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Laplace Transforms of Piecewise Continuous Functions
We'll need to consider initial value problems
\[ay''+by'+cy=f(t),\quad y(0)=k_0,\quad y'(0)=k_1,\nonumber \]
where \(a\), \(b\), and \(c\) are constants and \(f\) is piecewise continuous. Herewe’ll develop proceduresto find Laplace transforms of piecewise continuous functions, and to find the piecewise continuous inverses of Laplace transforms, which will allow us to solve these initial value problems..
Definition 9.5.1 Unit Step Function
For \(a>0\), the unit step functionis given by
\[\label{eq:8.4.4} {\cal U}(t-a)=\left\{\begin{array}{rl} 0,&t<a\\ 1,&t\ge a. \end{array}\right.\]
We’ll now developa systematic way to find the Laplace transform of a piecewise continuous function.
The step function enables us to represent piecewise continuous functions conveniently. For example, consider the function
\[\label{eq:8.4.5} f(t)=\left\{\begin{array}{rl} f_0(t),&0\le t<t_1,\\[4pt]f_1(t),&t\ge t_1, \end{array}\right.\]
where we assume that \(f_0\) and \(f_1\) are defined on \([0,\infty)\), even though they equal \(f\) only on the indicated intervals. This assumption enables us to rewrite Equation \ref{eq:8.4.5} as
\[\label{eq:8.4.6} f(t)=f_0(t)+ {\cal U} (t-t_1)\left(f_1(t)-f_0(t)\right).\]
To verify this, note that if \(t<t_1\) then \( {\cal U} (t-t_1)=0\) and Equation \ref{eq:8.4.6} becomes
\[f(t)=f_0(t)+(0)\left(f_1(t)-f_0(t)\right)=f_0(t). \nonumber\]
If \(t\ge t_1\) then \( {\cal U} (t-t_1)=1\) and Equation \ref{eq:8.4.6} becomes
\[f(t)=f_0(t)+(1)\left(f_1(t)-f_0(t)\right)=f_1(t). \nonumber\]
We need the next theorem to show how Equation \ref{eq:8.4.6} can be used to find \({\cal L}(f)\).
Theorem 9.5.1
Let \(g\) be defined on \([0,\infty).\) Suppose \(a\ge0\) and \({\cal L}\left(g(t+a)\right)\) exists for \(s>s_0.\) Then \({\cal L}\left({\cal U}(t-a)g(t)\right)\) exists for \(s>s_0\), and
\[{\cal L}({\cal U}(t-a)g(t))=e^{-sa}{\cal L}\left(g(t+a)\right).\nonumber\]
- Proof
-
By definition,
\[{\cal L}\left({\cal U}(t-a)g(t)\right)=\int_0^\infty e^{-st} {\cal U}(t-a)g(t)\, dt.\nonumber\]
From this and the definition of \({\cal U}(t-a)\),
\[{\cal L}\left({\cal U}(t-a)g(t)\right)=\int_0^ae^{-st}(0)\,dt+\int_{a}^\infty e^{-st}g(t)\,dt.\nonumber\]
The first integral on the right equals zero. Introducing the new variable of integration \(x=t-a\) in the second integral yields
\[{\cal L}\left({\cal U}(t-a)g(t)\right)=\int_0^\infty e^{-s(x+a)}g(x+a)\,dx =e^{-sa}\int_0^\infty e^{-sx} g(x+a)\,dx.\nonumber\]
Changing the name of the variable of integration in the last integral from \(x\) to \(t\) yields
\[{\cal L}\left({\cal U}(t-a)g(t)\right) =e^{-sa}\int_0^\infty e^{-st} g(t+a)\,dt=e^{-sa}{\cal L}(g(t+a)).\nonumber\]
Example 9.5.1
Find \[{\cal L}\left({\cal U}(t-1)(t^2+1)\right).\nonumber\]
Solution
Here \(a=1\) and \(g(t)=t^2+1\), so
\[g(t+1)=(t+1)^2+1=t^2+2t+2.\nonumber\]
Since
\[{\cal L}\left(g(t+1)\right)={2\over s^3}+{2\over s^2}+{2\over s},\nonumber\]
Theorem 9.5.1 implies that
\[{\cal L}\left({\cal U}(t-1)(t^2+1)\right) =e^{-s}\left({2\over s^3}+{2\over s^2}+{2\over s}\right).\nonumber\]
Example 9.5.2
Find the Laplace transform of the function
\[f(t)=\left\{\begin{array}{cl} 2t+1,&0\le t<2,\\[4pt]3t,&t\ge2, \end{array}\right. \nonumber\]
Solution
We first write \(f\) in the form Equation \ref{eq:8.4.6} as
\[f(t)=2t+1+{\cal U}(t-2)(t-1). \nonumber\]
Therefore
\[\begin{aligned} {\cal L}(f)&={\cal L}(2t+1) +{\cal L}\left({\cal U}(t-2)(t-1)\right)\\ &={\cal L}(2t+1) +e^{-2s}{\cal L}(t+1)\quad\mbox{ (from Theorem }\PageIndex{1})\\ &={2\over s^2}+{1\over s}+e^{-2s}\left({1\over s^2}+{1\over s}\right)\end{aligned}.\nonumber\]
Formula Equation \ref{eq:8.4.6} can be extended to more general piecewise continuous functions. For example, we can write
\[f(t)=\left\{\begin{array}{rl} f_0(t),&0\le t<t_1,\\[4pt]f_1(t),&t_1\le t<t_2,\\[4pt]f_2(t),&t\ge t_2, \end{array}\right.\nonumber\]
as
\[f(t)=f_0(t)+{\cal U}(t-t_1)\left(f_1(t)-f_0(t)\right)+ {\cal U}(t-t_2)\left(f_2(t)-f_1(t)\right) \nonumber\]
if \(f_0\), \(f_1\), and \(f_2\) are all defined on \([0,\infty)\).
This idea can be extended for any number of pieces of a piecewise continuous function.
Example 9.5.3
Find the Laplace transform of
\[\label{eq:8.4.7} f(t)=\left\{\begin{array}{cl} 1,&0\le t<2,\\[4pt]-2t+1,&2\le t<3,\\[4pt]3t,&3\le t<5,\\[4pt]t-1,&t\ge5 \end{array}\right.\]
(Figure 9.5.3).
Solution
In terms of step functions,
\[\begin{aligned} f(t)&=1+ {\cal U} (t-2)(-2t+1-1)+{\cal U}(t-3)(3t+2t-1)+ {\cal U} (t-5)(t-1-3t),\end{aligned}\nonumber\]
or
\[f(t)=1-2 {\cal U} (t-2)t+ {\cal U} (t-3)(5t-1)- {\cal U} (t-5)(2t+1). \nonumber\]
Now Theorem 9.5.1 implies that
\[\begin{aligned} {\cal L}(f)&={\cal L}(1)-2e^{-2s}{\cal L}(t+2)+e^{-3s}{\cal L}\left(5(t+3)-1\right)-e^{-5s}{\cal L}\left(2(t+5)+1\right)\\[4pt]&={\cal L}(1)-2e^{-2s}{\cal L}(t+2)+e^{-3s}{\cal L}(5t+14)-e^{-5s}{\cal L}(2t+11)\\[4pt]&={1\over s}-2e^{-2s}\left({1\over s^2}+{2\over s}\right)+ e^{-3s}\left({5\over s^2}+{14\over s}\right)-e^{-5s}\left({2\over s^2}+{11\over s}\right). \end{aligned}\nonumber\]
The trigonometric identities
\[\label{eq:8.4.8} \sin (A+B)=\sin A\cos B+\cos A\sin B\]
\[\label{eq:8.4.9} \cos (A+B)=\cos A\cos B-\sin A\sin B\]
are useful in problems that involve shifting the arguments of trigonometric functions. We’ll use these identities in the next example.
Example 9.5.4
Find the Laplace transform of
\[\label{eq:8.4.10} f(t)=\left\{\begin{array}{cl}{\sin t,}&{0\leq t<\frac{\pi }{2}}\\{\cos t-3\sin t,}&{\frac{\pi }{2}\leq t<\pi }\\{3\cos t,}&{t\geq \pi} \end{array} \right.\]
(Figure 9.5.4).
Solution
In terms of step functions,
\[f(t)=\sin t+ {\cal U} (t-\pi/2) (\cos t-4\sin t)+ {\cal U} (t-\pi) (2 \cos t+3\sin t). \nonumber\]
Now Theorem 9.5.1 implies that
\[\label{eq:8.4.11} \begin{array}{ccl} {\cal L}(f)&=&{\cal L}(\sin t)+e^{-{\pi\over 2}s}{\cal L} \left(\cos\left(t+{\pi\over2}\right)-4\sin\left(t+{\pi\over2}\right)\right) \\[4pt]&&\qquad+e^{-\pi s}{\cal L}\left(2\cos(t+\pi)+3\sin(t+\pi)\right). \end{array}\]
Since
\[\cos\left(t+{\pi\over 2}\right)-4\sin\left(t+{\pi\over 2}\right)=-\sin t-4\cos t \nonumber\]
and
\[2\cos (t+\pi)+3\sin (t+\pi)=-2\cos t-3\sin t, \nonumber\]
we see from Equation \ref{eq:8.4.11} that
\[\begin{align*} {\cal L}(f)&={\cal L}(\sin t)-e^{-\pi s/2}{\cal L}(\sin t+4\cos t) -e^{-\pi s}{\cal L}(2\cos t+3\sin t)\\[4pt]&={1\over s^2+1}-e^{-{\pi\over 2}s}\left({1+4s\over s^2+1}\right) -e^{-\pi s}\left({3+2s\over s^2+1}\right). \end{align*}\nonumber\]
The Second Shifting Theorem
Replacing \(g(t)\) by \(g(t-a)\) in Theorem 9.5.2 yields the next theorem.
Theorem 9.5.2 TheSecond Shifting Theorem
If \(a\ge0\) and \({\cal L}(g)\) exists for \(s>s_0\) then \({\cal L}\left( {\cal U} (t-a)g(t-a)\right)\) exists for \(s>s_0\) and
\[{\cal L}( {\cal U} (t-a)g(t-a))=e^{-sa}{\cal L}(g(t)),\nonumber\]
or, equivalently,
\[\label{eq:8.4.12} \mbox{if } g(t)\leftrightarrow G(s),\mbox{ then } {\cal U} (t-a)g(t-a)\leftrightarrow e^{-sa}G(s).\]
Note
Recall that the First Shifting Theoremstates that multiplying a function by \(e^{at}\) corresponds to shifting the argument of its transform by a units. The Second Shifting Theoremstates that multiplying a Laplace transform by the exponential \(e^{−as}\) corresponds to shifting the argument of the inverse transform by \(a\) units.
Example 9.5.5
Use Equation \ref{eq:8.4.12} to find
\[{\cal L}^{-1}\left(e^{-2s}\over s^2\right). \nonumber\]
Solution
To apply Equation \ref{eq:8.4.12} we let \(a=2\) and \(G(s)=1/s^2\). Then \(g(t)=t\) and Equation \ref{eq:8.4.12} implies that
\[{\cal L}^{-1}\left(e^{-2s}\over s^2\right)= {\cal U} (t-2)(t-2).\nonumber\]
Example 9.5.6
Find the inverse Laplace transform \(h\) of
\[H(s)={1\over s^2}-e^{-s}\left({1\over s^2}+{2\over s}\right)+ e^{-4s}\left({4\over s^3}+{1\over s}\right),\nonumber\]
and find distinct formulas for \(h\) on appropriate intervals.
Solution
Let
\[G_0(s)={1\over s^2},\quad G_1(s)={1\over s^2}+{2\over s},\quad G_2(s)={4\over s^3}+{1\over s}.\nonumber\]
Then
\[g_0(t)=t,\; g_1(t)=t+2,\; g_2(t)=2t^2+1.\nonumber\]
Hence, Equation \ref{eq:8.4.12} and the linearity of \({\cal L}^{-1}\) imply that
\[\begin{aligned} h(t)&={\cal L}^{-1}\left(G_0(s)\right)-{\cal L}^{-1}\left(e^{-s}G_1(s)\right)+{\cal L}^{-1}\left(e^{-4s}G_2(s)\right)\\[4pt]&=t- {\cal U} (t-1)\left[(t-1)+2\right]+ {\cal U} (t-4)\left[2(t-4)^2+1\right]\\[4pt]&=t- {\cal U} (t-1)(t+1)+ {\cal U} (t-4)(2t^2-16t+33),\end{aligned}\nonumber\]
which can also be written as
\[h(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\[4pt]-1,&1\le t<4,\\[4pt]2t^2-16t+32,&t\ge4. \end{array}\right.\nonumber \]
Example 9.5.7
Find the inverse transform of
\[H(s)={2s\over s^2+4}-e^{-{\pi\over 2}s} {3s+1\over s^2+9}+e^{-\pi s}{s+1\over s^2+6s+10}. \nonumber\]
Solution
Let
\[G_0(s)={2s\over s^2+4},\quad G_1(s)=-{(3s+1)\over s^2+9},\nonumber\]
and
\[G_2(s)={s+1\over s^2+6s+10}={(s+3)-2\over (s+3)^2+1}.\nonumber\]
Then
\[g_0(t)=2\cos 2t,\quad g_1(t)=-3\cos 3t-{1\over 3}\sin 3t,\nonumber\]
and
\[g_2(t)=e^{-3t}(\cos t-2\sin t).\nonumber\]
Therefore Equation \ref{eq:8.4.12} and the linearity of \({\cal L}^{-1}\) imply that
\[\begin{aligned} h(t)&=2\cos 2t- {\cal U} (t-\pi/2)\left[3\cos 3(t-\pi/2)+{1\over 3}\sin 3\left(t-{\pi\over 2}\right)\right]+{\cal U} (t-\pi)e^{-3(t-\pi)}\left[\cos (t-\pi)-2\sin (t-\pi)\right].\end{aligned}\nonumber\]
Using the trigonometric identities Equation \ref{eq:8.4.8} and Equation \ref{eq:8.4.9}, we can rewrite this as
\[\label{eq:8.4.13} \begin{align} h(t)&=2\cos 2t+ {\cal U} (t-\pi/2)\left(3\sin 3t- {1\over 3}\cos 3t\right)+{\cal U}(t-\pi)e^{-3(t-\pi)} (\cos t-2\sin t) \end{align}\]
(Figure 9.5.5).
We’ll now consider initial value problems of the form
\[\label{eq:8.5.1} ay''+by'+cy=f(t), \quad y(0)=k_0,\quad y'(0)=k_1,\]
where \(a\), \(b\), and \(c\) are constants (\(a\ne0\)) and \(f\) is piecewise continuous on \([0,\infty)\). Problems of this kind occur in situations where the input to a physical system undergoes instantaneous changes, as when a switch is turned on or off or the forces acting on the system change abruptly.
It can be shownthat the differential equation in Equation \ref{eq:8.5.1} has no solutions on an open interval that contains a jump discontinuity of \(f\). Therefore we must define what we mean by a solution of Equation \ref{eq:8.5.1} on \([0,\infty)\) in the case where \(f\) has jump discontinuities. The next theorem motivates our definition. We omit the proof.
Theorem 9.5.3
Suppose \(a,b\), and \(c\) are constants \((a\ne0),\) and \(f\) is piecewise continuous on \([0,\infty).\) with jump discontinuities at \(t_1,\) …, \(t_n,\) where
\[0<t_1<\cdots<t_n. \nonumber\]
Let \(k_0\) and \(k_1\) be arbitrary real numbers. Then there is a unique function \(y\) defined on \([0,\infty)\) with these properties:
- \(y(0)=k_0\) and \(y'(0)=k_1\).
- \(y\) and \(y'\) are continuous on \([0,\infty)\).
- \(y''\) is defined on every open subinterval of \([0,\infty)\) that does not contain any of the points \(t_1,\) …, \(t_n\), and \[ay''+by'+cy=f(t) \nonumber\] on every such subinterval.
- \(y''\) has limits from the right and left at \(t_1,\) …\(,\) \(t_n\).
We define the function \(y\) of Theorem 9.5.3 to be the solution of the initial value problem Equation \ref{eq:8.5.1}.
We begin by considering initial value problems of the form
\[\label{eq:8.5.2} ay''+by'+cy=\left\{\begin{array}{cl} f_0(t),&0\le t<t_1,\\[4pt]f_1(t),&t\ge t_1, \end{array}\right.\quad y(0)=k_0,\quad y'(0)=k_1,\]
where the forcing function has a single jump discontinuity at \(t_1\).
It can be shownthat \(y'\) exists and is continuous at \(t_1\). The next example illustrates this and shows how to solve this kind of intitial value problem without Laplace transforms.
Example 9.5.8
Solve the initial value problem
\[\label{eq:8.5.3} y''+y=f(t), \quad y(0)=2,\; y'(0)=-1,\]
where
\[f(t)=\left\{\begin{array}{rl} 1,&0\le t< {\pi\over2},\\[4pt] -1,&t\ge {\pi\over2}. \end{array}\right. \nonumber\]
Solution
The initial value problem when \(0\le t< {\pi\over2}\)is
\[y''+y=1, \quad y(0)=2,\quad y'(0)=-1. \nonumber\]
We leave it to you to verify that its solution is
\[y_0=1+\cos t-\sin t. \nonumber\]
To solve the initial value problem when \(t\ge {\pi\over2}\) we need to find new initial conditions, and we use \(t=\pi/2\) to get\(y_0(\pi/2)=0\) and \(y_0'(\pi/2)=-1\), so the second initial value problem is
\[y''+y=-1, \quad y\left({\pi\over2}\right)=0,\; y'\left({\pi\over 2}\right)=-1. \nonumber\]
We leave it to you to verify that the solution of this problem is
\[y_1=-1+\cos t+\sin t. \nonumber\]
Hence, the solution of Equation \ref{eq:8.5.3} is
\[\label{eq:8.5.4} y=\left\{\begin{array}{rl} 1+\cos t-\sin t,&0\le t< {\pi\over2}, \\[4pt] -1+\cos t+\sin t,&t\ge {\pi\over2} \end{array}\right.\]
This can quickly become unwieldy if we have many pieces, and it is this kind of initial value problem where Laplace transforms are particularly effective.
If \(f_0\) and \(f_1\) are defined on \([0,\infty)\), we can rewrite Equation \ref{eq:8.5.2} as
\[ay''+by'+cy=f_0(t)+ {\cal U} (t-t_1)\left(f_1(t)-f_0(t)\right), \quad y(0)=k_0,\quad y'(0)=k_1, \nonumber\]
and apply the method of Laplace transforms. We’ll now solve the problem considered in Example 9.5.8 by this method.
Example 9.5.9
Use the Laplace transform to solve the initial value problem
\[\label{eq:8.5.5} y''+y=f(t), \quad y(0)=2,\; y'(0)=-1,\]
where
\[f(t)=\left \{ \begin{array}{cl} \phantom{-}1,&0\le t< {\pi\over2},\\ -1,&t \ge \pi\over2. \end{array}\right. \nonumber\]
Solution
Here
\[f(t)=1-2 {\cal U} \left(t-{\pi\over2}\right), \nonumber\]
so Theorem 9.5.1 (with \(g(t)=1\)) implies that
\[{\cal L}(f)={1-2e^{-{\pi s/2}}\over s}. \nonumber\]
Therefore, transforming Equation \ref{eq:8.5.5} yields
\[(s^2+1)Y(s)={1-2e^{-{\pi s/ 2}}\over s}-1+2s, \nonumber\]
so
\[\label{eq:8.5.6} Y(s)=(1-2e^{-{\pi s/ 2}}) G(s)+{2s-1\over s^2+1},\]
with
\[G(s)={1\over s(s^2+1)}. \nonumber\]
The form for the partial fraction expansion of \(G\) is
\[\label{eq:8.5.7} {1\over s(s^2+1)}={A\over s}+{Bs+C\over s^2+1}. \]
Multiplying through by \(s(s^2+1)\) yields
\[A(s^2+1)+(Bs+C)s=1, \nonumber\]
or
\[(A+B)s^2+Cs+A=1. \nonumber\]
Equating coefficients of like powers of \(s\) on the two sides of this equation shows that \(A=1\), \(B=-A=-1\) and \(C=0\). Hence, from Equation \ref{eq:8.5.7},
\[G(s)={1\over s}-{s\over s^2+1}. \nonumber\]
Therefore
\[g(t)=1-\cos t. \nonumber\]
From this, Equation \ref{eq:8.5.6}, and Theorem 8.4.2,
\[y=1-\cos t-2 {\cal U} \left(t-{\pi\over2}\right)\left(1-\cos\left(t-{\pi \over2}\right)\right)+2\cos t-\sin t. \nonumber\]
Simplifying this (recalling that \(\cos (t-\pi/2)=\sin t)\) yields
\[y=1+\cos t-\sin t-2 {\cal U} \left(t-{\pi\over2}\right)(1-\sin t), \nonumber\]
or
\[y=\left\{\begin{array}{cl}{1+\cos t-\sin t,}&{0\leq t<\frac{\pi }{2}}\\{-1+\cos t+\sin t,}&{t\geq \frac{\pi }{2}} \end{array} \right.\nonumber \]
which is the result obtained in Example 9.5.8.
Note
It isn’t obvious that using the Laplace transform to solve Equation \ref{eq:8.5.2} as we did in Example 9.5.9 yields a function \(y\) with the properties stated in Theorem 9.5.3; that is, such that \(y\) and \(y'\) are continuous on \([0, ∞)\) and \(y''\) has limits from the right and left at \(t_{1}\). However, this is true if \(f_{0}\) and \(f_{1}\) are continuous and of exponential order on \([0, ∞)\).
Example 9.5.10
Solve the initial value problem
\[\label{eq:8.5.8} y''-y=f(t), \quad y(0)=-1,\; y'(0)=2,\]
where
\[f(t)=\left\{\begin{array}{cl} t,&0\le t<1,\\ 1,&t\ge 1. \end{array}\right.\nonumber\]
Solution
Here
\[f(t)=t- {\cal U} (t-1)(t-1),\nonumber\]
so
\[\begin{align*} {\cal L}(f)&={\cal L}(t)-{\cal L}\left( {\cal U} (t-1)(t-1)\right)\\[4pt] &=\cal L(t)-e^{-s}{\cal L}(t)\\[4pt] &={1\over s^2}-{e^{-s}\over s^2}.\end{align*}\nonumber\]
Since transforming Equation \ref{eq:8.5.8} yields
\[(s^2-1) Y(s)={\cal L}(f)+2-s,\nonumber\]
we see that
\[\label{eq:8.5.9} Y(s)=(1-e^{-s})H(s)+{2-s\over s^2-1},\]
where
\[H(s)={1\over s^2(s^2-1)}={1\over s^2-1}-{1\over s^2};\nonumber\]
therefore
\[\label{eq:8.5.10} h(t)=\sinh t-t.\]
Since
\[{\cal L}^{-1}\left({2-s\over s^2-1}\right)=2\sinh t-\cosh t,\nonumber\]
we conclude from Equation \ref{eq:8.5.9}, Equation \ref{eq:8.5.10}, and Theorem 9.5.2 that
\[y=\sinh t-t- {\cal U} (t-1)\left(\sinh (t-1)-t+1\right)+2\sinh t- \cosh t,\nonumber\]
or
\[\label{eq:8.5.11} y=3\sinh t-\cosh t-t- {\cal U} (t-1)\left(\sinh (t-1)-t+1\right)\]
We leave it to you to verify that \(y\) and \(y'\) are continuous and \(y''\) has limits from the right and left at \(t_1=1\).
Example 9.5.11
Solve the initial value problem
\[\label{eq:8.5.12} y''+y=f(t), \quad y(0)=0,\; y'(0)=0,\]
where
\[f(t)=\left\{\begin{array}{cl}{0,}&{0\leq t<\frac{\pi }{4}}\\{\cos 2t}&{\frac{\pi }{4}\leq t< \pi }\\{0,}&{t\geq \pi } \end{array} \right.\nonumber \]
Solution
Here
\[f(t)= {\cal U} (t-\pi/4)\cos2t-{\cal U}(t-\pi)\cos2t, \nonumber\]
so
\[\begin{align*} {\cal L}(f)&={\cal L}\left( {\cal U} (t-\pi/4)\cos2t\right)-{\cal L}\left( {\cal U} (t-\pi)\cos2t\right)\\[4pt] &=e^{-{\pi s/4}}{\cal L}\left(\cos2(t+\pi/4)\right)-e^{-\pi s} {\cal L}\left(\cos2(t+\pi)\right)\\[4pt] &=-e^{-{\pi s/4}}{\cal L}(\sin2t)-e^{-\pi s} {\cal L}(\cos2t)\\[4pt] &=-{2e^{-{\pi s/ 4}}\over s^2+4}-{se^{-\pi s}\over s^2+4}.\end{align*}\nonumber\]
Since transforming Equation \ref{eq:8.5.12} yields
\[(s^2+1)Y(s)={\cal L}(f),\nonumber\]
we see that
\[\label{eq:8.5.13} Y(s)=e^{-{\pi s/ 4}} H_1(s)+e^{-\pi s} H_2(s),\]
where
\[\label{eq:8.5.14} H_1(s)=-{2\over (s^2+1)(s^2+4)}\quad\mbox{ and }\quad H_2(s)=-{s \over (s^2+1)(s^2+4)}.\]
To simplify the required partial fraction expansions, we first write
\[{1\over (x+1)(x+4)}={1\over3}\left[{1\over x+1}-{1\over x+4}\right].\nonumber\]
Setting \(x=s^2\) and substituting the result in Equation \ref{eq:8.5.14} yields
\[H_1(s)=-{2\over3}\left[{1\over s^2+1}-{1\over s^2+4}\right] \quad\mbox{ and }\quad H_2(s)=-{1\over3}\left[{s\over s^2+1}-{s\over s^2+4}\right].\nonumber\]
The inverse transforms are
\[h_1(t)=-{2\over3}\sin t+{1\over3}\sin2t \quad\mbox{ and }\; h_2(t)=-{1\over3}\cos t+{1\over3}\cos2t.\nonumber\]
From Equation \ref{eq:8.5.13} and Theorem 9.5.2,
\[\label{eq:8.5.15} y= {\cal U} \left(t-{\pi\over4}\right) h_1\left(t-{\pi\over4}\right)+ {\cal U} (t-\pi) h_2(t-\pi).\]
Since
\[\begin{aligned} h_1\left(t-{\pi\over4}\right)&=-{2\over3}\sin\left(t-{\pi\over 4}\right)+{1\over3}\sin2\left(t-{\pi\over4}\right)\\ &=-{\sqrt{2}\over3} (\sin t-\cos t)-{1\over3}\cos2t\end{aligned}\nonumber\]
and
\[\begin{align*} h_2(t-\pi)&=-{1\over3}\cos (t-\pi)+{1\over3}\cos2(t-\pi)\\ &={1\over3}\cos t+{1\over3}\cos2t,\end{align*}\nonumber\]
Equation \ref{eq:8.5.15} can be rewritten as
\[y=-{1\over3} {\cal U} \left(t-{\pi\over4}\right)\left(\sqrt{2}(\sin t-\cos t)+\cos2t\right) + {1\over3} {\cal U} (t-\pi) (\cos t+\cos2t)\nonumber\]
or
\[\label{eq:8.5.16} \left\{\begin{array}{cl}{0,}&{0\leq t< \frac{\pi }{4}}\\{-\frac{\sqrt{2}}{3}(\sin t-\cos t)-\frac{1}{3}\cos 2t,}&{\frac{\pi }{4}\leq t<\pi ,}\\{-\frac{\sqrt{2}}{3}\sin t+\frac{1+\sqrt{2}}{3}\cos t,}&{t\geq\pi } \end{array} \right. \]
We leave it to you to verify that \(y\) and \(y'\) are continuous and \(y''\) has limits from the right and left at \(t_1=\pi/4\) and \(t_2=\pi\) (Figure 9.5.2).
