Inverse Laplace Transform Calculation

This solution details the process of finding the inverse Laplace transform, denoted as $f(t)$, for the given function $F(s)=\frac{s+1+\frac{1}{s^2}}{s^2-1}$. We will first simplify the expression for $F(s)$, then apply partial fraction decomposition, and finally compute the inverse Laplace transform of the resulting simpler terms.

Simplifying the Function F(s)

The initial step involves simplifying the numerator of $F(s)$ by finding a common denominator ($s^2$):

$F(s) = \frac{s+1+\frac{1}{s^2}}{s^2-1} = \frac{\frac{s \cdot s^2 + 1 \cdot s^2 + 1}{s^2}}{s^2-1}$

Combine the terms in the numerator:

$F(s) = \frac{\frac{s^3+s^2+1}{s^2}}{s^2-1}$

Now, express $F(s)$ as a single rational function:

$F(s) = \frac{s^3+s^2+1}{s^2(s^2-1)}$

Partial Fraction Decomposition of F(s)

To easily find the inverse Laplace transform, we break down $F(s)$ into simpler fractions using partial fraction decomposition. The denominator factors as $s^2(s^2-1) = s^2(s-1)(s+1)$. We express $F(s)$ in the following form:

$F(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s-1} + \frac{D}{s+1}$

To find the constants A, B, C, and D, we multiply both sides by the common denominator $s^2(s-1)(s+1)$:

$s^3+s^2+1 = As(s-1)(s+1) + B(s-1)(s+1) + Cs^2(s+1) + Ds^2(s-1)$

We determine the coefficients by substituting convenient values for $s$:

• Let $s=0$: Substitute $s=0$ into the equation.

  $0^3+0^2+1 = A(0) + B(-1)(1) + C(0) + D(0)$

  $1 = -B \implies B = -1$

• Let $s=1$: Substitute $s=1$.

  $1^3+1^2+1 = A(0) + B(0) + C(1)^2(1+1) + D(0)$

  $3 = 2C \implies C = \frac{3}{2}$

• Let $s=-1$: Substitute $s=-1$.

  $(-1)^3+(-1)^2+1 = A(0) + B(0) + C(0) + D(-1)^2(-1-1)$

  $1 = D(1)(-2) \implies 1 = -2D \implies D = -\frac{1}{2}$

To find $A$, we compare the coefficients of the $s^3$ term on both sides of the expanded equation:

$s^3+s^2+1 = A(s^3-s) + B(s^2-1) + C(s^3+s^2) + D(s^3-s^2)$

$s^3+s^2+1 = (A+C+D)s^3 + (B+C-D)s^2 - As - B$

Equating the coefficients of $s^3$:

$1 = A+C+D$

Substitute the values $C = \frac{3}{2}$ and $D = -\frac{1}{2}$:

$1 = A + \frac{3}{2} - \frac{1}{2}$

$1 = A + \frac{2}{2} \implies 1 = A + 1 \implies A = 0$

Thus, the partial fraction decomposition of $F(s)$ is:

$F(s) = \frac{0}{s} + \frac{-1}{s^2} + \frac{\frac{3}{2}}{s-1} + \frac{-\frac{1}{2}}{s+1}$

Simplifying this gives:

$F(s) = -\frac{1}{s^2} + \frac{3}{2}\frac{1}{s-1} - \frac{1}{2}\frac{1}{s+1}$

Calculating the Inverse Laplace Transform

Now, we find the inverse Laplace transform $f(t)$ using the linearity property and standard inverse Laplace transform pairs:

• $L^{-1}\left\{\frac{1}{s^2}\right\} = t$
• $L^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}$

Applying the inverse transform to each term of $F(s)$:

$f(t) = L^{-1}\left\{-\frac{1}{s^2}\right\} + L^{-1}\left\{\frac{3}{2}\frac{1}{s-1}\right\} - L^{-1}\left\{\frac{1}{2}\frac{1}{s+1}\right\}$

$f(t) = -1 \cdot L^{-1}\left\{\frac{1}{s^2}\right\} + \frac{3}{2} \cdot L^{-1}\left\{\frac{1}{s-1}\right\} - \frac{1}{2} \cdot L^{-1}\left\{\frac{1}{s+1}\right\}$

Substitute the known inverse transforms:

$f(t) = -t + \frac{3}{2}e^{1 \cdot t} - \frac{1}{2}e^{-1 \cdot t}$

$f(t) = -t + \frac{3}{2}e^t - \frac{1}{2}e^{-t}$

Expressing the Result in Terms of Hyperbolic Sine

The calculated inverse Laplace transform is $f(t) = \frac{3}{2}e^t - t - \frac{1}{2}e^{-t}$.

We can rewrite this expression using the definition of the hyperbolic sine function, $\sinh t = \frac{e^t - e^{-t}}{2}$.

Let's rearrange the terms of $f(t)$:

$f(t) = e^t + \frac{1}{2}e^t - t - \frac{1}{2}e^{-t}$

Group the exponential terms and the hyperbolic sine terms:

$f(t) = e^t - t + \left(\frac{1}{2}e^t - \frac{1}{2}e^{-t}\right)$

The term in the parenthesis is exactly $\sinh t$:

$f(t) = e^t - t + \sinh t$

Rearranging the terms to match the option format:

$f(t) = e^t + \sinh t - t$

This matches the expression given in option 2.