352 - Topic 15 - Laplace Transforms and Differential Equations -- Sage

var('t,s') # f=t^30 f = e^(4*t+8) L=laplace(f,t,s) html.table([['$f(t) =$',f],['$F(s) =$',L]])

$f(t) =$	$e^{\left(4 \, t + 8\right)}$
$F(s) =$	$\frac{e^{8}}{s - 4}$

F = 14/((3*s+2)*(s-4)) IL=inverse_laplace(F,s,t) html.table([['$F(s) =$',F],['$f(t) =$',IL]])

$F(s) =$	$\frac{14}{{\left(3 \, s + 2\right)} {\left(s - 4\right)}}$
$f(t) =$	$e^{\left(4 \, t\right)} - e^{\left(-\frac{2}{3} \, t\right)}$

# If one wants to do this inverse transform by hand, then partial fractions may need to be used # to break up the term into linear denominators # var('s') # F = (15*s^2-12*s+19)/((s^2+4)^2*s^3) F.partial_fraction(s).show()

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{11 \, s - 24}{32 \, {\left(s^{2} + 4\right)}} - \frac{41 \, s - 48}{16 \, {\left(s^{2} + 4\right)}^{2}} + \frac{11}{32 \, s} - \frac{3}{4 \, s^{2}} + \frac{19}{16 \, s^{3}}$

# Using these with differential equations var('t,s,y,y0') y = function('y')(t) eqn = diff(y,t) - 5*y == e^(-t) # eqn = diff(y,t) + 4*y == 8*sin(4*t) ICS =[0,y0] print 'Differential Equation:' eqn.show() print 'With initial condition',ICS desolve_laplace(eqn,y,ICS).show()

Differential Equation:

$\newcommand{\Bold}[1]{\mathbf{#1}}-5 \, y\left(t\right) + \frac{\partial}{\partial t}y\left(t\right) = e^{\left(-t\right)}$

With initial condition [0, y0]

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{6} \, {\left(6 \, y_{0} + 1\right)} e^{\left(5 \, t\right)} - \frac{1}{6} \, e^{\left(-t\right)}$

Differential Equation:

With initial condition [0, y0]

# to quickly solve (or to check your manual solution) var('t,y,C') y=function('y')(t) @interact def _(eq = input_box(diff(y,t) + 5*y == exp(-t),label='DE:'),y0 = input_box(2,width=2,label='$y_0$')): html('Remember to enter diff(y,t) for any derivative.') ics = [0,y0] eq.show() desolve_laplace(eq,y,ics).show()

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# Notice the Sage commands can handle the translation theorem forward # but cannot deal with the translation theorem in reverse var('t,v') u = unit_step(v) plot(u(v=t-3),t,0,6).show() F = laplace(u(v=t-3),t,s) F.show() f = inverse_laplace(F,s,t) f.show()

$\newcommand{\Bold}[1]{\mathbf{#1}}\frac{e^{\left(-3 \, s\right)}}{s}$

$\newcommand{\Bold}[1]{\mathbf{#1}}\mathcal{L}^{-1}\left(\frac{e^{\left(-3 \, s\right)}}{s}, s, t\right)$

# We can have Sage help with parts along the way and then apply the theorems ourselves # Below, consider e^(-2s)*F(s) var('t,s') F = 4/(s*(s+3)) f =inverse_laplace(F,s,t) show(f(t = t-2)*u(v = t-2))

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{4}{3} \, {\left(e^{\left(-3 \, t + 6\right)} - 1\right)} \mathrm{u}\left(t - 2\right)$

var('t,s') F = s/((s^2+1)*(s^2+2*s+2)) IL=inverse_laplace(F,s,t) # html.table([['$F(s) =$',F],['$f(t) =$',IL]]) F.partial_fraction(s).show()

$\newcommand{\Bold}[1]{\mathbf{#1}}-\frac{s + 4}{5 \, {\left(s^{2} + 2 \, s + 2\right)}} + \frac{s + 2}{5 \, {\left(s^{2} + 1\right)}}$

# periodic function var('t') f1 = 2*unit_step(t-2) f2 = 1*unit_step(t-3) f3 = -3*unit_step(t-5) f4 = 2*unit_step(t-7) f5 = 1*unit_step(t-8) f6 = -3*unit_step(t-10) f = f1+f2+f3+f4+f5+f6 G = plot(0,t,0,2,thickness = 5,ticks=[[1,2,3,4,5,6,7,8,9,10,11,12],[1,2,3,4]])+plot(2,t,2,3,thickness = 5)+plot(3,t,3,5,thickness = 5)+plot(0,t,5,7,thickness = 5)+plot(2,t,7,8,thickness = 5)+plot(3,t,8,10,thickness = 5) G += circle((10.5,1.5),.1,fill=true)+circle((11,1.5),.1,fill=true)+circle((11.5,1.5),.1,fill=true) G.show()

352 - Topic 15 - Laplace Transforms and Differential Equations

8 days ago by Professor352

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