The Laplace transform of a function f(t) is given by Evaluate the integral to sh

Question

The Laplace transform of a function f(t) is given by Evaluate the integral to show that the Laplace transform of f(t) - sinh t can be written as 1/s2-1'. Use your working to explain what condition must be satisfied by s so that the integral is convergent. Consider the first order ordinary differential equation with P(t) = Find the general solution for 0 t 2. Find the constant in the general solution for 0 t 2 so that the initial condition is satisfied. Find a general solution for t > 2. Find the constant in the general solution for t > 2 so that the solution for t > 2 and the solution for 0 t 2 match at t = 2. Use Maple, Matlab, or MS-excel to plot the solution for 0 t 5. Explain, with reference to your expression for y(t) and plot of y(t), whether or not the solution is continuous at t = 2. Explain, with reference to your expression for y(t) and plot of y(t), whether or not the solution is differentiable at t = 2.

Explanation / Answer

1.

a)

F(s) = integral(0 to infinity)[e^-st * sinht ]dt

we can write sinht = (e^t-e^-t) /2

=integral(0 to infinity)[e^-st *(e^t-e^-t) /2]dt

=integral(0 to infinity)[e^(t-st) - e^-(t+st)] dt/2

=integral(0 to infinity)[e^(t-st) - e^-(t+st)] dt/2

=integral(0 to infinity)[e^(-(s-1)t) - e^-((1+s)t)] dt/2

= {[e^(-(s-1)t)/(-s+1) + e^-((1+s)t)/(s+1)](0, infinity) }/2

= {1/(s-1) - 1/(s+1) }/2

= 1/(s-1)(s+1)

= 1/(s^2 - 1)

2.

a)

for 0 <= t <= 2

here P(t) = 1

dy/dt + y = t

here Q = t

I.F. = e^integral P dt = e^t

so the solution is

y * e^t = integral[e^t * t]dt + C

or

y = (t - 1) + Ce^-t

b)

y = (t - 1) + Ce^-t

put x = 0 , y = 1

we get

1 = -1 + C

or

C = 2

y = (t - 1) + 2e^-t

c)

for t>2

P(t) = 3

I.F. = e^3t

solution is

y*e^3t = integral[e^3t * t]dt + C

or

y = (t-1)/3 + Ce^-3t

here t = 2

y = 1

1 = 1/3 + Ce^-6

or

C = (2/3)e^6

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