Laplace transforms may be used to find solutions to some linear systems of diffe

Question

Laplace transforms may be used to find solutions to some linear systems of differential equations. Consider the linear system of differential equations: x%u2032 =x+y y%u2032 =4x+y with initial conditions x(0) = 0 and y(0) = 2. (%u2217)
(a) Let X(s) = L{x(t)} and Y (s) = L{y(t)} be the Laplace transforms of the functions x(t) and y(t), respectively. Take the Laplace transform of each of the differential equations in (%u2217) and solve for X(s) (i.e., eliminate Y (s)). (b) Using the function X(s) from (a), determine x(t). (c) Use the expression for x(t) and the first equation in (%u2217) to determine y(t). Laplace transforms may be used to find solutions to some linear systems of differential equations. Consider the linear system of differential equations: x%u2032 =x+y y%u2032 =4x+y with initial conditions x(0) = 0 and y(0) = 2. (%u2217)
(a) Let X(s) = L{x(t)} and Y (s) = L{y(t)} be the Laplace transforms of the functions x(t) and y(t), respectively. Take the Laplace transform of each of the differential equations in (%u2217) and solve for X(s) (i.e., eliminate Y (s)). (b) Using the function X(s) from (a), determine x(t). (c) Use the expression for x(t) and the first equation in (%u2217) to determine y(t).

Explanation / Answer

x'=x+y

y'=4x+y

a.)Now applying Laplace Transform to both equations, we get

sX(s)-0=X(s)+Y(s)

sY(s)-2=4X(s) + Y(s)

Y(s)(s-1)=4X(s)+2

Y(s)=(4X(s)+2)/(s-1)

X(s)(s-1)=Y(s)

So X(s)(s-1)=(4X(s)+2)/(s-1)

X(s)(s-1)^2=4X(s)+2

X(s)(s^2-2s+1)=4X(s)+2

X(s)(s^2-2s-3)=2

X(s)=2/(s^2-2s-3)

b.)X(s)=2/(s-3)(s+1)

X(s)=[1/(2(s-3))] - [1/(2(s+1))]

Now taking Laplace inverse, we get

x(t)=(e^(3t))/2 - (e^(-t))/2

c.)x'(t)=x(t)+y(t)

So y(t)=x'(t) - x(t)

y(t)=3e^(3t)/2 + e^(-t)/2 - e^(3t)/2 + e^(-t)/2

=e^(3t)+e^(-t)

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