Question

Find the inverse Laplace transforms, as a function of x, of the following functions:

Answer 1

Answer:  The required answer is

$f(x)=e^x+\cos x+\sin x.$

Step-by-step explanation:  We are given to find the inverse Laplace transform of the following function as a function of x :

$F(s)=\dfrac{2s^2}{(s-1)(s^2+1)}.$

We will be using the following formulas of inverse Laplace transform :

$(i)~L^{-1}\{\dfrac{1}{s-a}\}=e^{ax},\\\\\\(ii)~L^{-1}\{\dfrac{s}{s^2+a^2}\}=\cos ax,\\\\\\(iii)~L^{-1}\{\dfrac{1}{s^2+a^2}\}=\dfrac{1}{a}\sin ax.$

By partial fractions, we have

$\dfrac{s^2}{(s-1)(s^2+1)}=\dfrac{A}{s-1}+\dfrac{Bs+C}{s^2+1},$

where A, B and C are constants.

Multiplying both sides of the above equation by the denominator of the left hand side, we get

$2s^2=A(s^2+1)+(Bs+C)(s-1).$

If s = 1, we get

$2\times 1=A(1+1)\\\\\Rightarrow A=1.$

Also,

$2s^2=A(s^2+1)+(Bs^2-Bs+Cs-C)\\\\\Rightarrow 2s^2=(A+B)s^2+(-B+C)s+(A-C).$

Comparing the coefficients of x² and 1, we get

$A+B=2\\\\\Rightarrow B=2-1=1,\\\\\\A-C=0\\\\\Rightarrow C=A=1.$

So, we can write

$\dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s+1}{s^2+1}\\\\\\\Rightarrow \dfrac{2s^2}{(s-1)(s^2+1)}=\dfrac{1}{s-1}+\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}.$

Taking inverse Laplace transform on both sides of the above, we get

$L^{-1}\{\dfrac{2s^2}{(s-1)(s^2+1)}\}=L^{-1}\{\dfrac{1}{s-1}\}+L^{-1}\{\dfrac{s}{s^2+1}+\dfrac{1}{s^2+1}\}\\\\\\\Rightarrow f(x)=e^{1\times x}+\cos (1\times x)+\dfrac{1}{1}\sin(1\times x)\\\\\\\Rightarrow f(x)=e^x+\cos x+\sin x.$

Thus, the required answer is

$f(x)=e^x+\cos x+\sin x.$