Respuesta :
Answer: The required solution of the given IVP is
[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
Step-by-step explanation: We are given to use Laplace transforms to solve the following initial value problem :
[tex]x^{\prime\prime}+9x=1,~~~x(0)=0=x^\prime(0).[/tex]
We will be using the following formula for the Laplace transform :
[tex](i)~L\{t^n\}=\dfrac{n!}{s^{n+1}},\\\\\\(ii)~L\{\coskt\}=\dfrac{s}{s^2+k^2}.[/tex]
Applying Laplace transform on both sides of the above equation, we have
[tex]L\{x^{\prime\prime}+9x\}=L\{1\}\\\\\\\Rightarrow s^2X(s)-sx(0)-x^\prime(0)+9X(s)=\dfrac{1}{s}\\\\\\\Rightarrow s^2X(s)-s\times0-0=\dfrac{1}{s}\\\\\\\Rightarrow (s^2+9)X(s)=\dfrac{1}{s}\\\\\\\Rightarrow X(s)=\dfrac{1}{s(s^2+9)}\\\\\\\Rightarrow X(s)=\dfrac{1}{9s}-\dfrac{s}{9(s^2+9)}.[/tex]
Taking inverse Laplace transform on both sides of the above equation, we get
[tex]L^{-1}\{X(s)\}=L^{-1}\left(\dfrac{1}{9s}\right)-L^{-1}\left(\dfrac{s}{9(s^2+9)}\right)\\\\\\\Rightarrow x(t)=\dfrac{1}{9}\times1-\dfrac{1}{9}\times \cos3t\\\\\\\Rightarrow x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
Thus, the required solution of the given IVP is
[tex]x(t)=\dfrac{1}{9}(1-\cos3t).[/tex]
