sql server reporting services barcode font Ordinary Differential Equations in Software

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Ordinary Differential Equations
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11.18 Solve the equation
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d2y + y = sin t with initial conditions y(0) = 0, y '(0) = 2 . dt 2
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equation = y''[t]+ y[t] Sin[t]; temp = LaplaceTransform[equation, t, s] LaplaceTransform[y[t],t,s]+ 1 1 + s2 temp2 = Solve[temp, LaplaceTransform[y[t], t, s]] /. {y'[0] 2, y[0] 0} s2 LaplaceTransform[y[t],t,s] s y[0] y'[0]
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2 LaplaceTransform[y[t],t,s] 3 + 2 s 2 (1 + s2)
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temp3 = temp2[[1, 1, 2]]; InverseLaplaceTransform[temp3, s, t] 1 ( t Cos[t]+ 5 Sin[t]) 2
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11.19 Solve
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d 2 y dy + + y = e t , y(0) = 3, y '(0) = 2. dt 2 dt
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SOLUTION
equation = y''[t]+ y'[t] + y[t] Exp[t] ; temp = LaplaceTransform[equation, t, s] LaplaceTransform[y[t],t,s]+ s LaplaceTransform[y[t],t,s]+ n 1 1 + s temp2 = Solve[temp, LaplaceTransform[y[t], t, s]] /. {y'[0] 2, y[0] 3} s2 LaplaceTransform[y[t],t,s] y[0] s y[0] y'[0] {{LaplaceTransform[y[t],t,s] temp3 = temp2[[1, 1, 2]]; InverseLaplaceTransform[temp3, s, t] 1 t/2 3t/2 + 8 Cos 3t + 6 3 Sin 3t 2 2 3 4 + 2 s + 3 s2 }} ( 1 + s)(1 + s + s2)
d2y dy t if 0 t 1 11.20 Solve the equation 2 2 + y = g(t ), y(0) = y '(0) = 0 , where g(t ) = 2 and plot dt dt t if t > 1 the solution for 0 x 4.
SOLUTION
equation = y''[t] 2y'[t] + y[t] t UnitStep[1 t] + t2 UnitStep[t 1]; temp = LaplaceTransform[equation, t, s] LaplaceTransform[y[t],t,s]+ s2 LaplaceTransform[y[t],t,s] 2(s LaplaceTransform[y[t], t, s] y[0]) s y[0] y'[0] s (2 + 2 s + s2) 1 Cosh[s]- s Cosh[s]+ Sinh[s]+ s Sinh[s] + s2 s3 temp2 = Solve[temp, LaplaceTransform[y[t], t, s]] /.{y'[0] 0, y[0] 0} {{LaplaceTransform[y[t], t, s]
s h (2 + 2 s + s s + s2 s s Cosh[s] s s2 Cosh[s] + s s Sinh[s] + s s2 Sinh[s]) }} s3(1 2s + s2)
Ordinary Differential Equations
temp3 = temp2[[1, 1, 2]] s(2 + 2 s + s s + s2 s s Cosh[s] s s2 Cosh[s] + s s Sinh[s] + s s2 Sinh[s]) ] }} s3(1 2s + s2) f[t_] = InverseLaplaceTransform[temp3, s, t] (2 + t( 2 + t)+ t)+( t( 11 + 3 t)+ (4 + 3 t + t2)) HeavisideTheta[ 1 + t] Plot[f[t], {t, 0, 4}]
100 80 60 40 20 1 2 3 4
11.21 Solve the system dx + y = t sin t dt dy x + = t cos t dt
SOLUTION
x ( 0 ) = y( 0 ) = 0
system = {x'[t] + y[t] t Sin[t], x[t]+ y'[t] t Cos[t]}; temp = LaplaceTransform[system, t, s] {s LaplaceTransform[x[t], t, s]+ LaplaceTransform[y[t], t, s] 2s x[0] 2 , LaplaceTransform[x[t], t, s]+ (1 + s2) 1+ s2 } s LaplaceTransform [y[t], t, s] y[0] 2 (1 + s2) temp2 = Solve[temp, {LaplaceTransform[x[t], t, s], LaplaceTransform[y[t], t, s]}] /. {x[0] 0, y[0] 0} 1 {{LaplaceTransform[x[t], t, s] , ( 1 + s2)(1 + s2) 3 3s s }} LaplaceTransform[y[t], t, s] 2 ( 1 + s2)(1 + s2) temp3a = temp2[[1, 1, 2]]; temp3b = temp2[[1, 2, 2]]; InverseLaplaceTransform[temp3a, s, t] //Simplify 1 ( t + t 2 Sin [t]) 4 InverseLaplaceTransform[temp3b, s, t] //Simplify 1 ( t t + 2 Cos [t]+ 4 t Sin[t]) 4 The solution of the system is x = y = 1 ( e t + e t 2 sin t ) 4 1 ( e t e t + 2 cos t + 4t sin t ) 4
Ordinary Differential Equations
11.22 The equation governing the amount of current, I, flowing through a simple resistance-inductance circuit when an EMF (voltage) E is applied is L dI + RI = E . The units for E, I, and L are, respectively, dt volts, amperes, and henries. Suppose L = 1 and R = 10. If 1 volt is applied at time t = 0 and removed 1 sec later, plot the current in the circuit during the first 2 seconds.
SOLUTION
e[t_] = UnitStep[1 t]; Plot[e[t], {t, 0, 2}, PlotStyle Thickness[.01]];
1.0 0.8 0.6 0.4 0.2
l = 1; r = 10; equation = l i'[t]+ r i[t] e[t]; temp = LaplaceTransform[equation, t, s] i[0]+10 LaplaceTransform[i[t], t, s]+ s LaplaceTransform[i [t], t, s] 1 Cosh[s]+ Sinh[s] s temp2 = Solve[temp, LaplaceTransform[i[t], t, s]] /. i[0] 0 1 cosh[s]+ Sinh[s] LaplaceTransform[i[t], t, s s(10 + s) temp3 = temp2[[1, 1, 2]] // Expand 1 Cosh[s]+ Sinh[s] s(10 + s) f[t_] = InverseLaplaceTransform[temp3, s, t] 1 10t( 1 + 10t +( 10 10t) HeavisideTheta[ 1 + t]) 10 Plot[f[t], {t, 0, 2}, PlotStyle Thickness[.01], AxesLabel {"Time","Current"}]
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