Respuesta :
Answer: The required answer is
[tex]f(x)=e^x+\cos x+\sin x.[/tex]
Step-by-step explanation: We are given to find the inverse Laplace transform of the following function as a function of x :
[tex]F(s)=\dfrac{2s^2}{(s-1)(s^2+1)}.[/tex]
We will be using the following formulas of inverse Laplace transform :
[tex](i)~L^{-1}\{\dfrac{1}{s-a}\}=e^{ax},\\\\\\(ii)~L^{-1}\{\dfrac{s}{s^2+a^2}\}=\cos ax,\\\\\\(iii)~L^{-1}\{\dfrac{1}{s^2+a^2}\}=\dfrac{1}{a}\sin ax.[/tex]
By partial fractions, we have
[tex]\dfrac{s^2}{(s-1)(s^2+1)}=\dfrac{A}{s-1}+\dfrac{Bs+C}{s^2+1},[/tex]
where A, B and C are constants.
Multiplying both sides of the above equation by the denominator of the left hand side, we get
[tex]2s^2=A(s^2+1)+(Bs+C)(s-1).[/tex]
If s = 1, we get
[tex]2\times 1=A(1+1)\\\\\Rightarrow A=1.[/tex]
Also,
[tex]2s^2=A(s^2+1)+(Bs^2-Bs+Cs-C)\\\\\Rightarrow 2s^2=(A+B)s^2+(-B+C)s+(A-C).[/tex]
Comparing the coefficients of x² and 1, we get
[tex]A+B=2\\\\\Rightarrow B=2-1=1,\\\\\\A-C=0\\\\\Rightarrow C=A=1.[/tex]
So, we can write
[tex]\dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s+1}{s^2+1}\\\\\\\Rightarrow \dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}.[/tex]
Taking inverse Laplace transform on both sides of the above, we get
[tex]L^{-1}\{\dfrac{2s^2}{(s-1)(s^2+1)}\}=L^{-1}\{\dfrac{1}{s-1}\}+L^{-1}\{\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\}\\\\\\\Rightarrow f(x)=e^{1\times x}+\cos (1\times x)+\dfrac{1}{1}\sin(1\times x)\\\\\\\Rightarrow f(x)=e^x+\cos x+\sin x.[/tex]
Thus, the required answer is
[tex]f(x)=e^x+\cos x+\sin x.[/tex]
