Y(t) satisfies the initial value problem: dy/dt - 4y = 5 H(t - 5), y(0) = - 3 Fi

Question

Y(t) satisfies the initial value problem: dy/dt - 4y = 5 H(t - 5), y(0) = - 3 Find the solution of the initial value problem using Laplace transforms. Your answer should be expressed as a function of t using the correct syntax For example: 3*H(t - 1)*exp (-2*(t - 1)) + 5* exp(3*t) Solution is y(t) = y(t) satisfies the initial value problem: d^2y/dt^2 - 6 dy/dt - 7y = - 168 H(t - 1), y(0) = - 5, y'(0) = - 35. Find the solution of the initial value problem using Laplace transforms. Your answer should be expressed as a function of t using the correct syntax For example: 3*H(t - 1)*(3 + exp(- 2*(t - 1))) + 5*exp(3*t) the correct answer is: y(t) = 3H(t - 1)[8 - e^7(t - 1) - 7e^-(t - 1)] + (-5)e^7t in the correct syntax: 3*H(t - 1)*(8 - exp(7*(t - 1)) - 7*exp(- (t - 1))) + (- 5)*exp(7*t) You should review the Operational Theorem for Derivatives and the Second Shift Theorem.

Explanation / Answer

Proof:

we first prove this theorem for the case when f'(t) is continuous on[0,) then

L{f'(t)}=e-stf'(t)dt

=[e-stf(t)]+se-stf(t)dt

=-f(0)+sL{f(t)},s>y

Let f(t)=sin2t

Then f(0)=0,f(t)=2 sint cost=sin2t

and hence

L{f(t)}=2/s2+4

s L{f(t)}=2/s2+4or L{f(t)}=2/s(s2+4)

or |y|=e(2t+c)=eCe2t=c1e2T

where C1=eC is an arbitrary but always positive constants

to simplify one step father we can drop the absolute value sign and

rela% the restriction c1,c1 can now be any positive or negative constant hence

y(t)=c1e2T,c1#0

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