\magnification 1800 \parskip 10pt \parindent = 0pt \hsize 7truein \hoffset -0.4truein \vsize 9.7truein \def\u{\vskip -10pt} \def\v{\vskip 10pt} $g(t) = \cases{ 0 & $t < 4$ \cr 2 & $ 4 \leq t < 10$ \cr t & $t \geq 10$} Hence $g(t) = 2u_4(t) + (t - 2) u_{10}(t)$ \vskip 10pt \centerline{Solve $3y'' + y' + y = 2u_4(t) + (t - 2) u_{10}(t)$, } \centerline{$y(0) = 0$, $y'(0) = 0$.} $3{\cal L}(y'') + {\cal L}(y') + {\cal L}(y) $ $= {\cal L}(2u_4(t)) + {\cal L}( (t - 2) u_{10}(t))$} $3[s^2{\cal L}(y) - sy(0) - y'(0)] + s{\cal L}(y) - y(0) + {\cal L}(y) $ \rightline{$= {\cal L}(2u_4(t)) + {\cal L}( (t - 10 + 8) u_{10}(t))$} \v \centerline{Thm: ${\cal L}(u_c(t)f(t-c))= e^{-cs} {\cal L}(f(t)).$} \v \centerline{If $f(t) = \underline{\hskip 0.7in}$, then $f(t - \underline{\hskip 0.2in}) = \underline{\hskip 0.7in}$} \v \centerline{If $f(t) = \underline{\hskip 0.7in}$, then $f(t - \underline{\hskip 0.2in}) = \underline{\hskip 0.7in}$} $3[s^2{\cal L}(y)] + s{\cal L}(y) + {\cal L}(y) $ $= e^{-4s}{\cal L}(2) + e^{-10s}{\cal L}( (t + 8) )$} ${\cal L}(y) [3s^2 + s + 1] = 2e^{-4s}{\cal L}(1) + e^{-10s}{\cal L}(t) + 8e^{-10s}{\cal L}(1) $ ${\cal L}(y) [3s^2 + s + 1] = e^{-4s}{2 \over s} + e^{-10s}{1 \over s^2} + e^{-10s} {8 \over s} $ ${\cal L}(y) = e^{-4s}{2 \over s [3s^2 + s + 1] } + e^{-10s} {1 \over s^2[3s^2 + s + 1] } + 8e^{-10s} {1 \over s[3s^2 + s + 1] } $ $y = 2{\cal L}^{-1}(e^{-4s}{1 \over s [3s^2 + s + 1] }) + {\cal L}^{-1}(e^{-10s} {1 \over s^2[3s^2 + s + 1] }) $ \rightline{$ + 8{\cal L}^{-1}(e^{-10s} {1 \over s[3s^2 + s + 1] } ) $} \vskip 0.5in \vfill $y = u_4(t) f(t-4) + u_{10}h(t-10) + 8u_{10}f(t-10)$ where $f(t) = {\cal L}^{-1}({1 \over s [3s^2 + s + 1] }) $ and $h(t) = {\cal L}^{-1}({1 \over s^2 [3s^2 + s + 1] }) $ \v \hrule ${1 \over s [3s^2 + s + 1] } = {A \over s} + {Bs + C \over 3s^2 + s + 2} $ $1 = A(3s^2 + s + 1) + (Bs + C)s$ $0s^2 + 0s + 1 = (3A + B)s^2 + (A + C)s + A$ $0 = 3A + B$, $0 = A + C$, $1 = A$ Hence $A = 1$, $B = -3A = -3$, $C = -A = -1$ \v \hrule $f(t) = {\cal L}^{-1}({1 \over s [3s^2 + s + 1] }) $ $~~~~~~= {\cal L}^{-1}({1 \over s} + {-3s - 1 \over 3s^2 + s + 1} )$ $~~~~~~= {\cal L}^{-1}({1 \over s} + {-3s - 1 \over 3s^2 + s + 1} )$ $~~~~~~ = 1 + {\cal L}^{-1}({-3s - 1 \over 3[s^2 + {1 \over 3}s + {1 \over 3}]}) )$ $ ~~~~~~= 1 + {\cal L}^{-1}({-3s - 1 \over 3[(s^2 + {1 \over 3}s + \underline{\hskip 0.3in}) - \underline{\hskip 0.3in}+ {1 \over 3}]}) $ $~~~~~~ = 1 + {\cal L}^{-1}({-3s - 1 \over 3[(s + {1 \over 6})^2 - {1 \over 36} + {1 \over 3}]} )$ $ ~~~~~~= 1 +{\cal L}^{-1}( {-3(s + {1 \over 3}) \over 3[(s + {1 \over 6})^2 + {11 \over 36}]} )$ $ ~~~~~~= 1 + {\cal L}^{-1}({-(s + {1 \over 6}- {1 \over 6}+ {1 \over 3}) \over [(s + {1 \over 6})^2 + {11 \over 36}]}) $ $ ~~~~~~= 1 + {\cal L}^{-1}({-(s + {1 \over 6} + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]}) $ $ ~~~~~~= 1 + {\cal L}^{-1}({-(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} + {-{1 \over 6} \over [(s + {1 \over 6})^2 + {11 \over 36}]} )$ $~~~~~~ = 1 + {\cal L}^{-1}({-(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} + {-{1 \over 6} {6 \over \sqrt{11}} {\sqrt{11} \over 6} \over [(s + {1 \over 6})^2 + {11 \over 36}]} )$ $~~~~~~ = 1 + {\cal L}^{-1}({-(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} + {- {1 \over \sqrt{11}} {\sqrt{11} \over 6} \over [(s + {1 \over 6})^2 + {11 \over 36}]} )$ $~~~~~~ = 1 - e^{-{1 \over 6}t}cos {\sqrt{11} \over 6}t - {1 \over \sqrt{11}} e^{-{1 \over 6}t} sin{\sqrt{11} \over 6}t$ \hrule $h(t) = {\cal L}^{-1}({1 \over s^2 [3s^2 + s + 1] }) $ ${1 \over s^2 [3s^2 + s + 1] } = {As + D \over s^2} + {Bs + C \over 3s^2 + s + 2} $ $1 = (As + D)(3s^2 + s + 1) + (Bs + C)s^2$ $0 s^3 + 0s^2 + 0s + 1 = (3A + B)s^3 + (A + 3D + C)s^2 + (A + D)s + D$ $0 = 3A + B$, $0 = A + 3D + C$, $0 = A + D$, $1 = D$. Hence $D = 1$, $A = -D = -1$, $C = -A -3D = 1 - 3 = -2$, $B = -3A = 3$. \v \hrule ${1 \over s^2 [3s^2 + s + 1] } = {-s + 1 \over s^2} + {3s - 2 \over 3s^2 + s + 1} $ $ = {-s \over s^2} + {1 \over s^2} + {3(s - {2 \over 3}) \over 3[(s + {1 \over 6})^2 + {11 \over 36}]} } $ $ = {-1 \over s} + {1 \over s^2} + {(s + {1 \over 6} - {1 \over 6}- {2 \over 3}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } $ $ = {-1 \over s} + {1 \over s^2} + {(s + {1 \over 6} - {5 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } $ $ = {-1 \over s} + {1 \over s^2} + {(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } - {({5 \over 6})({6 \over \sqrt{11}} {\sqrt{11} \over 6} ) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } $ $ = {-1 \over s} + {1 \over s^2} + {(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } - {({5 \over 6})({6 \over \sqrt{11}} {\sqrt{11} \over 6} ) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } $ $ = {-1 \over s} + {1 \over s^2} + {(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } - {{5 \over \sqrt{11}} ({\sqrt{11} \over 6} ) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } $ $h(t) = {\cal L}^{-1}({1 \over s^2 [3s^2 + s + 1] }) $ $~~~~~~ = {\cal L}^{-1}({-1 \over s} + {1 \over s^2} + {(s + {1 \over 6}) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } - {{5 \over \sqrt{11}} ( {\sqrt{11} \over 6} ) \over [(s + {1 \over 6})^2 + {11 \over 36}]} } )$ $~~~~~~ = -1 + t + e^{-{1 \over 6}t}cos {\sqrt{11} \over 6}t - {5 \over \sqrt{11}} e^{-{1 \over 6}t}sin {\sqrt{11} \over 6}t $ Hence the final answer is $y = u_4(t) f(t-4) + u_{10}h(t-10) + 8u_{10}f(t-10)$ $~~~=u_4(t) [ 1 - e^{-{1 \over 6}(t-4)}cos {\sqrt{11} \over 6}(t-4) - {1 \over \sqrt{11}} e^{-{1 \over 6}(t-4)} sin{\sqrt{11} \over 6}(t-4)] + $ \line{$u_{10}[ -1 + t -10 + e^{-{1 \over 6}(t-10)}cos {\sqrt{11} \over 6}(t-10) - {5 \over \sqrt{11}} e^{-{1 \over 6}(t-10)}sin {\sqrt{11} \over 6}(t-10) ]$} \line{$+ 8u_{10}[ 1 - e^{-{1 \over 6}(t-10)}cos {\sqrt{11} \over 6}(t-10) - {1 \over \sqrt{11}} e^{-{1 \over 6}(t-10)} sin{\sqrt{11} \over 6}(t-10) ] $} \v \hrule \v\v HW due 4/8: 6.4: 2, 3, 4, 10; 6.5 \#2, 5, 8, 11, 12 Exam 2: Friday April 8 over 3.9, 6.1 - 6.5 \end