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6.2 Solutions of Initial-Value Problems

Use LT to solve:

\[ y'' + p(t)y' + q(t)y = g(t) \]\[ y(0) = y_0, \quad y'(0) = y'_0 \]

Definition of LT:

\[ \mathcal{L} \{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt \]

We often look up a Table for commonly used LT (Table 6.2.1 p. 321).

Inverse LT:

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

Formula is complicated and is not practical. Use Table.

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Example

\[ F(s) = \frac{2s - 3}{s^2 - 4} \quad \text{find } f(t) \]

Note denominator: \( s^2 - a^2 \) (\( a \): constant)

Closest match:

  • #7 \( \mathcal{L} \{\sinh at\} = \frac{a}{s^2 - a^2} \)
  • #8 \( \mathcal{L} \{\cosh at\} = \frac{s}{s^2 - a^2} \)
\[ F(s) = \frac{2s}{s^2 - 4} - \frac{3}{s^2 - 4} \quad (a \text{ is } 2) \]\[ = 2 \cdot \frac{s}{s^2 - 4} - \frac{3}{2} \frac{2}{s^2 - 4} \]

\[ f(t) = 2 \cosh 2t - \frac{3}{2} \sinh 2t \]

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Example: Inverse Laplace Transform

\[ F(s) = \frac{3s}{s^2 - s - 6} \]
\[ = \frac{3s}{(s-3)(s+2)} = \frac{A}{s-3} + \frac{B}{s+2} \]

#2 Laplace Transform Property

\[ \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \]

To find the constants \( A \) and \( B \), we multiply by the common denominator:

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

Method 1: Substitution

  • Let \( s = 3 \): \( 9 = 5A \implies A = \frac{9}{5} \)
  • Let \( s = -2 \): \( -6 = -5B \implies B = \frac{6}{5} \)

Method 2: Equating Coefficients

Alternatively, expand the equation:

\[ 3s = (A+B)s + (2A - 3B) \]

Then solve the system of equations:

  • \( A + B = 3 \)
  • \( 2A - 3B = 0 \)

Final Solution

\[ F(s) = \frac{9}{5} \frac{1}{s-3} + \frac{6}{5} \frac{1}{s+2} \]
\[ f(t) = \frac{9}{5} e^{3t} + \frac{6}{5} e^{-2t} \]
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If \( y \) is not specified, then:

\[ \mathcal{L}\{y\} = Y \]

Laplace Transform of a Derivative

Find \( \mathcal{L}\{y'\} = ? \)

Use the definition of the Laplace Transform:

\[ \mathcal{L}\{y'\} = \int_{0}^{\infty} e^{-st} y' dt \]

Using integration by parts:

  • \( u = e^{-st} \)
  • \( dv = y' dt \)
  • \( du = -s e^{-st} dt \)
  • \( v = y \)
\[ \mathcal{L}\{y'\} = \lim_{A \to \infty} \left( y e^{-st} \Big|_0^A + s \int_{0}^{A} e^{-st} y dt \right) \]
\[ = \lim_{A \to \infty} [y(A) e^{-sA} - y(0)] + s \int_{0}^{\infty} e^{-st} y dt \]

Note that as \( A \to \infty \), \( y(A) e^{-sA} \to 0 \). Also, the integral term is the definition of \( \mathcal{L}\{y\} = Y \).

\[ \mathcal{L}\{y'\} = sY - y(0) \]
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Laplace Transform of Higher-Order Derivatives

\[ \mathcal{L} \{y''\} = \mathcal{L} \{(y')'\} = s \mathcal{L} \{y'\} - y'(0) \]

\[ = s (sY - y(0)) - y'(0) \]

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

Similarly,

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

\[ \mathcal{L} \{y^{(4)}\} = s^4 Y - s^3 y(0) - s^2 y'(0) - s y''(0) - y'''(0) \]

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Example: Solving a Differential Equation

Solve: \[ y'' - 2y' + 2y = 0 \quad y(0)=0, \ y'(0)=1 \]

LT both sides

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

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

Solve for Y

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

\[ (s^2 - 2s + 2) Y = 1 \]

characteristic polynomial: \( s^2 - 2s + 2 \)

\[ Y = \frac{1}{s^2 - 2s + 2} \]

Find \( \mathcal{L}^{-1} \{Y\} = y \)

Denominator not factorable \( \rightarrow \) complete the square.

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Completing the Square for Inverse Laplace Transforms

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

Note: The term \((\frac{-2}{2})^2\) is added and subtracted to complete the square, where \(-2\) is the coefficient of \(s\).

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

#9 Laplace Transform Property

\[ \mathcal{L} \{ e^{at} \sin bt \} = \frac{b}{(s-a)^2 + b^2} \]

Applying the property to find the inverse transform:

so \( y(t) = e^t \sin t \)

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Example: Solving a Third-Order Differential Equation

Example: \( y''' + y' = 1 \)

Initial conditions: \( y(0) = y'(0) = y''(0) = 0 \)

Taking the Laplace transform of both sides:

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

Applying initial conditions (all zero):

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

This is a case of repeated linear factors.

Partial Fraction Decomposition

\[ \frac{1}{s \cdot s \cdot (s^2 + 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 1} \]
\[ 1 = As(s^2 + 1) + B(s^2 + 1) + (Cs + D)s^2 \]
\[ 1 = (A + C)s^3 + (B + D)s^2 + As + B \]

Solving for Coefficients:

\( B = 1 \)

\( A = 0 \)

\( B + D = 0 \implies D = -1 \)

\( A + C = 0 \implies C = 0 \)

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\[ Y = \frac{1}{s^2} + \frac{-1}{s^2+1} \]

\( y = t - \sin t \)

Differential Equation Analysis

\[ y''' + y' = 1 \]

Characteristic equation and roots:

\[ r^3 + r = 0 \quad \implies \quad r = 0, \pm i \]

General solution form:

\[ y = C_1 + C_2 \cos t + C_3 \sin t + At \]

LT doesn't care about duplication.

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Laplace Transforms and Discontinuous Functions

LT can handle discontinuous forcing functions (just needs to be piecewise continuous).

Example

\[ y'' + y = \begin{cases} t & 0 \le t < 1 \\ 0 & 1 \le t < \infty \end{cases} \quad y(0) = y'(0) = 0 \]
Coordinate graph of f(t) vs t. It shows a line segment from the origin to t=1, then a jump down to zero.

Applying the Laplace Transform:

\[ s^2 Y - s y(0) - y'(0) + Y = \mathcal{L} \{ f(t) \} \]
\[ \begin{aligned} \mathcal{L} \{ f(t) \} &= \int_{0}^{\infty} e^{-st} f(t) dt = \int_{0}^{1} e^{-st} t dt + \int_{1}^{\infty} e^{-st} 0 dt \\ &= 1 - \frac{1}{s} e^{-s} - \frac{1}{s^2} e^{-s} \end{aligned} \]
\[ (s^2 + 1) Y = 1 - \frac{1}{s} e^{-s} - \frac{1}{s^2} e^{-s} \]
\[ Y = \frac{1 - \frac{1}{s} e^{-s} - \frac{1}{s^2} e^{-s}}{s^2 + 1} \]

\( \mathcal{L}^{-1} \{ Y \} \) discussed in 6.3