Use laplace transform to solve the given system (a)*- 2x -- y = 6et,— %- y = 6e3t,2 - 3x + - 3y = 6e3;dxdydtdtx(0)y(0) = 0.

Given, dxdt-2x-dydt-y=6e3t2dxdt-3x+dydt-3y=6e3t, x0=0, y0=0 We use the Laplace transforms…

Answered: dy 2х dx - y = 6e3t, dt dt dy 3x + dt…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Use laplace transform to solve the given system

(a)
*- 2x -- y = 6et,
— %
- y = 6e3t,
2 - 3x + - 3y = 6e3;
dx
dy
dt
dt
x(0)
y(0) = 0.

Expert Solution

Step 1

$Given, \frac{d x}{d t} - 2 x - \frac{d y}{d t} - y = 6 e^{3 t} 2 \frac{d x}{d t} - 3 x + \frac{d y}{d t} - 3 y = 6 e^{3 t}, x (0) = 0, y (0) = 0$

$We use the Laplace transforms initially, L (x (t)) = \bar{x} (s) L (y (t)) = \bar{y} (s) L (x' (t)) = s \bar{x} (s) - x (0) L (y' (t)) = s \bar{y} (s) - y (0) And in the latter stage, we use the known inverse Laplace transforms to get the required solution .$

Step 2

So, we have

$s \bar{x} (s) - x (0) - 2 \bar{x} (s) - s \bar{y} (s) + y (0) + \bar{y} (s) = \frac{6}{s - 3} 2 s \bar{x} (s) - 2 x (0) - 3 \bar{x} (s) + s \bar{y} (s) - y (0) - 3 \bar{y} (s) = \frac{6}{s - 3}$

Simplifying and applying the initial conditions,

$(s - 2) \bar{x} (s) + (1 - s) \bar{y} (s) = \frac{6}{s - 3} + 3 - - - (1) (2 s - 3) \bar{x} (s) + (x - 3) \bar{y} (s) = \frac{6}{s - 3} + 3 - - - (2)$

Using Cramer's rule,

$\bar{x} (s) = \frac{|\begin{matrix} \frac{6}{s - 3} + 3 & 1 - s \\ \frac{6}{s - 3} + 6 & s - 3 \end{matrix}|}{|\begin{matrix} s - 2 & 1 - s \\ 2 s - 3 & s - 3 \end{matrix}|} = \frac{3 s - 3 - \frac{6 (s - 2) (1 - s)}{s - 3}}{(s - 2) (s - 3) - (2 s - 3) (1 - s)} = \frac{(s - 1) (9 s - 21)}{3 s^{2} - 10 s + 9} - - - - - - - - (3) = 3 - \frac{6}{3 s^{2} - 10 s + 9} = 3 - \frac{2}{{(s - \frac{10}{6})}^{2} + {(\frac{\sqrt{8}}{6})}^{2}} x (t) = 3 δ (t) - \frac{6}{\sqrt{8}} e^{\frac{10 t}{6}} \sin (\frac{\sqrt{8} t}{6}) Where, δ (t) is the delta function$

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