91Ó°ÊÓ

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right),$

where,$f={\mathcal{L}}^{-1}\left\{F\right\}$.Use this equation in Problems 33-36 to compute${\mathcal{L}}^{-1}\left\{F\right\}.$$F\left(s\right)=ln\left(\frac{s+2}{s-5}\right)$

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right),$

Where$f={\mathcal{L}}^{-1}\left\{F\right\}$.Use this equation in Problems 33-36 to compute${\mathcal{L}}^{-1}\left\{F\right\}$.$F\left(s\right)=ln\left(\frac{s-4}{s-3}\right)$.

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right)$,

Whererole="math" localid="1664423939060" $f={\mathcal{L}}^{-1}\left(F\right)$.Use this equationin Problems 33-36 to compute${\mathcal{L}}^{-1}\left(F\right).$$F\left(s\right)=ln\left(\frac{s^2+9}{s^2+1}\right)$

Theorem 6 in Section 7.3 on page 364 can be expressed in terms of the inverse Laplace transform as

${\mathcal{L}}^{-1}\left\{\frac{d^{n}F}{d{s}^n}\right\}\left(t\right)={\left(-t\right)}^{n}f\left(t\right)$

Where $f={\mathcal{L}}^{-1}\left\{F\right\}$.Use this equation in Problems 33-36 to compute

${\mathcal{L}}^{-1}\left\{F\right\}$.role="math" localid="1664422930667" $F\left(s\right)=arctan\left(\frac{1}{s}\right)$

Prove Theorem 7, page 368, on the linearity of the inverse transform. [Hint: Show that the right-hand side of equation (3) is a continuous function on $[0,∞)$ whose Laplace transform is$\left[F_1\left(s\right)+F_2\left(s\right)\right]$

Residue Computation. Let $\frac{P\left(s\right)}{Q\left(s\right)}$ be a rational function with deg$P<degQ$ and suppose $s-r$ is a non-repeated linear factor of $Q\left(s\right)$ . Prove that the portion of the partial fraction expansion of $\frac{P\left(s\right)}{Q\left(s\right)}$ corresponding to $s-r$ is $\frac{A}{s-r}$, where $A$ (called the residue) is given by the formula

role="math" localid="1664365632102" $A=\underset{s→r}{lim}\frac{\left(s-r\right)P\left(s\right)}{Q\left(s\right)}=\frac{P\left(r\right)}{Q^{\text{'}}\left(r\right)}$

Use the residue computation formula derived in Problem 38 to determine quickly the partial fraction expansion for$F\left(s\right)=\frac{2s+1}{s\left(s-1\right)\left(s+2\right)}$

In Problems $1-14$ , solve the given initial value problem using the method of Laplace transforms.

$\mathbf{3}\mathbf{.}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{0}\right)\mathbf{=}\mathbf{6}$

Heaviside's Expansion Formula. Let $P\left(s\right)$ and $Q\left(s\right)$ be polynomials with the degree of $P\left(s\right)$ less than the degree of $Q\left(s\right)$ . Let

role="math" localid="1664358171256" $Q\left(s\right)=\left(s-r_1\right)\left(s-r_2\right)...\left(s-r_n\right)$role="math" localid="1664359866810" $,{\mathcal{L}}^{-1}\left\{\frac{P}{Q}\right\}\left(t\right)=\underset{i=1}{\overset{n}{∑}}\frac{P\left(r_i\right)}{P\left(r_i\right)}{e}^{r_{i}t}$

Use the residue formulas derived in Problems 38 and 42 to determine the partial fraction expansion for

$F\left(s\right)=\frac{6s^2+28}{\left(s^2-2s+5\right)(s+2)}$