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Problems #6 and #12 are on pg.320 are graded.

**#6,pg.320:** Find the inverse Laplace transform of $\dfrac{2s-3}{s^2-4}$.

**Solution:** We will use partial fractions to expand this function:
$$\dfrac{2s-3}{s^2-4} = \dfrac{A}{s-2} + \dfrac{B}{s+2},$$
multiply by the denominator to get
$$2s-3 = A(s+2)+B(s-2)=(A+B)s+(2A-2B),$$
and equating coefficients yields the following system of equations:
$$\left\{ \begin{array}{ll}
A+B=2 \\
2A-2B=-3,
\end{array} \right.$$
which has solution $A=\dfrac{1}{4}$ and $B=\dfrac{7}{4}$. Thus using the formula $\mathscr{L}^{-1}\left\{\dfrac{1}{s-a} \right\}(t)=e^{at}$ we compute
$$\mathscr{L}^{-1} \left\{\dfrac{2s-3}{s^2-4} \right\}(t)= \dfrac{1}{4} \mathscr{L}^{-1} \left\{ \dfrac{1}{s-2} \right\}+ \dfrac{7}{4} \mathscr{L}^{-1} \left\{ \dfrac{1}{s+2} \right\}(t)=\dfrac{e^{2t}}{4} + \dfrac{7e^{-2t}}{4}.$$

**#12,pg.320:** Use Laplace transforms to solve the IVP
$$y''+3y'+2y=0; y(0)=1,y'(0)=0.$$

**Solution:** Take the Laplace transform of both sides of the differential equation to get
$$(s^2\mathscr{L}\{y\}-sy(0)-y'(0))+3(s\mathscr{L}\{y\}-y(0))+2\mathscr{L}\{y\}=0$$
and apply the initial conditions to get
$$(s^2+3s+2)\mathscr{L}\{y\}=s+3$$
divide by $s^2+3s+2$ to get
$$\mathscr{L}\{y\} = \dfrac{s+3}{s^2+3s+2}.$$
Partial fractions shows us that
$$\dfrac{s+3}{s^2+3s+2} = \dfrac{2}{s+1} - \dfrac{1}{s+2}.$$
So we solve the IVP by computing
$$y(t) = \mathscr{L}^{-1} \left\{ \dfrac{s+3}{s^2+3s+2} \right\}(t) = \mathscr{L}^{-1} \left\{ \dfrac{2}{s+1} \right\}(t) - \mathscr{L}^{-1} \left\{ \dfrac{1}{s+2} \right\}(t)=2e^{-t}-e^{-2t}.$$