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Differential Equations: Piecewise Forcing Functions

1. Initial Value Problem

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

Note: The solution form may involve \( R \cos(\omega_0 t - \delta) \).

Laplace Transform Solution

After applying the Laplace transform and solving for \( Y(s) \):

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

Taking the inverse Laplace transform:

\[ y = \sin t + 1 - \cos t - u_{3\pi}(t) (1 - \cos(t - 3\pi)) \]

Recall the identity: \( \cos(t - \pi) = -\cos t \)

Figure: A hand-drawn graph showing a periodic wave starting at t=0, with points marked at pi/2, pi, 3pi/2, and 2pi.

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Differential Equations: Ramp Forcing Function

9. Initial Value Problem

\[ y'' + y = \begin{cases} t/2 & 0 \le t < 6 \\ 3 & t \ge 6 \end{cases} \]

Initial conditions: \( y(0) = 0, y'(0) = 1 \)

Figure: A graph showing a linear ramp from the origin to (6, 3), where it becomes a horizontal line.

Step Function Representation

Expressing the forcing function using the unit step function \( u_6(t) \):

\[ = \frac{1}{2}t + u_6(t) \left( -\frac{1}{2}t + 3 \right) \]

Shifting by 6 units: \( -\frac{1}{2}t + 3 \rightarrow -\frac{1}{2}(t+6) + 3 = -\frac{1}{2}t \)

Solving the Equation

Applying the Laplace transform:

\[ s^2 Y - s y(0) - y'(0) + Y = \frac{1}{2s^2} + e^{-6s} \left( -\frac{1}{2s^2} \right) \]

\[ (s^2 + 1) Y = 1 + \frac{1}{2s^2} + e^{-6s} \left( -\frac{1}{2s^2} \right) \]

\[ Y = \frac{1}{s^2+1} + \frac{1}{2s^2(s^2+1)} - e^{-6s} \frac{1}{2s^2(s^2+1)} \]

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Differential Equation Solution

12. \( y^{(4)} - y = u_1(t) - u_2(t) \) with initial conditions \( y(0) = y'(0) = y''(0) = y'''(0) = 0 \)

Taking the Laplace transform:

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

Solving for \( Y \):

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

Note on partial fraction decomposition term:

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

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6.5 Impulse Function

Impulse: short-acting force (e.g. bat hitting baseball)

Model with Step Functions

Consider a rectangular pulse defined by:

Step up at \( t = -a \)
Step down at \( t = a \)
Rectangle area = 1

Figure: Graph of a rectangular pulse f(t) centered at the origin with height 1/(2a) and width from -a to a.

\[ f(t) = \frac{1}{2a} u_{-a}(t) - \frac{1}{2a} u_a(t) \]

Limiting Case

Now let \( a \to 0 \), keeping area 1.

Figure: Sequence of narrowing and taller rectangular pulses, all with area 1, as a approaches 0.

Figure: The limit as a goes to 0: an infinitely tall and thin spike at the origin, representing the impulse function.

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The Unit Impulse Function

This is called the unit impulse function or Dirac delta function.

\[ \delta(t) = 0, \quad t \neq 0 \]

but

\[ \int_{-\infty}^{\infty} \delta(t) \, dt = 1 \]

Time-Shifted Impulse

\( \delta(t - t_0) \) puts the impulse at \( t = t_0 \).

\[ \delta(t - t_0) = 0, \quad t \neq t_0 \]\[ \int_{-\infty}^{\infty} \delta(t - t_0) \, dt = 1 \]

Figure: Graph of f(t) vs t showing a vertical arrow at t=t0 representing the Dirac delta impulse.

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\[ \mathcal{L} \{ \delta(t - t_0) \} = e^{-t_0 s} \]

Note similarity and difference to \( \mathcal{L} \{ u_c(t) \} = \frac{e^{-cs}}{s} \)

Example

Solve the initial value problem:

\[ y'' + 4y = \delta(t - \pi), \quad y(0) = y'(0) = 0 \]

Taking the Laplace transform of both sides:

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

Note that the inverse transform of \( \frac{1}{s^2+4} \) is \( \frac{1}{2} \sin 2t \). Applying the second shifting theorem:

\[ y(t) = u_{\pi}(t) \left[ \frac{1}{2} \sin 2(t - \pi) \right] \]\[ y(t) = \begin{cases} 0 & t < \pi \\ \frac{1}{2} \sin 2(t - \pi) = \frac{1}{2} \sin 2t & t \geq \pi \end{cases} \]

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Differential Equations: Impulse Response Example

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Example: Second-Order ODE with Dirac Delta Functions

Consider the following initial value problem involving two impulse functions:

\[ y'' + 4y = \delta(t - \pi) - \delta(t - 2\pi), \quad y(0) = y'(0) = 0 \]

Figure: Graph of y(t) vs t. It is zero until t=pi, then a sine wave pulse between pi and 2pi, then zero again.

Laplace Transform Solution

Taking the Laplace transform of both sides:

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

Solving for \( Y \):

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

Inverse Laplace Transform

Note on periodicity (\( \text{period} = \pi \)):

\( \sin 2(t - \pi) = \sin(2t - 2\pi) = \sin 2t \)
\( \sin 2(t - 2\pi) = \sin(2t - 4\pi) = \sin 2t \)

Applying the second shifting theorem:

\[ y = u_{\pi}(t) \cdot \frac{1}{2} \sin 2(t - \pi) - u_{2\pi}(t) \cdot \frac{1}{2} \sin 2(t - 2\pi) \] \[ = u_{\pi}(t) \cdot \frac{1}{2} \sin 2t - u_{2\pi}(t) \cdot \frac{1}{2} \sin 2t \]

Expressed as a piecewise function:

\[ y = \begin{cases} 0 & t < \pi \\ \frac{1}{2} \sin 2t & \pi \leq t < 2\pi \\ 0 & t \geq 2\pi \end{cases} \]

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Differential Equations: Forced Response with Impulse

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Example: Forced Damped Oscillator

Solve the following ODE with a sinusoidal forcing term and a delayed impulse:

\[ y'' + 2y' + 3y = \sin t + \delta(t - 3\pi), \quad y(0) = y'(0) = 0 \]

Figure: Graph of y(t) vs t. Shows a damped oscillation starting at t=pi and another smaller one at t=2pi.

Laplace Transform and Partial Fractions

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

Expanding the first term using partial fraction decomposition:

\[ Y = \frac{1}{4} \frac{1}{s^2 + 1} - \frac{1}{4} \frac{s}{s^2 + 1} + \frac{1}{4} \frac{s + 1}{s^2 + 2s + 3} + e^{-3\pi s} \frac{1}{s^2 + 2s + 3} \]

Inverse Transform Components

Term 1 & 2:

\( \frac{1}{s^2 + 1} \rightarrow \sin t \)

\( \frac{s}{s^2 + 1} \rightarrow \cos t \)

Term 3 (Completing the Square):

\( \frac{s + 1}{s^2 + 2s + 3} = \frac{s + 1}{(s + 1)^2 + 2} \rightarrow e^{-t} \cos \sqrt{2}t \)

Term 4 (Impulse Response):

\( \frac{1}{s^2 + 2s + 3} = \frac{1}{(s + 1)^2 + 2} \rightarrow \frac{1}{\sqrt{2}} e^{-t} \sin \sqrt{2}t \)

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Solution Decomposition and Piecewise Form

The solution for \( y \) is composed of three distinct parts based on the input functions and initial conditions:

\[ y = \underbrace{\left( \frac{1}{4} \sin t - \frac{1}{4} \cos t \right)}_{\text{particular due to } \sin t} + \underbrace{\left( \frac{1}{4} e^{-t} \cos \sqrt{2} t \right)}_{\text{homogeneous w/ IC's}} + \underbrace{u_{3\pi}(t) \frac{1}{\sqrt{2}} e^{-(t-3\pi)} \sin \sqrt{2}(t-3\pi)}_{\text{particular due to } \delta(t-3\pi)} \]

Piecewise Representation

Expressing the solution as a piecewise function for different time intervals:

\[ y = \begin{cases} \frac{1}{4} \sin t - \frac{1}{4} \cos t + \frac{1}{4} e^{-t} \cos \sqrt{2} t & t < 3\pi \\ \\ \frac{1}{4} \sin t - \frac{1}{4} \cos t + \frac{1}{4} e^{-t} \cos \sqrt{2} t + \frac{1}{\sqrt{2}} e^{-(t-3\pi)} \sin \sqrt{2}(t-3\pi) & t \geq 3\pi \end{cases} \]

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Graphical Analysis of the Solution

The following graph illustrates the behavior of the solution \( y(t) \) over time. Note the transition point at \( t = 3\pi \approx 9.42 \), where the impulse function \( \delta(t-3\pi) \) introduces a sudden change in the oscillation pattern.

The red dot on the graph highlights this critical junction, showing the impact of the particular solution component triggered by the unit step function.

Figure: A plot of an oscillating function with a red dot at t ≈ 9.42, showing a sharp change in the wave's amplitude.