Question

2.26 Solve the following first-order differential equations by the Laplace transform: (a) $2\dot{x} + x = 4 + t$, $x(0) = 1$ (b) $\dot{x} + 5x = 8t \sin(3t)$, $x(0) = -1$ (c) $\dot{x} + 6x = 3e^{-t} \cos(2t)$, $x(0) = -2$ (d) $\dot{x} + 3x = 2te^{-t} \sin(3t)$, $x(0) = \frac{37}{169}$ (e) $2\dot{x} - x = 4t^2 \sin(t/2)$, $x(0) = 4$ (f) $\dot{x} - 2x = 7t^2e^{-5t} + t \cos(2t)$, $x(0) = \frac{5}{49}$

          2.26 Solve the following first-order differential equations by the Laplace transform:
(a) $2\dot{x} + x = 4 + t$, $x(0) = 1$
(b) $\dot{x} + 5x = 8t \sin(3t)$, $x(0) = -1$
(c) $\dot{x} + 6x = 3e^{-t} \cos(2t)$, $x(0) = -2$
(d) $\dot{x} + 3x = 2te^{-t} \sin(3t)$, $x(0) = \frac{37}{169}$
(e) $2\dot{x} - x = 4t^2 \sin(t/2)$, $x(0) = 4$
(f) $\dot{x} - 2x = 7t^2e^{-5t} + t \cos(2t)$, $x(0) = \frac{5}{49}$

2.26 Solve the following first-order differential equations by the Laplace transform:
(a) 2ẋ + x = 4 + t, x(0) = 1
(b) ẋ + 5x = 8t sin(3t), x(0) = -1
(c) ẋ + 6x = 3e^-tcos(2t), x(0) = -2
(d) ẋ + 3x = 2te^-tsin(3t), x(0) = (37)/(169)
(e) 2ẋ - x = 4t^2 sin(t/2), x(0) = 4
(f) ẋ - 2x = 7t^2e^-5t + t cos(2t), x(0) = (5)/(49)

Added by Jeffrey H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Keondre Parker

07/10/2024

Step 1

Step 1: Taking the Laplace transform of both sides of the equation in (a), we get: $$2\mathcal{L}\{x\}+\mathcal{L}\{x\}=\mathcal{L}\{4\}+\mathcal{L}\{t\}$$ Show more…

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2.26 Solve the following first-order differential equations by the Laplace transform: (a) 2x+x=4+t, x(0) = 1 (b) x + 5x = 8t sin(3t), x(0) = -1 (c) x + 6x = 3e⁻ᵗ cos(2t), x(0) = -2 (d) x + 3x = 2te⁻ᵗ sin(3t), x(0) = 37/169 (e) 2x-x = 4t² sin(t/2), x(0) = 4 (f) x - 2x = 7t²e⁻⁵ᵗ + t cos(2t), x(0) = 5/49