (t) ==³e² +²e² -2t 1 8 1 b) f(t) ==e+e-²t +²e²t е d)f(t) = ²³² e: B Q9) The solution of y" + y' = 0 by using power series method, is: x3 x² a) y(x) = a₁ + a₁( 1 −²+2 .-...) b) y(x) = ao + a₁(1+² 41 5! x5 c) y(x) = a₁ + α₁ ( x −₁+3 + .) d)y(x) = ao+a₁(x+² 4! 51 3! IT ·+...) - ...)
Transcribed Image Text:Q1) One of the following is correct for functions the f(x) = 1, g(x) = x³ and h(x) = ln(x): a) f(x), g(x) and h(x) are linearly dependent c) W (f(x), g(x), h(x)) = 0 b) f(x), g(x) and h(x) are linearly independent d) W (f(x), g(x), h(x)) = 9 Q2) L-1 -2s S²+7²) = a) -2coshπt b) -2sinhnt c)-2сost d) -2sinnt Q3) The Integrating factor which make (3x²y + 2xy + y³) dx + (x² + y²)dy = 0 exact, is: a) e-3x b) ex c)ex d) e 3x Q4) The linear form of nonlinear ODE y' - 2y = 2y, is: a) u' + 6u = -6 b) u' - 6u = -6 c) u' - 6u = 6 d) u' + 6u = 6 Q5) The general solution of 2x2y" + 3xy' - 15y = 0, is: a) y(x) = C₁x² + ₂x³ b) y(x) = ₂x + ₂x-³ 5 c) y(x) = C₁x² + ₂x−3 d) y(x) = C₁xz+C₂x³ Q6) Evaluate (2 e-2t sin4t -0.5 cos3t): 8 S b) 8 (S-2)²+16 2s²+18 4 (s+2)² +16 2s²+18 c) (8+2)²+16 28² +18 (S-2)²+16 2s²+18 Q7) The general solution of y"-4y' +9y = 0, is: a) y(t) = c₁e²t cos(5 t) + c₂e²t sin (5 t) c)y(t)=c₁e²t cos(√5 t) + c₂e²tsin (√5 t) b)y(t)=c₁et cos(√5 t) + c₂etsin (√5t) d) y(t) = c₁et cos(5 t) + c₂esin (5 t) Q8) The inverse Laplace transform of H(s): = is: I (3s+2)(S-2) e2t |a) ƒ (t) = ¹ + ²t | c) ƒ (t) ==³e² +²e² -2t 1 8 1 b) f(t) ==e+e-²t +²e²t е d)f(t) = ²³² e: B Q9) The solution of y" + y' = 0 by using power series method, is: x3 x² a) y(x) = a₁ + a₁( 1 −²+2 .-...) b) y(x) = ao + a₁(1+² 41 5! x5 c) y(x) = a₁ + α₁ ( x −₁+3 + .) d)y(x) = ao+a₁(x+² 4! 51 3! IT ·+...) - ...)
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