# L^{-1}bigg(frac{5}{s^{2}(s^{2}-4s+5)}bigg)=5int_{0}^{infty}(t-T)e^{2T}sin 2T dT Select one: True or False

Laplace transform
$$L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)=5\int_{0}^{\infty}(t-T)e^{2T}\sin 2T\ dT$$
Select one: True or False

2021-03-10

Step 1
Given:
Let us consider,
$$L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)$$
Step 2
$$L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)$$
$$=L^{-1}\bigg(\frac{5}{s^{2}}\cdot\frac{1}{s^2-4s+5}\bigg)$$
$$\text{Let us consider,}$$
$$H(s)=\frac{5}{s^{2}}$$
$$\text{By Applying the inverse laplace transform as below,}$$
$$L^{-1}(H(s))=5L^{-1}\bigg(\frac{1}{s^{2}}\bigg)$$
$$h(t)=5t$$
$$\text{Step 3}$$
$$\text{Let us consider,}$$
$$G(s)=\frac{1}{s^{2}-4s+5}$$
$$G(s)=\frac{1}{s^{2}-4s+4+1}$$
$$G(s)=\frac{1}{(s-2)^{2}+1}$$
$$\text{By Applying the inverse Laplace Transform, }$$
$$L^{-1}(G(s))=L^{-1}\bigg(\frac{1}{(s-2)^{2}+1}\bigg)$$
$$=e^{2t}\sin t$$
$$\text{Step 4}$$
$$L^{-1}(H(s)G(s))=\int_0^\infty h(t-T)\cdot g(T)$$
$$\text{Therefore, }$$
$$L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)=5\int_{0}^{\infty}(t-T)e^{2T}\sin T$$
Hence ,
The given Equation is False.