**Laplace Transforms. Please answer part b and c. Thanks. **

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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/4step: 1Please follow the below explanation:To solve the ODE, we begin by taking the Laplace transform of both sides:\( \ce{s^2 X(s)+2αsX(s)+(α^2 +ω^2 )X(s)=L[F(t)]= {1−e^(−sT)}/s .} \)Simplifying, we obtain,\( \ce{X(s)= {1−e^( −sT)}} \)/{ \( \ce{s(s^2 +2αs+α^2 +ω^2 ) .} \)we can write denominator as :\( \ce{s^2 +2αs+α^2 +ω^2 =(s+α)^2 +(ω^2 −α^2 ), } \)which suggests that we complete the square and use the formula for the inverse Laplace transform of a damped harmonic oscillator:\( \ce{L^(-1) [ 1/{(s+α) 2 +(ω^2 −α^2 )} ]= 1/ω} \)\( \ce{sin(ω(t−t_0 ))e^(−α(t−t_0 )) ,} \)Explanation:Please refer to step in this solutionExplanation:Please refer to solution in this step.Step2/4where t_0 is a constant of integration. Then we can use the shifting theorem to find the inverse Laplace transform of \( \ce{1/(s(s^2+2\alpha s + \alpha^2+\omega^2)) :} \)\( \ce{L^(−1) [ 1/{ s(s^2 +2αs+α^2 +ω^2 )} ]= 1/ω^2 } \)\( \ce{ (1−e ^(−α(t−t_0 ) ) cos(ω(t−t_0 ))−α(t−t_0} \)\( \ce{)e^(−α(t−t_0 ) ) sin(ω(t−t_0 ))). } \)Finally, we can use the Laplace transform of a step function to obtain the inverse Laplace transform of F(s) : Putting this altogether we get\( \ce{x(t)= 1/ω^2 int_0^t (1−e^(−α(t−τ))} \)\( \ce{cos(ω(t−τ))−α(t−τ)e −α(t−τ) sin(ω(t−τ)))[u(τ)} \)\( \ce{−u(τ−T)e −T(t−τ) ]dτ. To show that} \)\( \ce{x(t)=0 for t>T , we note that for t>T , we} \)\( \ce{have u(t-T) = 0 , so the integral ... See the full answer