Question: Question 4(a) Compute the Laplace transforms of the following functions.(i) g(t)=∫0te-2ucos3(t-u)du(ii) h(t)={3,0≤t<2t2+1,t≥2(b) Use Laplace

Question 4(

a)

Compute the Laplace transforms of the following functions.(

i) g(t)=∫0te-2ucos3(t-u)du(

ii) h(t)={3,0≤t<2t2+1,t≥2(

b)

Use Laplace transform to solve the following initial value problem.y”-4y’+3y=8f(t)

with y'(0)=1,y(0)=0

and f(t)={0,0≤t<1e5t,t≥1

ANS:

a. Compute the Laplace transforms of the following functions.

i. g(t)=∫0te-2ucos3(t-u)du

Using the convolution theorem, we can write the Laplace transform of g(t) as: L[g(t)]=e^(-2s)L{cos3t}=e^(-2s)(s^2+9)^(-1/2) = (e^(-2s)/(√(s^2+9)))

ii. h(t)={3,0≤t<2t2+1,t≥2

Using the piecewise definition of h(t), we can write its Laplace transform as: L[h(t)]=e^(-2s)L{3}+e^(-2s)L{t^2+1}=e^(-2s)(3+1/(s^2+1)) = (3e^(-2s) + (e^(-2s)/(s^2+1)))

b. Use Laplace transform to solve the following initial value problem.

y”-4y’+3y=8f(t) with y'(0)=1,y(0)=0 and f(t)={0,0≤t<1e5t,t≥1

Taking the Laplace transform of both sides of the equation, we get: s^2Y(s)-4sY(s)+3Y(s)=8F(s)

where Y(s) is the Laplace transform of y(t) and F(s) is the Laplace transform of f(t). Substituting the given values of f(t), we get: s^2Y(s)-4sY(s)+3Y(s)=8(1/(s+5))(1-e^(-st))

Solving for Y(s), we get: Y(s)=8(s+5)/(s^2-4s+3)(s+5)+1/(s+5)

Using partial fractions, we can decompose Y(s) as: Y(s)=(s+1)/(s-3)+(4/(s-3))+(2/(s+5))

Taking the inverse Laplace transform of Y(s), we get: y(t)=(e^(3t)+4e^(3t)-2e^(-5t))u(t-1)

where u(t-1) is the unit step function. Therefore, the solution to the initial value problem is: y(t)=(e^(3t)+4e^(3t)-2e^(-5t))u(t-1)

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