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Problems #4 on pg.350 and #8 on pg.360 are graded.

#4,pg.350: Find the Laplace transform of the function $$f(t) = \displaystyle\int_0^t (t-\tau)^2 \cos(2\tau) d\tau.$$
Solution: Recall the Convolution Theorem says that $$\mathscr{L}\{f*g\}=\mathscr{L}\{f\}\mathscr{L}\{g\}.$$ Notice that we may express this integral as $(f*g)(t)$ where $f(t)=t^2$ and $g(t)=\cos(2t)$. We may compute $\mathscr{L}\{f\}=\dfrac{2}{s^3}$ and $\mathscr{L}\{g\}=\dfrac{s}{s^2+4}.$ Thus using the convolution theorem, we see $$\mathscr{L}\{f*g\} = \mathscr{L}\{f\} \mathscr{L}\{g\} = \dfrac{2}{s^3} \dfrac{s}{s^2+4} = \dfrac{2}{s^4+4s^2}.$$

#8, pg.360: Transform the following system of differential equations into a second order equation to solve the system: $$\left\{ \begin{array}{ll} y_1' = 3y_1-2y_2&; y_1(0)=3 \\ y_2' = 2y_1-2y_2 &; y_2(0) = \dfrac{1}{2}. \end{array} \right.$$
Solution: From the first equation we have $$y_2 = \dfrac{3}{2}y_1 - \dfrac{1}{2}y_1'.$$ Calculate $$y_2' = \dfrac{3}{2}y_1'-\dfrac{1}{2}y_1''.$$ Plug these into the second equation in the system to see $$\dfrac{3}{2}y_1'-\dfrac{1}{2}y_1'' = 2y_1 - 2 \left( \dfrac{3}{2}y_1 - \dfrac{1}{2} y_1' \right),$$ distribute the $2$ to get $$\dfrac{3}{2}y_1'-\dfrac{1}{2}y_1''=2y_1 -3y_1+y_1'$$ multiply by $-2$ to get $$-3y_1'+y_1''=-4y_1+6y_1-2y_1'$$ simplify to get $$y_1''-y_1'-2y_1=0.$$ This 2nd order ODE has characteristic equation $r^2-r-2=(r-2)(r+1)=0$ with roots $r=-1,2$. Hence $$y_1(t)=c_1e^{-t} + c_2 e^{2t}.$$ The initial condition applied to this yields $$3=y_1(0)=c_1+c_2.$$ Differentiate the formula for $y_1$ to get $$y_1'(t) = -c_1e^{-t}+2c_2 e^{2t}.$$ Plug our formulas for $y_1$ and $y_1'$ into the equation defining $y_2(t)$ to get $$y_2(t)=\dfrac{3}{2} \left( c_1 e^{-t} +c_2e^{2t} \right) - \dfrac{1}{2} \left( -c_1e^{-t}+2c_2e^{2t} \right) = 2c_1e^{-t} +\dfrac{1}{2}c_2e^{2t}.$$ Apply the other initial condition $y_2(0)=\dfrac{1}{2}$ to get $$\dfrac{1}{2} = y_2(0) = 2c_1 + \dfrac{1}{2} c_2.$$ Hence we have the system of equations $$\left\{ \begin{array}{ll} c_1+c_2 &= 3 \\ 2c_1 + \dfrac{1}{2}c_2 &= \dfrac{1}{2} \end{array} \right.$$ which has solution $c_1=-\dfrac{2}{3}$ and $c_2=\dfrac{11}{3}$. Hence we see that the solution is given by $$y_1(t)=-\dfrac{2}{3}e^{-t}+\dfrac{11}{3}e^{2t}$$ and $$y_2(t)=-\dfrac{4}{3}e^{-t}+\dfrac{11}{6}e^{2t}.$$