\magnification 2200 \parindent 0pt \parskip 10pt \pageno=1 \hsize 7truein \vsize 9.2truein \def\u{\vskip -5pt} \def\v{\vskip 10pt} The LaPlace Transform is a method to change a differential equation to a linear equation. Example: Solve $y'' + 3y' + 4y = 0$, $y(0) = 5$, $y'(0) = 6$ 1.) Take the LaPlace Transform of both sides of the equation: ${\cal L}(y'' + 3y' + 4y) = {\cal L}(0)$ 2.) Use the fact that the LaPlace Transform is linear: ${\cal L}(y'') + 3{\cal L}(y') + 4{\cal L}(y) = 0$ 3.) Use thm to change this equation into an algebraic equation: $s^2{\cal L}(y) - sy(0) - y'(0) + 3[s{\cal L}(y) - y(0)] + 4{\cal L}(y) = 0$ 3.5) Substitute in the initial values: $s^2{\cal L}(y) - 5s - 6 + 3[s{\cal L}(y) - 5] + 4{\cal L}(y) = 0$ \eject 4.) Solve the algebraic equation for ${\cal L}(y)$ $s^2{\cal L}(y) - 5s - 6 + 3s{\cal L}(y) - 15 + 4{\cal L}(y) = 0$ $[s^2 + 3s + 4]{\cal L}(y) = 5s + 21$ ${\cal L}(y) = {5s + 21 \over s^2 + 3s + 4}$ Some algebra implies ${\cal L}(y) = {5s + 21 \over s^2 + 3s + 4}$ 5.) Solve for $y$ by taking the inverse LaPlace transform of both sides (use a table): ${\cal L}^{-1}({\cal L}(y)) = {\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4})$ $y = {\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4})$ \eject Find the inverse LaPlace transform of ${5s + 21 \over s^2 + 3s + 4}$ Look at the denominator first to determine if it is of the form $ s^2 \pm a^2$ or $(s - a)^{n+1}$ or $(s - a)^2 + b^2$ OR if you should factor and use partial fractions $s^2 + 3s + 4$: $b^2 - 4ac = 3^2 - 4(1)(4) = 9 - 16 < 0$ Hence $s^2 + 3s + 4$ does not factor over the reals. Hence to avoid complex numbers, we won't factor it. $s^2 + 3s + 4$ is not an $s^2 - a^2$ or an $s^2 + a^2$ or an $(s - a)^2$, so it must be an $(s-a)^2 + b^2$. Hence we will complete the square: $s^2 + 3s + \underline{\hskip 0.2in} - \underline{\hskip 0.2in} + 4 = (s + \underline{\hskip 0.2in})^2- \underline{\hskip 0.2in} + 4$ Hence ${5s + 21 \over s^2 + 3s + 4}$ = ${5s + 21 \over (s + {3 \over 2})^2 + {7 \over 4}}$ \eject Must now consider the numerator. We need it to look like $s - a = s + {3 \over 2}$ or $b = \sqrt{7 \over 4}$ in order to use ${\cal L}^{-1}({s-a\over (s -a)^2 + b^2}) = e^{at} cos bt$ \hfil \break and/or ${\cal L}^{-1}({b \over (s -a)^2 + b^2}) = e^{at} sin bt$ $5s + 21 = 5(s + {3 \over 2}) - {15 \over 2} + 21 = 5(s + {3 \over 2}) + {27 \over 2}$ $ = 5(s + {3 \over 2}) + [{27 \over 2}\sqrt{4 \over 7}] \sqrt{7 \over 4} = 5(s + {3 \over 2}) + [{27 \over \sqrt{7}}] \sqrt{7 \over 4} $ Hence ${5s + 21 \over s^2 + 3s + 4}$ = ${5(s + {3 \over 2}) + [{27 \over\sqrt{7}}] \sqrt{7 \over 4} \over (s + {3 \over 2})^2 + {7 \over 4}}$ \hskip 0.95in= $5[{s + {3 \over 2} \over (s + {3 \over 2})^2 + {7 \over 4}}] + {27 \over\sqrt{7}}[{\sqrt{7 \over 4} \over (s + {3 \over 2})^2 + {7 \over 4}}]$ Thus ${\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4}) $ =${\cal L}^{-1}(5[{s + {3 \over 2} \over (s + {3 \over 2})^2 + {7 \over 4}}] + {27 \over\sqrt{7}}[{\sqrt{7 \over 4} \over (s + {3 \over 2})^2 + {7 \over 4}}]$) \vfill \rightline{=$5{\cal L}^{-1}({s + {3 \over 2} \over (s + {3 \over 2})^2 + {7 \over 4}}) + {27 \over\sqrt{7}} {\cal L}^{-1}( {\sqrt{7 \over 4} \over (s + {3 \over 2})^2 + {7 \over 4}})$} \vfill \rightline{$=5e^{-{3 \over 2}t} cos \sqrt{7 \over 4}t + {27 \over\sqrt{7}}e^{-{3 \over 2}t} sin \sqrt{7 \over 4}t$~~~~~} \vfill Hence $y(t) = 5e^{-{3 \over 2}t} cos \sqrt{7 \over 4}t + {27 \over\sqrt{7}}e^{-{3 \over 2}t} sin \sqrt{7 \over 4}t$. \end Since $y(t) ={\cal L}^{-1}({5s + 21 \over s^2 + 3s + 4})$, 6.3: Step functions. $u_c(t) = \cases{0 & $t < c$ \cr 1 & $t \geq c$}$ 1.) Graph $u_c(t)$: \eject 2.) Given $f$, graph $u_c(t) f(t-c)$: \vskip 0.8in 3.) Calculate ${\cal L}(u_c(t) f(t-c))$ in terms of ${\cal L}(f(t))$: \vfill Example: Find the LaPlace transform of 1.) \centerline{$g(t) = \cases{0 & $t < 3$ \cr e^{t-3} & $t \geq 3$}$} 2.) \centerline{$f(t) = \cases{0 & $t < 3$ \cr 5 & $3 \leq t < 4$ \cr t - 5 & $t \geq 4$}$} Example: Find the inverse Laplace transform of ${e^{-8s} \over s^3 } 4.) Calculate ${\cal L}(e^{ct} f(t))$ in terms of $F(s) = {\cal L}(f(t))$ \vfill Example: Use formula 6 (p. 304) to find the inverse LaPlace transform of ${s - c \over (s-c)^2 + a^2}$. \end \end To find the inverse LaPlace transform, you may need to use that the inverse LaPlace transform in linear. You may also need to use partial fractions or other methods in order to right the righthand side of (*) as a sum of functions whose inverse LaPlace transform is known. \end \eject Calculus pre-requisites you must know. Derivative = slope of tangent line = rate. Integral = area between curve and x-axis (where area can be negative). Integration by parts: Derivative of a product: $(uv)' = uv' + vu'$ \centerline{$uv' = (uv)' - vu'$} \centerline{$\int uv' = \int (uv)' - \int vu'$} \centerline{$\int uv' = (uv) - \int vu'$} Example: $\int e^{2x} sin(3x)$ Let $u = sin(3x)$, $dv = e^{2x}$ then $du = 3cos(3x)$, $v = {1 \over 2}e^{2x}$ then $d^2u = -9sin(3x)$, $\int v = {1 \over 4}e^{2x}$ $\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - \int {3 \over 2}e^{2x}cos(3x)$ \hskip 0.4in $= {1 \over 2}sin(3x)e^{2x} - [{3 \over 4}cos(3x)e^{2x} - \int {-9 \over 4}sin(3x)e^{2x}$ $\int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x} - {9 \over 4}\int sin(3x)e^{2x}$ ${13 \over 4} \int e^{2x} sin(3x) = {1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}$ $\int e^{2x} sin(3x) = {4 \over 13}[{1 \over 2}sin(3x)e^{2x} - {3 \over 4}cos(3x)e^{2x}]$ Exercise: Calculate $\int e^{x} cos(2x)$ \end