12 d3x(t) 5 dt3 d²x(t) dt2 + 5x(t) = 3cos(2t) d²x(0) = 2, dx(0) 1, x(0) = 1 dt2 dt

Q: Find the Laplace transform

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Q: t* + at - 4os t + t - 41 + ta

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Q: 8s – 18 Find the inverse Laplace transformf(t) = L¯' {F(s)} of the function F(s) : s2 – 4s + 20

A: Consider the provided the question,

Q: If L{f"(t)} = arctan ), f(0) = 2 and f'(0) = 1, find L{f(t)}. %3D

A: Given - If ℒf''t = arctan12 , f0 = 2 and f'0 = 1  To find - ℒft       Formula used - ℒfnt = snFs -…

Q: (a) f(t) = cosh² t (b) f(t) = 2 sinh? 2t 2

A: To find the Laplace transforms of the given functions we will use the following table of Laplace…

Q: Problem 5 Using Laplace transform, solve y" - 4y + 7y = 0 y(0) = 3, y'(0) = 7

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Q: f (t) = e(-3t) sin(2t) cos(3t) cosh(t)

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Q: f(t) = [1 – u2m (t)](2 sin t – sin 2t)

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Q: Solve the inverse Laplace transform: 2s-4 s2+8s+13

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Q: 1-e g(t) t

A: Given the given function is gt=1-ett. Apply formula Lftt=∫s∞fsds L1=1sLeat=1s-a log a-log b=logab

Q: f(t) = t5et 120 А. (s-4)6 24 C. (s-5)6 120 В. (s-4)5 24 D. (s-5)5

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Q: 2s2+5s-4 Find the inverse transform: F(s) s3+s2-2s A. 2 – et – 2e-2t C. 1+ 2et – e-2t B. 2 + et –…

A:

Q: dt (te-¹u (t))?

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Q: 6, 6 01 2 Lt

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Q: f(t) = e-2tsin²tcos²t

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Q: f(t) = te¯3t cos² 2t -3t cos 2t

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Q: f(t) = [1– u2m(t)](2 sin t – sin 2t) U2n

A: Some of the properties of Laplace transformation Lsinat=as2+a2Luc(t)=e-cssLuc(t)g(t)=e-csL[g(t+c)]

Q: f(t) = [1– uzn(t)](2 sin t – sin 2t

A: Solution

Q: F(s)= (s +1)es» s²(s + 2) -5s

A:

Q: on the function fcs) 35 - 17 s- 65+25 %3D

A: Inverse Laplace transform of function is : 12e3t(6cos4t-sin4t

Q: 2. t5/2 sinh 4t

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Q: Find the Laplace transform

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Q: Solve the laplace transform 3x-6x= Sin 2t %3D given when t=0

A: Use the laplace transformation table and derivative formula

Q: e at -e bt

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Q: Find the Laplace transform

A: Given,

Q: f (t) 7t2 te-4t

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Q: cos 4t-cos 2t 3. f(t) = t

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Q: Find L (t*ett)

A: You asked to find inverse Laplace transform.but your image is for Laplace transform.

Q: trigonometric identities to find the Laplace Transform of the given function

A: We know the the identity, cos(2x)=1-sin2(x) so we have,            sin2(4x)=1-cos(8x) now we can use…

Q: f (t) = 3e " cosh (In 2r)

A: This is a problem of Laplace Transform.