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Differential Equations: Laplace Transform Examples

Problem 24

Solve the following initial value problem using Laplace transforms:

\[ y'' + 4y = f(t) = \begin{cases} 1 & 0 \le t < \pi \\ 0 & \pi \le t < \infty \end{cases} \quad y(0)=1, \ y'(0)=0 \]

Step 1: Find the Laplace Transform of the Forcing Function

The Laplace transform of the piecewise function \( f(t) \) is calculated as follows:

\[ \mathcal{L} \{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) dt \]\[ = \int_{0}^{\pi} e^{-st} dt = -\frac{1}{s} e^{-st} \Big|_{0}^{\pi} \]\[ = -\frac{1}{s} e^{-s\pi} + \frac{1}{s} = \frac{1 - e^{-\pi s}}{s} \]

Step 2: Apply Laplace Transform to the Differential Equation

Taking the Laplace transform of both sides of the equation:

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

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

\[ s^2 Y + 4Y = s + \frac{1 - e^{-\pi s}}{s} \]\[ (s^2 + 4) Y = s + \frac{1 - e^{-\pi s}}{s} \]

Solving for \( Y \):

\[ Y = \frac{s}{s^2 + 4} + \frac{1 - e^{-\pi s}}{s(s^2 + 4)} \]

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Problem 21

Solve the following initial value problem:

\[ y'' - 2y' + 2y = \cos t \quad y(0)=1, \ y'(0)=0 \]

Step 1: Apply Laplace Transform

Transforming the equation into the s-domain:

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

Substituting initial conditions and grouping terms:

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

Solving for \( Y \):

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

Step 2: Partial Fraction Decomposition and Simplification

Simplifying the second term:

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

Setting up partial fractions for the first term:

\[ \frac{s}{(s^2 + 1)(s^2 - 2s + 2)} = \frac{As + B}{s^2 + 1} + \frac{Cs + D}{s^2 - 2s + 2} \]

Solve for constants \( A, B, C, D \).

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6.3 Step Functions

Define a unit step function:

\[ u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]

Example:

\[ u_2(t) = \begin{cases} 0 & t < 2 \\ 1 & t \geq 2 \end{cases} \]

Figure: Graph of the unit step function u_2(t) with a jump from 0 to 1 at t=2.

Laplace Transform of the Unit Step Function

\[ \mathcal{L} \{ u_c(t) \} = \int_c^{\infty} e^{-st} dt = \lim_{A \to \infty} \int_c^A e^{-st} dt \]\[ = \lim_{A \to \infty} -\frac{1}{s} e^{-st} \Big|_c^A = \lim_{A \to \infty} -\frac{1}{s} e^{-sA} + \frac{1}{s} e^{-cs} \]

Note: The first term vanishes as A approaches infinity for s > 0.

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\[ \mathcal{L} \{ u_c(t) \} = \frac{e^{-cs}}{s} \]

\[ u_c(t) = \begin{cases} 0 & t < c \\ 1 & t \geq c \end{cases} \]

Example

Find LT of \[ f(t) = \begin{cases} 2 & -\infty < t < 3 \\ -3 & 3 \leq t < \infty \end{cases} \]

Figure: Graph of f(t) showing a constant value of 2 for t < 3 and a jump to -3 for t > 3.

Expressing in terms of step functions:

\[ f(t) = 2 - 5u_3(t) \]

\[ F(s) = \frac{2}{s} - \frac{5}{s} e^{-3s} \]

Generalize:

\[ u_c(t) \cdot g(t) = \begin{cases} 0 & t < c \\ g(t) & t \geq c \end{cases} \]

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\[ \mathcal{L} \{ u_c(t) \cdot f(t-c) \} = e^{-cs} \mathcal{L} \{ f(t) \} = e^{-cs} F(s) \]

NOT LT of this

Example

\[ f(t) = u_1(t) (t-1)^2 \]

\[ = \begin{cases} 0 & t < 1 \\ (t-1)^2 & t \geq 1 \end{cases} \]

Figure: Coordinate graph showing a parabola f(t) starting at t=1 and its leftward shift to the origin.

\[ F(s) = e^{-s} \mathcal{L} \left\{ \text{what } (t-1)^2 \text{ would look like if it is shifted to start at } t=0 \right\} \]

Shift \( (t-1)^2 \) LEFT one unit

Replace \( t \) with \( t+1 \)

So \( (t-1)^2 \) becomes \( t^2 \)

transform THIS after shift NOT this

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\[ f(t) = u_1(t) (t-1)^2 \]

\[ F(s) = e^{-s} \mathcal{L} \{ t^2 \} = e^{-s} \frac{2!}{s^{2+1}} \]

\[ = e^{-s} \frac{2}{s^3} \]

Example

\[ f(t) = u_1(t) (t-2) - u_2(t) (t-1) \]

Figure: Piecewise linear graph on t-axis showing segments at t=1 and t=2 with a downward shift to -1.

\( t=2 \):

\[ t-2 - (t-1) \]

\[ -1 \]

\( t \geq 2 \):

\[ (t-2) - (t-1) \]

\[ -1 \text{ for all } t \]

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\[ F(s) = e^{-s} \mathcal{L} \{t-1\} - e^{-2s} \mathcal{L} \{t+1\} \]

Note: \(t-1\) shift LEFT 2

\[ = e^{-s} \left( \frac{1}{s^2} - \frac{1}{s} \right) - e^{-2s} \left( \frac{1}{s^2} + \frac{1}{s} \right) \]

Inverse LT

\[ \mathcal{L} \{ u_c(t) f(t-c) \} = e^{-cs} \mathcal{L} \{ f(t) \} \]

Example

\[ \mathcal{L}^{-1} \left\{ \frac{2(s-1)}{s^2 - 2s + 2} e^{-2s} \right\} \]

discontinuity at \(t=2\)

\[ f(t) = \mathcal{L}^{-1} \left\{ \frac{2(s-1)}{s^2 - 2s + 2} \right\} = 2e^t \cos t \]

shift RIGHT two units

\[ = 2 u_2(t) e^{t-2} \cos(t-2) \]

Reverse ALL steps we took to do LT