Respuesta :
The inverse Laplace transform of the given Laplace transform will be given below.
[tex]\rm L^{-1} \left [ \dfrac{1}{s^2 (s^2 + a^2 )} \right ] = \dfrac{1}{a^3} \left [ at - \sin at \right ][/tex]
What is the inverse Laplace transform?
The transformation of Laplace. The change of a Laplace transform into a variable of the period is called the inverse Laplace transform.
Let the Laplace equation will be
[tex]\rm F(s) = \dfrac{1}{s^2 + a^2} \ and \ G(s) = \dfrac{1}{s^2}[/tex]
So the given function will be
= F(s)G(s)
We know that
[tex]\rm L^{-1} [F(s)] = L^{-1}\left [ \dfrac{1}{s^2 + a^2 } \right ] = \dfrac{\sin at }{a} = f(t)[/tex]
And
[tex]\rm L^{-1} [G(s)] = L^{-1}\left [ \dfrac{1}{s^2 } \right ] = t = g(t)[/tex]
According to the Convolution Theorem, performing their overlap and then Laplace is equivalent to getting the Laplace first and then combining the two Laplace Transforms.
Now using the convolution theorem, then we get
[tex]\begin{aligned} L^{-1}\left [ \dfrac{1}{s^2(s^2 +a^2)} \right ] &= L^{-1}\left [ F(s)G(s) \right ]\\\\&= \int_{0}^{t} f(z)g(t-z) dz\\\\&= \int_{0}^{t} \frac{\sin az}{a}(t-z)dz\\\\&= \frac{1}{a} \int_{0}^{t} \sin az (t-z)dz \end[/tex]
On integration, we get
[tex]= \dfrac{1}{a}\left [ \left ( -\dfrac{\cos az}{a} (t-z) \right )_{0}^{t} - \int_{0}^{t}-\dfrac{\cos az}{a}(-1)dz \right ]\\\\\\= \dfrac{1}{a} \left [ \dfrac{-1}{a} \cos at (t-t) + \dfrac{1}{a} \cos 0 (t -0)\right ] - \frac{-1}{a}\left ( \dfrac{\sin az}{a} \right )_0^t \\\\\\= \dfrac{1}{a}\left [ 0 + \dfrac{1}{a}t- \dfrac{1}{a^2} \left ( \sin at - 0 \right ) \right ] \\\\\\= \dfrac{1}{a} \left [ \dfrac{at - \sin at}{a^2} \right ] \\\\\\=\dfrac{1}{a^3} \left [ at - \sin at \right ][/tex]
Then the inverse of the Laplace transform will be
[tex]\rm L^{-1} \left [ \dfrac{1}{s^2 (s^2 + a^2 )} \right ] = \dfrac{1}{a^3} \left [ at - \sin at \right ][/tex]
More about the inverse Laplace transform link is given below.
https://brainly.com/question/1675085
#SPJ4
