Find L[e3t + e⁻³t] and L⁻¹[1/(s(s+1))].

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6. Find $L[e^{3t} + e^{-3t}]$

Use the linearity property of Laplace transforms.
$L[e^{3t} + e^{-3t}] = L[e^{3t}] + L[e^{-3t}]$
Apply the standard Laplace transform formula $L[e^{at}] = \frac{1}{s-a}$ .
$L[e^{3t}] = \frac{1}{s-3}$ $L[e^{-3t}] = \frac{1}{s-(-3)} = \frac{1}{s+3}$
Combine the terms and simplify.
$L[e^{3t} + e^{-3t}] = \frac{1}{s-3} + \frac{1}{s+3}$ $L[e^{3t} + e^{-3t}] = \frac{(s+3) + (s-3)}{(s-3)(s+3)}$ $L[e^{3t} + e^{-3t}] = \frac{2s}{s^2 - 3^2}$ $L[e^{3t} + e^{-3t}] = \boxed{\frac{2s}{s^2 - 9}}$

7. Find $L^{-1}\left[\frac{1}{s(s+1)}\right]$
Use partial fraction decomposition for $\frac{1}{s(s+1)}$ .
Let $\frac{1}{s(s+1)} = \frac{A}{s} + \frac{B}{s+1}$ .
Multiplying by $s(s+1)$ gives $1 = A(s+1) + Bs$ .
Setting $s=0 \implies 1 = A(1) \implies A=1$ .
Setting $s=-1 \implies 1 = B(-1) \implies B=-1$ .
So, $\frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1}$ .
Apply the inverse Laplace transform to each term.
$L^{-1}\left[\frac{1}{s(s+1)}\right] = L^{-1}\left[\frac{1}{s} - \frac{1}{s+1}\right]$ $L^{-1}\left[\frac{1}{s(s+1)}\right] = L^{-1}\left[\frac{1}{s}\right] - L^{-1}\left[\frac{1}{s+1}\right]$
Use the standard inverse Laplace transform formulas $L^{-1}\left[\frac{1}{s}\right] = 1$ and $L^{-1}\left[\frac{1}{s+a}\right] = e^{-at}$ .
$L^{-1}\left[\frac{1}{s(s+1)}\right] = 1 - e^{-t}$ $L^{-1}\left[\frac{1}{s(s+1)}\right] = \boxed{1 - e^{-t}}$

8. Evaluate $L^{-1}\left[\log\left(\frac{s^2+1}{s(s+1)}\right)\right]$
Let $F(s) = \log\left(\frac{s^2+1}{s(s+1)}\right)$ .
Use logarithm properties to expand $F(s)$ .
$F(s) = \log(s^2+1) - \log(s) - \log(s+1)$

Find L[e3t + e⁻³t] and L⁻¹[1/(s(s+1))].

Step-by-step solution

Use the linearity property of Laplace transforms.

Apply the standard Laplace transform formula $L[e^{at}] = \frac{1}{s-a}$ .

Combine the terms and simplify.

Use partial fraction decomposition for $\frac{1}{s(s+1)}$ .

Apply the inverse Laplace transform to each term.

Use the standard inverse Laplace transform formulas $L^{-1}\left[\frac{1}{s}\right] = 1$ and $L^{-1}\left[\frac{1}{s+a}\right] = e^{-at}$ .

Let $F(s) = \log\left(\frac{s^2+1}{s(s+1)}\right)$ .

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Find L[e3t + e⁻³t] and L⁻¹[1/(s(s+1))].

Step-by-step solution

Use the linearity property of Laplace transforms.

Apply the standard Laplace transform formula L[eat]=1s−aL[e^{at}] = \frac{1}{s-a}L[eat]=s−a1​.

Combine the terms and simplify.

Use partial fraction decomposition for 1s(s+1)\frac{1}{s(s+1)}s(s+1)1​.

Apply the inverse Laplace transform to each term.

Use the standard inverse Laplace transform formulas L−1[1s]=1L^{-1}\left[\frac{1}{s}\right] = 1L−1[s1​]=1 and L−1[1s+a]=e−atL^{-1}\left[\frac{1}{s+a}\right] = e^{-at}L−1[s+a1​]=e−at.

Let F(s)=log⁡(s2+1s(s+1))F(s) = \log\left(\frac{s^2+1}{s(s+1)}\right)F(s)=log(s(s+1)s2+1​).

Still stuck on this one?

More Mathematics Questions

Apply the standard Laplace transform formula $L[e^{at}] = \frac{1}{s-a}$ .

Use partial fraction decomposition for $\frac{1}{s(s+1)}$ .

Use the standard inverse Laplace transform formulas $L^{-1}\left[\frac{1}{s}\right] = 1$ and $L^{-1}\left[\frac{1}{s+a}\right] = e^{-at}$ .

Let $F(s) = \log\left(\frac{s^2+1}{s(s+1)}\right)$ .