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== 6.2#h(0.6em)Transforms of derivatives and ODEs


    
=== Transforms of derivatives


    Let us see how the Laplace transform is used for differential equations. First let us try to find the Laplace transform of a function that is a derivative. Suppose #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] is a differentiable function of exponential order, that is, #math.equation(block: false, alt: "| g open parenthesis t close parenthesis | less than or equal to M e to the power c t")[$| g ( t ) | ≤ M e^(c t)$] for some #math.equation(block: false, alt: "M")[$M$] and #math.equation(block: false, alt: "c")[$c$]. So #math.equation(block: false, alt: "ℒ { g open parenthesis t close parenthesis }")[$ℒ \{ g ( t ) \}$] exists, and what is more, #math.equation(block: false, alt: "lim sub t → infinity e to the power minus s t g open parenthesis t close parenthesis equals 0")[$lim_(t → ∞) e^(− s t) g ( t ) = 0$] when #math.equation(block: false, alt: "s greater than c")[$s > c$]. Then


    #math.equation(block: true, alt: "ℒ { g prime open parenthesis t close parenthesis } equals ∫ 0 infinity e to the power minus s t g prime open parenthesis t close parenthesis d t equals [ e to the power minus s t g open parenthesis t close parenthesis ] t equals 0 infinity minus ∫ 0 infinity open parenthesis minus s close parenthesis e to the power minus s t g open parenthesis t close parenthesis d t equals minus g open parenthesis 0 close parenthesis plus s ℒ { g open parenthesis t close parenthesis } .")[$ℒ \{ g^(′) ( t ) \} = ∫_(0)^(∞) e^(− s t) g^(′) ( t ) d t = attach([ e^(− s t) g ( t ) ], t: ∞, b: t = 0) − ∫_(0)^(∞) ( − s ) e^(− s t) g ( t ) d t = − g ( 0 ) + s ℒ \{ g ( t ) \} .$]


    We repeat this procedure for higher derivatives. The results are listed in Table #math.equation(block: false, alt: "1")[$1$]. The procedure also works for piecewise smooth functions, that is functions that are piecewise continuous with a piecewise continuous derivative. The fact that the function is of exponential order is used to show that the limits appearing above exist. We will not worry much about this fact.


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))


    
==== Solving ODEs with the Laplace Transform


    Notice that the Laplace transform turns differentiation into multiplication by #math.equation(block: false, alt: "s")[$s$]. Let us see how to apply this fact to differential equations.


    #examplebox("Example 1")[][
Take the equation


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus x open parenthesis t close parenthesis equals cos open parenthesis 2 t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 1 .")[$x^(″) ( t ) + x ( t ) = cos ( 2 t ) , " " " " " " " " " " " " " " x ( 0 ) = 0 , " " " " " " " " " " " " " " x^(′) ( 0 ) = 1 .$]


        We will take the Laplace transform of both sides. By #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$] we will, as usual, denote the Laplace transform of #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$].


        #math.equation(block: true, alt: "ℒ { x double prime open parenthesis t close parenthesis plus x open parenthesis t close parenthesis } equals ℒ { cos open parenthesis 2 t close parenthesis } ,; s squared X open parenthesis x close parenthesis minus s x open parenthesis 0 close parenthesis plus x prime open parenthesis 0 close parenthesis plus X open parenthesis s close parenthesis equals the fraction s over s squared plus 4 .")[$ℒ \{ x^(″) ( t ) + x ( t ) \} = ℒ \{ cos ( 2 t ) \} , \ s^(2) X ( x ) − s x ( 0 ) + x^(′) ( 0 ) + X ( s ) = frac(s, s^(2) + 4) .$]


        We plug in the initial conditions now—this makes the computations more streamlined—to obtain


        #math.equation(block: true, alt: "s squared X open parenthesis s close parenthesis minus 1 plus X open parenthesis s close parenthesis equals the fraction s over s squared plus 4 .")[$s^(2) X ( s ) − 1 + X ( s ) = frac(s, s^(2) + 4) .$]


        We solve for #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$],


        #math.equation(block: true, alt: "X open parenthesis s close parenthesis equals the fraction s over open parenthesis s squared plus 1 close parenthesis open parenthesis s squared plus 4 close parenthesis plus the fraction 1 over s squared plus 1 .")[$X ( s ) = frac(s, ( s^(2) + 1 ) ( s^(2) + 4 )) + frac(1, s^(2) + 1) .$]


        We use partial fractions (exercise) to write


        #math.equation(block: true, alt: "X open parenthesis s close parenthesis equals the fraction 1 over 3 the fraction s over s squared plus 1 minus the fraction 1 over 3 the fraction s over s squared plus 4 plus the fraction 1 over s squared plus 1 .")[$X ( s ) = frac(1, 3) frac(s, s^(2) + 1) − frac(1, 3) frac(s, s^(2) + 4) + frac(1, s^(2) + 1) .$]


        Now take the inverse Laplace transform to obtain


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals the fraction 1 over 3 cos open parenthesis t close parenthesis minus the fraction 1 over 3 cos open parenthesis 2 t close parenthesis plus sin open parenthesis t close parenthesis .")[$x ( t ) = frac(1, 3) cos ( t ) − frac(1, 3) cos ( 2 t ) + sin ( t ) .$]
]


    The procedure for linear constant coefficient equations is as follows. We take an ordinary differential equation in the time variable #math.equation(block: false, alt: "t")[$t$]. We apply the Laplace transform to transform the equation into an algebraic (non differential) equation in the frequency domain. All the #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$], #math.equation(block: false, alt: "x prime open parenthesis t close parenthesis")[$x^(′) ( t )$], #math.equation(block: false, alt: "x double prime open parenthesis t close parenthesis")[$x^(″) ( t )$], and so on, will be converted to #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$], #math.equation(block: false, alt: "s X open parenthesis s close parenthesis minus x open parenthesis 0 close parenthesis")[$s X ( s ) − x ( 0 )$], #math.equation(block: false, alt: "s squared X open parenthesis s close parenthesis minus s x open parenthesis 0 close parenthesis minus x prime open parenthesis 0 close parenthesis")[$s^(2) X ( s ) − s x ( 0 ) − x^(′) ( 0 )$], and so on. We solve the equation for #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$]. Then taking the inverse transform, if possible, we find #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$].


    It should be noted that since not every function has a Laplace transform, not every equation can be solved in this manner. Also if the equation is not a linear constant coefficient ODE, then by applying the Laplace transform we may not obtain an algebraic equation.


    
==== Using the Heaviside Function


    Before we move on to more general equations than those we could solve before, we want to consider the Heaviside function. See Figure #math.equation(block: false, alt: "1")[$1$] for the graph.


    #math.equation(block: true, alt: "0 i f t less than 0 ,; 1 i f t greater than or equal to 0 .")[$0 upright(i) upright(f) " " t < 0 , \ 1 upright(i) upright(f) " " t ≥ 0 .$]


    #figure(figph[Graph that includes the negative x-axis and the horizontal right directional ray starting at (0,1)], alt: "Graph that includes the negative x-axis and the horizontal right directional ray starting at (0,1)", caption: [Figure #math.equation(block: false, alt: "1")[$1$]: Plot of the Heaviside (unit step) function #math.equation(block: false, alt: "u open parenthesis t close parenthesis")[$u ( t )$].])


    This function is useful for putting together functions, or cutting functions off. Most commonly it is used as #math.equation(block: false, alt: "u open parenthesis t minus a close parenthesis")[$u ( t − a )$] for some constant #math.equation(block: false, alt: "a")[$a$]. This just shifts the graph to the right by #math.equation(block: false, alt: "a")[$a$]. That is, it is a function that is 0 when #math.equation(block: false, alt: "less than a")[$< a$] and 1 when #math.equation(block: false, alt: "t greater than or equal to a")[$t ≥ a$]. Suppose for example that #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is a “signal” and you started receiving the signal #math.equation(block: false, alt: "sin t")[$sin t$] at time #math.equation(block: false, alt: "t equals π")[$t = π$]. The function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] should then be defined as


    #math.equation(block: true, alt: "0 i f t less than π ,; sin t i f t greater than or equal to π .")[$0 upright(i) upright(f) " " t < π , \ sin t upright(i) upright(f) " " t ≥ π .$]


    Using the Heaviside function, #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] can be written as


    #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals u open parenthesis t minus π close parenthesis sin t")[$f ( t ) = u ( t − π ) sin t$]


    Similarly the step function that is #math.equation(block: false, alt: "1")[$1$] on the interval #math.equation(block: false, alt: "[ 1 , 2 close parenthesis")[$[ 1 , 2 )$] and zero everywhere else can be written as


    #math.equation(block: true, alt: "u open parenthesis t minus 1 close parenthesis minus u open parenthesis t minus 2 close parenthesis .")[$u ( t − 1 ) − u ( t − 2 ) .$]


    The Heaviside function is useful to define functions defined piecewise. If you want to define #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] such that #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$] when #math.equation(block: false, alt: "t")[$t$] is in #math.equation(block: false, alt: "[ 0 , 1 ]")[$[ 0 , 1 ]$], #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals minus t plus 2")[$f ( t ) = − t + 2$] when #math.equation(block: false, alt: "t")[$t$] is in #math.equation(block: false, alt: "[ 1 , 2 close parenthesis")[$[ 1 , 2 )$] and #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 0")[$f ( t ) = 0$] otherwise, you can use the expression


    #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals t open parenthesis u open parenthesis t close parenthesis minus u open parenthesis t minus 1 close parenthesis close parenthesis plus open parenthesis minus t plus 2 close parenthesis open parenthesis u open parenthesis t minus 1 close parenthesis minus u open parenthesis t minus 2 close parenthesis close parenthesis .")[$f ( t ) = t ( u ( t ) − u ( t − 1 ) ) + ( − t + 2 ) ( u ( t − 1 ) − u ( t − 2 ) ) .$]


    Hence it is useful to know how the Heaviside function interacts with the Laplace transform. We have already seen that


    #math.equation(block: true, alt: "ℒ { u open parenthesis t minus a close parenthesis } equals the fraction e to the power minus a s over 2 .")[$ℒ \{ u ( t − a ) \} = frac(e^(− a s), 2) .$]


    Shifting Property
    This can be generalized into a shifting property or second shifting property.


    #math.equation(block: true, alt: "ℒ { f open parenthesis t minus a close parenthesis u open parenthesis t minus a close parenthesis } equals e to the power minus a s ℒ { f open parenthesis t close parenthesis } .")[$ℒ \{ f ( t − a ) u ( t − a ) \} = e^(− a s) ℒ \{ f ( t ) \} .$]


    #examplebox("Example 2")[][
Suppose that the forcing function is not periodic. For example, suppose that we had a mass-spring system


        #math.equation(block: true, alt: "x double prime open parenthesis t close parenthesis plus x open parenthesis t close parenthesis equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 ,")[$x^(″) ( t ) + x ( t ) = f ( t ) , " " " " " " " " " " " " x ( 0 ) = 0 , " " " " " " " " " " " " x^(′) ( 0 ) = 0 ,$]


        where #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 1")[$f ( t ) = 1$] if #math.equation(block: false, alt: "1 less than or equal to t less than 5")[$1 ≤ t < 5$] and zero otherwise. We could imagine a mass-spring system, where a rocket is fired for 4 seconds starting at #math.equation(block: false, alt: "t equals 1")[$t = 1$]. Or perhaps an RLC circuit, where the voltage is raised at a constant rate for 4 seconds starting at #math.equation(block: false, alt: "t equals 1")[$t = 1$], and then held steady again starting at #math.equation(block: false, alt: "t equals 5")[$t = 5$].


        We can write #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals u open parenthesis t minus 1 close parenthesis minus u open parenthesis t minus 5 close parenthesis")[$f ( t ) = u ( t − 1 ) − u ( t − 5 )$]. We transform the equation and we plug in the initial conditions as before to obtain


        #math.equation(block: true, alt: "s squared X open parenthesis s close parenthesis plus X open parenthesis s close parenthesis equals the fraction e to the power minus s over s minus the fraction e to the power minus 5 s over s .")[$s^(2) X ( s ) + X ( s ) = frac(e^(− s), s) − frac(e^(− 5 s), s) .$]


        We solve for #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$] to obtain


        #math.equation(block: true, alt: "X open parenthesis s close parenthesis equals the fraction e to the power minus s over s open parenthesis s squared plus 1 close parenthesis minus the fraction e to the power minus 5 s over s open parenthesis s squared plus 1 close parenthesis .")[$X ( s ) = frac(e^(− s), s ( s^(2) + 1 )) − frac(e^(− 5 s), s ( s^(2) + 1 )) .$]


        We leave it as an exercise to the reader to show that


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction 1 over s open parenthesis s squared plus 1 close parenthesis } equals 1 minus cos t .")[$ℒ^(− 1) \{ frac(1, s ( s^(2) + 1 )) \} = 1 − cos t .$]


        In other words #math.equation(block: false, alt: "ℒ { 1 minus cos t } equals the fraction 1 over s open parenthesis s squared plus 1 close parenthesis")[$ℒ \{ 1 − cos t \} = frac(1, s ( s^(2) + 1 ))$]. So using (6.2.1) we find


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction e to the power minus s over s open parenthesis s squared plus 1 close parenthesis } equals ℒ to the power minus 1 { e to the power minus s ℒ { 1 minus cos t } } equals open parenthesis 1 minus cos open parenthesis t minus 1 close parenthesis close parenthesis u open parenthesis t minus 1 close parenthesis .")[$ℒ^(− 1) \{ frac(e^(− s), s ( s^(2) + 1 )) \} = ℒ^(− 1) \{ e^(− s) ℒ \{ 1 − cos t \} \} = ( 1 − cos ( t − 1 ) ) u ( t − 1 ) .$]


        Similarly


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction e to the power minus 5 s over s open parenthesis s squared plus 1 close parenthesis } equals ℒ to the power minus 1 { e to the power minus 5 s ℒ { 1 minus cos t } } equals open parenthesis 1 minus cos open parenthesis t minus 5 close parenthesis close parenthesis u open parenthesis t minus 5 close parenthesis .")[$ℒ^(− 1) \{ frac(e^(− 5 s), s ( s^(2) + 1 )) \} = ℒ^(− 1) \{ e^(− 5 s) ℒ \{ 1 − cos t \} \} = ( 1 − cos ( t − 5 ) ) u ( t − 5 ) .$]


        Hence, the solution is


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals open parenthesis 1 minus cos open parenthesis t minus 1 close parenthesis close parenthesis u open parenthesis t minus 1 close parenthesis minus open parenthesis 1 minus cos open parenthesis t minus 5 close parenthesis close parenthesis u open parenthesis t minus 5 close parenthesis .")[$x ( t ) = ( 1 − cos ( t − 1 ) ) u ( t − 1 ) − ( 1 − cos ( t − 5 ) ) u ( t − 5 ) .$]


        The plot of this solution is given in Figure #math.equation(block: false, alt: "2")[$2$].


        
          The section's own solution of x″ + x = f(t), x(0) = x′(0) = 0, namely x(t) = (1 − cos(t−1))u(t−1) − (1 − cos(t−1−d))u(t−1−d), with time t on the horizontal axis and the burn now lasting d seconds — the book's case is d = 4, a rocket fired for 4 seconds starting at t = 1. Drag d to π (≈ 3.14) for the loudest possible aftermath, amplitude exactly 2; then on to 2π (≈ 6.28), where the two cosine pieces cancel and the mass returns to rest and stays there. The leftover ringing is 2|sin(d/2)|, so it depends only on the burn length modulo the natural period 2π — not on how long ago the burn ended.
          {"functions":\[{"color":"\#b8442a","domain":\[0,20\],"expression":"(1-cos(x-1))\*((x-1)/abs(x-1)+1)/2 - (1-cos(x-1-d))\*((x-1-d)/abs(x-1-d)+1)/2","expression2":"","id":"f-x","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":1,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"p-d","label":"Burn duration","max":8,"min":0.5,"name":"d","step":0.01,"unit":"s","value":4}\],"title":"Rocket burn duration and residual motion","version":1,"viewport":{"centerX":8.5,"centerY":0,"scale":44}}
        
        #figure(figph[Graph that starts on the x-axis flat and then after a bit becomes the cos curve.], alt: "Graph that starts on the x-axis flat and then after a bit becomes the cos curve.", caption: [Figure #math.equation(block: false, alt: "2")[$2$]: Plot of #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$].])
]


    
==== Transfer Functions


    Laplace transform leads to the following useful concept for studying the steady state behavior of a linear system. Suppose we have an equation of the form


    #math.equation(block: true, alt: "L x equals f open parenthesis t close parenthesis ,")[$L x = f ( t ) ,$]


    where #math.equation(block: false, alt: "L")[$L$] is a linear constant coefficient differential operator. Then #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] is usually thought of as input of the system and #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$] is thought of as the output of the system. For example, for a mass-spring system the input is the forcing function and output is the behavior of the mass. We would like to have an convenient way to study the behavior of the system for different inputs.


    Let us suppose that all the initial conditions are zero and take the Laplace transform of the equation, we obtain the equation


    #math.equation(block: true, alt: "A open parenthesis s close parenthesis X open parenthesis s close parenthesis equals F open parenthesis s close parenthesis .")[$A ( s ) X ( s ) = F ( s ) .$]


    Solving for the ratio #math.equation(block: false, alt: "the fraction X open parenthesis s close parenthesis over F open parenthesis s close parenthesis")[$frac(X ( s ), F ( s ))$] we obtain the so-called #emph[transfer function] #math.equation(block: false, alt: "H open parenthesis s close parenthesis equals the fraction 1 over A open parenthesis s close parenthesis")[$H ( s ) = frac(1, A ( s ))$].


    #math.equation(block: true, alt: "H open parenthesis s close parenthesis equals the fraction X open parenthesis s close parenthesis over F open parenthesis s close parenthesis")[$H ( s ) = frac(X ( s ), F ( s ))$]


    In other words, #math.equation(block: false, alt: "X open parenthesis s close parenthesis equals H open parenthesis s close parenthesis F open parenthesis s close parenthesis")[$X ( s ) = H ( s ) F ( s )$]. We obtain an algebraic dependence of the output of the system based on the input. We can now easily study the steady state behavior of the system given different inputs by simply multiplying by the transfer function.


    #examplebox("Example 3")[][
Given #math.equation(block: false, alt: "x double prime plus ω 0 2 x equals f open parenthesis t close parenthesis")[$x^(″) + ω_(0)^(2) x = f ( t )$], let us find the transfer function (assuming the initial conditions are zero).


        First, we take the Laplace transform of the equation.


        #math.equation(block: true, alt: "s squared X open parenthesis s close parenthesis plus ω 0 2 X open parenthesis s close parenthesis equals F open parenthesis s close parenthesis .")[$s^(2) X ( s ) + ω_(0)^(2) X ( s ) = F ( s ) .$]


        Now we solve for the transfer function #math.equation(block: false, alt: "the fraction X open parenthesis s close parenthesis over F open parenthesis s close parenthesis")[$frac(X ( s ), F ( s ))$].


        #math.equation(block: true, alt: "H open parenthesis s close parenthesis equals the fraction X open parenthesis s close parenthesis over F open parenthesis s close parenthesis equals the fraction 1 over s squared plus ω 0 2 .")[$H ( s ) = frac(X ( s ), F ( s )) = frac(1, s^(2) + ω_(0)^(2)) .$]


        Let us see how to use the transfer function. Suppose we have the constant input #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 1")[$f ( t ) = 1$]. Hence #math.equation(block: false, alt: "F open parenthesis s close parenthesis equals the fraction 1 over s")[$F ( s ) = frac(1, s)$], and


        #math.equation(block: true, alt: "X open parenthesis s close parenthesis equals H open parenthesis s close parenthesis F open parenthesis s close parenthesis equals the fraction 1 over s squared plus ω 0 2 the fraction 1 over s .")[$X ( s ) = H ( s ) F ( s ) = frac(1, s^(2) + ω_(0)^(2)) frac(1, s) .$]


        Taking the inverse Laplace transform of #math.equation(block: false, alt: "X open parenthesis s close parenthesis")[$X ( s )$] we obtain


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals the fraction 1 minus cos open parenthesis ω sub 0 t close parenthesis over ω 0 2 .")[$x ( t ) = frac(1 − cos ( ω_(0) t ), ω_(0)^(2)) .$]
]


    
==== Transforms of Integrals


    A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is, equations in which integrals rather than derivatives of functions appear. The basic property, which can be proved by applying the definition and doing integration by parts, is


    #math.equation(block: true, alt: "ℒ { ∫ 0 t f open parenthesis τ close parenthesis d τ } equals the fraction 1 over s F open parenthesis s close parenthesis .")[$ℒ \{ ∫_(0)^(t) f ( τ ) d τ \} = frac(1, s) F ( s ) .$]


    It is sometimes useful (e.g. for computing the inverse transform) to write this as


    #math.equation(block: true, alt: "∫ 0 t f open parenthesis τ close parenthesis d τ equals ℒ to the power minus 1 { the fraction 1 over s F open parenthesis s close parenthesis } .")[$∫_(0)^(t) f ( τ ) d τ = ℒ^(− 1) \{ frac(1, s) F ( s ) \} .$]


    #examplebox("Example 4")[][
To compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction 1 over s open parenthesis s squared plus 1 close parenthesis }")[$ℒ^(− 1) \{ frac(1, s ( s^(2) + 1 )) \}$] we could proceed by applying this integration rule.


        #math.equation(block: true, alt: "ℒ to the power minus 1 { the fraction 1 over 2 the fraction 1 over s squared plus 1 } equals ∫ 0 t ℒ to the power minus 1 { the fraction 1 over s squared plus 1 } equals ∫ 0 t sin τ d τ equals 1 minus cos t .")[$ℒ^(− 1) \{ frac(1, 2) frac(1, s^(2) + 1) \} = ∫_(0)^(t) ℒ^(− 1) \{ frac(1, s^(2) + 1) \} = ∫_(0)^(t) sin τ d τ = 1 − cos t .$]
]


    #examplebox("Example 5")[][
An equation containing an integral of the unknown function is called an #emph[integral equation]. For example, take


        #math.equation(block: true, alt: "t squared equals ∫ 0 t e to the power τ x open parenthesis τ close parenthesis d τ")[$t^(2) = ∫_(0)^(t) e^(τ) x ( τ ) d τ$]


        where we wish to solve for #math.equation(block: false, alt: "x open parenthesis t close parenthesis")[$x ( t )$]. We apply the Laplace transform and the shifting property to get


        #math.equation(block: true, alt: "the fraction 2 over s cubed equals the fraction 1 over s ℒ { e to the power t x open parenthesis t close parenthesis } equals the fraction 1 over s X open parenthesis s minus 1 close parenthesis ,")[$frac(2, s^(3)) = frac(1, s) ℒ \{ e^(t) x ( t ) \} = frac(1, s) X ( s − 1 ) ,$]


        where #math.equation(block: false, alt: "X open parenthesis s close parenthesis equals ℒ { x open parenthesis t close parenthesis }")[$X ( s ) = ℒ \{ x ( t ) \}$]. Thus


        #math.equation(block: true, alt: "X open parenthesis s minus 1 close parenthesis equals the fraction 2 over s squared or X open parenthesis s close parenthesis equals the fraction 2 over open parenthesis s plus 1 close parenthesis squared .")[$X ( s − 1 ) = frac(2, s^(2)) #h(1em) "or" #h(1em) X ( s ) = frac(2, ( s + 1 )^(2)) .$]


        We use the shifting property again to get


        #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals 2 e to the power minus t t .")[$x ( t ) = 2 e^(− t) t .$]
]