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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 $$y(t) = \mathscr{L}^{-1} \left\{ \dfrac{s+3}{s^2+3s+2} \right\}(t) = \mathscr{L}^{-1} \left\{ \dfrac{2}{s+1} \right\}(t) - \mathscr{L}^{-1} \left\{ \dfrac{1}{s+2} \right\}(t)=2e^{-t}-e^{-2t}.$$