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Q1: Solve the following IVPs by using the table of Laplace Transform.\\ a) $y' - y = te^t \sin t$, $y(0) = 0$\ b) $ty'' + (t - 3)y' + y = 0$, $y(0) = 0$, $y'(0) = 0$\ c) $y'' + 16y = f(t)$, $y(0) = 0$, $y'(0) = 1$, where\\$f(t) = \begin{cases} \cos 4t, & 0 \le t < \pi\\0, & t \ge \pi \end{cases}$

          Q1: Solve the following IVPs by using the table of Laplace Transform.\\
a) $y' - y = te^t \sin t$, $y(0) = 0$\
b) $ty'' + (t - 3)y' + y = 0$, $y(0) = 0$, $y'(0) = 0$\
c) $y'' + 16y = f(t)$, $y(0) = 0$, $y'(0) = 1$, where\\$f(t) = \begin{cases} \cos 4t, & 0 \le t < \pi\\0, & t \ge \pi \end{cases}$

Q1: Solve the following IVPs by using the table of Laplace Transform.

a) y' - y = te^t sin t, y(0) = 0b) ty” + (t - 3)y' + y = 0, y(0) = 0, y'(0) = 0c) y” + 16y = f(t), y(0) = 0, y'(0) = 1, where
f(t) = cos 4t, 0 ≤ t < π
0, t ≥π

Added by Cole A.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ishana K

05/31/2022

Step 1

Taking the Laplace Transform of both sides of the equation, we get: sY(s) - y(0) - Y(s) = L{tet sin t} Using the Laplace Transform table, we have: L{tet sin t} = -2/(s^2 - 1)^2 Substituting this into the equation, we get: sY(s) - Y(s) = -2/(s^2 - Show more…

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Q1: Solve the following IVPs by using the table of Laplace Transform. a) y' - y = tet sin t, y(0) = 0 b) ty" + (t^3)y' + y = 0, y(0) = 0, y'(0) = 0 c) y" + 16y = f(t), y(0) = 0, y'(0) = 1, where S = cos 4t, 0 ≤ t ≤ 10