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6.2 Solution of Initial-Value Problems

last time: \( f(t), t \ge 0 \quad \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty} f(t) e^{-st} dt \)

\[ f(t) \rightarrow \boxed{\mathcal{L}} \rightarrow F(s) \]

we derived some common transforms, but in practice we normally do a table look up

\( F(s) \), want \( f(t) \)

\[ F(s) \rightarrow \boxed{\mathcal{L}^{-1}} \rightarrow f(t) \]

inverse Laplace transform

\[ f(t) = \mathcal{L}^{-1}\{F(s)\} = \frac{1}{2\pi i} \lim_{T \to \infty} \int_{\gamma-iT}^{\gamma+iT} e^{st} F(s) ds \]

not usually used to find inverse LT
normally also a table look up

\[ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+4} \right\} = \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2^2} \right\} \quad \text{looks like} \quad \mathcal{L}\{\cos at\} = \frac{s}{s^2+a^2} \]

\( = \cos(2t) \)

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how about \( F(s) = \frac{1}{s^2+3s+2} \quad f(t) = ? \)

no exact fit, table entries have \( s-a, s^2+a^2, \) or \( s^2-a^2 \) in denominator

\[ \frac{1}{s^2+3s+2} = \frac{1}{(s+1)(s+2)} \quad \text{no product of } s-a \text{ on the table} \]

would be nice if

\[ \frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \quad \text{partial fraction decomposition} \]

multiply by \( (s+1)(s+2) \)

\[ 1 = A(s+2) + B(s+1) \]\[ 0s + 1 = (A+B)s + (2A+B) \]
\[ \left. \begin{aligned} A+B &= 0 \\ 2A+B &= 1 \end{aligned} \right\} \dots A = -1, B = 1 \]
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\[ \begin{aligned} \mathcal{L}^{-1} \left\{ \frac{1}{s^2+3s+2} \right\} &= \mathcal{L}^{-1} \left\{ -\frac{1}{s+1} + \frac{1}{s+2} \right\} \\ &= -\mathcal{L}^{-1} \left\{ \frac{1}{s+1} \right\} + \mathcal{L}^{-1} \left\{ \frac{1}{s+2} \right\} \\ &= -e^{-t} + e^{-2t} \end{aligned} \]
table:
\( \mathcal{L} \{ e^{at} \} = \frac{1}{s-a} \)

Goal: solve something like \( y'' + 3y' + 2y = 0 \), \( y(0)=1 \), \( y'(0)=0 \) using LT.

basic idea: LT of eq, solve in s domain, then go back to t domain

if \( y \) has an LT, then \( \mathcal{L} \{ y(t) \} = Y(s) \) even if \( y \) is unknown

What is \( \mathcal{L} \{ y' \} = ? \)

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\[ \begin{aligned} \mathcal{L} \{ y'(t) \} &= \int_{0}^{\infty} y'(t) e^{-st} dt \\ &= \lim_{a \to \infty} \int_{0}^{a} y'(t) e^{-st} dt \quad \text{by parts } \begin{cases} u = e^{-st} & dv = y' dt \\ du = -se^{-st} dt & v = y \end{cases} \\ &= \lim_{a \to \infty} \left( y(t) e^{-st} \Big|_0^a - \int_{0}^{a} -se^{-st} y(t) dt \right) \\ &= \lim_{a \to \infty} (y(t) e^{-st}) \Big|_0^a + s \underbrace{\lim_{a \to \infty} \int_{0}^{a} y(t) e^{-st} dt}_{\mathcal{L} \{ y(t) \} = Y(s)} \\ &= \underbrace{\lim_{a \to \infty} (y(a) e^{-sa} - y(0))}_{\to 0 \text{ if } s > 0} + s Y(s) \end{aligned} \]
\[ \mathcal{L} \{ y'(t) \} = s Y(s) - y(0) \]

Similarly,

\[ \mathcal{L} \{ y''(t) \} = s^2 Y(s) - sy(0) - y'(0) \]

on table

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Solving a Second-Order Homogeneous ODE with Laplace Transforms

Now we solve the following initial value problem:

\[y'' + 3y' + 2y = 0, \quad y(0) = 1, \quad y'(0) = 0\]

Step 1: Apply Laplace Transform (LT) to both sides

\[\mathcal{L}\{y'' + 3y' + 2y\} = \mathcal{L}\{0\}\]
\[\mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{0\}\]

Using the differentiation property of Laplace transforms and substituting the given initial conditions:

\[s^2 Y - s y(0) - y'(0) + 3(s Y - y(0)) + 2Y = 0\]

Note: \(y(0) = 1\) and \(y'(0) = 0\) are given.

Step 2: Solve for \(Y(s)\)

Grouping the terms with \(Y\) on the left side:

\[(s^2 + 3s + 2)Y = s + 3\]

The term \(s^2 + 3s + 2\) is the characteristic equation.

\[Y = \frac{s+3}{s^2 + 3s + 2}\]

This is the solution in the \(s\)-domain. We want the solution in the \(t\)-domain.

Step 3: Inverse Laplace Transform and Partial Fraction Expansion

\[y = \mathcal{L}^{-1} \left\{ \frac{s+3}{s^2 + 3s + 2} \right\}\]

Performing partial fraction expansion:

\[\frac{s+3}{(s+2)(s+1)} = \frac{A}{s+2} + \frac{B}{s+1} = \dots = \frac{2}{s+1} - \frac{1}{s+2}\]

Finally, applying the inverse transform:

\[y = \mathcal{L}^{-1} \left\{ \frac{2}{s+1} - \frac{1}{s+2} \right\} = 2e^{-t} - e^{-2t}\]
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Solving a Non-Homogeneous ODE with Laplace Transforms

Consider the non-homogeneous equation:

\[y'' + 3y' + 2y = 1, \quad y(0) = 1, \quad y'(0) = 0\]

Step 1: Apply Laplace Transform

LT both sides:

\[\mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{1\}\]
\[s^2 Y - s y(0) - y'(0) + 3(s Y - y(0)) + 2Y = \frac{1}{s}\]

Substituting initial conditions \(y(0)=1\) and \(y'(0)=0\):

\[(s^2 + 3s + 2)Y = \frac{1}{s} + s + 3\]

Step 2: Solve for \(Y(s)\)

\[Y = \frac{1}{s(s^2 + 3s + 2)} + \frac{s+3}{s^2 + 3s + 2}\]

The first term corresponds to the particular solution, and the second term is from the last example (complementary).

Step 3: Partial Fraction Expansion

Expanding the particular part:

\[\frac{1}{s(s+2)(s+1)} = \frac{A}{s} + \frac{B}{s+2} + \frac{C}{s+1} = \dots = \frac{1}{2} \frac{1}{s} + \frac{1}{2} \frac{1}{s+2} - \frac{1}{s+1}\]

Step 4: Inverse Laplace Transform

Combining the particular and complementary parts in the time domain:

\[y(t) = \frac{1}{2} + \frac{1}{2}e^{-2t} - e^{-t} + 2e^{-t} - e^{-2t}\]

Simplifying the expression:

\[y(t) = \frac{1}{2} - \frac{1}{2}e^{-2t} + e^{-t}\]
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Laplace Transforms with Discontinuous Forcing Functions

LT can handle discontinuous right side

\[y'' + 3y' + 2y = \begin{cases} 0 & 0 \le t < \pi \\ 1 & \pi \le t < \infty \end{cases} \quad y(0) = y'(0) = 0\]

problem if done the "old way"

LT both sides

\[s^2 Y - sy(0) - y'(0) + 3(sY - y(0)) + 2Y = \int_{0}^{\pi} 0 \cdot e^{-st} dt + \int_{\pi}^{\infty} 1 \cdot e^{-st} dt\]

given initial conditions

\[(s^2 + 3s + 2)Y = 0 + \frac{1}{s} e^{-\pi s}\]
\[Y = e^{-\pi s} \frac{1}{s(s+2)(s+1)}\]

find out how to deal w/ this in 6.3

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258 CHAPTER 6 The Laplace Transform

TABLE 6.2.1 Elementary Laplace Transforms

No.\(f(t) = \mathcal{L}^{-1}\{F(s)\}\)\(F(s) = \mathcal{L}\{f(t)\}\)Notes
1.1\(\frac{1}{s}, s > 0\)Sec. 6.1; Ex. 4
2.\(e^{at}\)\(\frac{1}{s-a}, s > a\)Sec. 6.1; Ex. 5
3.\(t^n\), \(n\) a positive integer\(\frac{n!}{s^{n+1}}, s > 0\)Sec. 6.1; Prob. 24
4.\(t^p, p > -1\)\(\frac{\Gamma(p+1)}{s^{p+1}}, s > 0\)Sec. 6.1; Prob. 24
5.\(\sin(at)\)\(\frac{a}{s^2+a^2}, s > 0\)Sec. 6.1; Ex. 7
6.\(\cos(at)\)\(\frac{s}{s^2+a^2}, s > 0\)Sec. 6.1; Prob. 5
7.\(\sinh(at)\)\(\frac{a}{s^2-a^2}, s > |a|\)Sec. 6.1; Prob. 7
8.\(\cosh(at)\)\(\frac{s}{s^2-a^2}, s > |a|\)Sec. 6.1; Prob. 6
9.\(e^{at}\sin(bt)\)\(\frac{b}{(s-a)^2+b^2}, s > a\)Sec. 6.1; Prob. 10
10.\(e^{at}\cos(bt)\)\(\frac{s-a}{(s-a)^2+b^2}, s > a\)Sec. 6.1; Prob. 11
11.\(t^n e^{at}\), \(n\) a positive integer\(\frac{n!}{(s-a)^{n+1}}, s > a\)Sec. 6.1; Prob. 14
12.\(u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \ge c \end{cases}\)\(\frac{e^{-cs}}{s}, s > 0\)Sec. 6.3
13.\(u_c(t)f(t-c)\)\(e^{-cs}F(s)\)Sec. 6.3
14.\(e^{ct}f(t)\)\(F(s-c)\)Sec. 6.3
15.\(f(ct)\)\(\frac{1}{c}F\left(\frac{s}{c}\right), c > 0\)Sec. 6.3; Prob. 17
16.\((f * g)(t) = \int_0^t f(t-\tau)g(\tau)d\tau\)\(F(s)G(s)\)Sec. 6.6
17.\(\delta(t-c)\)\(e^{-cs}\)Sec. 6.5
18.\(f^{(n)}(t)\)\(s^n F(s) - s^{n-1}f(0) - \dots - f^{(n-1)}(0)\)Sec. 6.2; Cor. 6.2.2
19.\((-t)^n f(t)\)\(F^{(n)}(s)\)Sec. 6.2; Prob. 21

Frequently, a Laplace transform \(F(s)\) is expressible as a sum of several terms

\[F(s) = F_1(s) + F_2(s) + \dots + F_n(s). \tag{17}\]

Suppose that \(f_1(t) = \mathcal{L}^{-1}\{F_1(s)\}, \dots, f_n(t) = \mathcal{L}^{-1}\{F_n(s)\}\). Then the function

\[f(t) = f_1(t) + \dots + f_n(t)\]

has the Laplace transform \(F(s)\). By the uniqueness property stated previously, no other continuous function \(f\) has the same transform. Thus

\[\mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{F_1(s)\} + \dots + \mathcal{L}^{-1}\{F_n(s)\}; \tag{18}\]

that is, the inverse Laplace transform is also a linear operator.