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Differential Equations: Variation of Parameters

Problem 10

\[ y'' - 2y' + y = \frac{e^t}{1+t^2} \quad y_1 = e^t, \quad y_2 = te^t \]

Assume a particular solution of the form:

\[ y = u_1 e^t + u_2 te^t \]

The system of equations for \( u_1' \) and \( u_2' \) is:

\[ \begin{aligned} u_1' y_1 + u_2' y_2 &= 0 \\ u_1' y_1' + u_2' y_2' &= g(t) \end{aligned} \]

Substituting the known functions:

\[ \begin{aligned} u_1' e^t + u_2' te^t &= 0 \\ u_1' e^t + u_2' (te^t + e^t) &= \frac{e^t}{1+t^2} \end{aligned} \]

Solving for \( u_2' \) and \( u_1' \):

\[ \begin{aligned} u_2' e^t &= \frac{e^t}{1+t^2} \implies u_2' = \frac{1}{1+t^2} \\ u_1' &= -u_2' t = -\frac{t}{1+t^2} \end{aligned} \]

Integrating to find \( u_1 \) and \( u_2 \):

\[ \begin{aligned} u_2 &= \tan^{-1}(t) + C_2 \\ u_1 &= -\frac{1}{2} \ln |1+t^2| + C_1 \end{aligned} \]
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Differential Equations: Initial Value Problem

Problem 5

\[ t^2 y'' - 4ty' + 4y = -2t^2 \quad y_1 = t, \quad y_2 = t^4 \]

Initial conditions:

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

Assume a particular solution of the form:

\[ y = u_1 y_1 + u_2 y_2 \]

The system of equations for \( u_1' \) and \( u_2' \) is:

\[ \begin{aligned} u_1' y_1 + u_2' y_2 &= 0 \\ u_1' y_1' + u_2' y_2' &= g(t) \end{aligned} \]

Substituting the known functions (note: divide by \( t^2 \) to get standard form):

\[ \begin{aligned} u_1' t + u_2' t^4 &= 0 \\ u_1' + 4u_2' t^3 &= -2 \end{aligned} \]

Multiplying the second equation by \( t \):

\[ u_1' t + 4u_2' t^4 = -2t \]

Subtracting the first equation from this result:

\[ 3u_2' t^4 = -2t \implies u_2' = -\frac{2}{3} t^{-3} \]

Solving for \( u_1' \):

\[ u_1' = -u_2' t^3 = -\left(-\frac{2}{3} t^{-3}\right) t^3 = \frac{2}{3} \]

Integrating to find \( u_1 \) and \( u_2 \):

\[ \begin{aligned} u_2 &= +\frac{1}{3} t^{-2} + C_2 \\ u_1 &= \frac{2}{3} t + C_1 \end{aligned} \]
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3.7 Mechanical Vibrations

Spring with natural length of \( l \), spring constant \( k \).

Attach object mass \( m \).

At equilibrium \( (u = 0) \):

Force due to spring \( F_s = -kL \)

and is equal to weight: \( mg \)

so \( mg - kL = 0 \)

Two diagrams: one showing a spring of natural length l, and another with a mass m attached, stretching it by L to equilibrium u=0.

Down is positive direction.

If \( u > 0, F_s < 0 \)

If \( u < 0, F_s > 0 \)

\[ F_s = -k(L + u) \]

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Now attach a damper that exerts a force proportional to speed in the opposite direction of velocity (like drag).

\[ F_d = -\gamma u' \]

\( \gamma \) is the damping constant

Newton's 2nd Law: \( F = ma \)

\[ mu'' = \underbrace{-k(L+u)}_{\text{spring}} + \underbrace{mg}_{\text{gravity}} - \underbrace{\gamma u'}_{\text{damper}} \]

\[ = -kL - ku + mg - \gamma u' \]

Note: \( mg - Lk = 0 \)

Equation of Motion

\[ mu'' + \gamma u' + ku = 0 \]

\( m, \gamma, k > 0 \) constants

Linear, constant coeff, homogeneous

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Roots of Characteristic Equation

The roots of the characteristic equation are derived from:

\[mr^2 + \gamma r + k = 0\]
\[r = \frac{-\gamma \pm \sqrt{\gamma^2 - 4km}}{2m}\]
\[= -\frac{1}{2} \frac{\gamma}{m} \pm \frac{\sqrt{\gamma^2 - 4km}}{2m}\]

Case 1: Undamped Free Vibration

If \(\gamma = 0\) (undamped free vibration):

\[r = \pm \frac{\sqrt{-4km}}{2m} = \pm i \sqrt{\frac{k}{m}}\]
\[u = C_1 \cos \sqrt{\frac{k}{m}} t + C_2 \sin \sqrt{\frac{k}{m}} t\]
  • \(\sqrt{\frac{k}{m}}\) : circular frequency (\(\omega_0\))
  • \(\frac{2\pi}{\sqrt{\frac{k}{m}}}\) : period (\(T\))
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Case 2: Overdamped

If \(\gamma^2 > 4km\) (overdamped \(\rightarrow\) no oscillation because roots are real and distinct):

\[u = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]

Need \(\lim_{t \to \infty} u = 0\) from physical intuition.

\(\gamma, m, k > 0\) so \(\gamma^2 > \gamma^2 - 4km\)

\[r = -\frac{1}{2} \frac{\gamma}{m} \pm \frac{\sqrt{\gamma^2 - 4km}}{2m}\]
Note: \(\sqrt{\gamma^2 - 4km}\) is less than \(\gamma\)

Case 3: Critically Damped

If \(\gamma^2 = 4km\) (critically damped \(\rightarrow\) no oscillation) repeated roots:

\[u = C_1 e^{rt} + C_2 t e^{rt} \quad r = -\frac{\gamma}{2m} < 0\]

Critically damped \(\rightarrow\) return to \(u = 0\) in shortest possible time.

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Underdamped Oscillations

If \( \gamma^2 < 4km \) (underdamped), the system has oscillations.

\[ u = e^{-\gamma/2m t} \left( C_1 \cos \frac{\sqrt{\gamma^2 - 4km}}{2m} t + C_2 \sin \frac{\sqrt{\gamma^2 - 4km}}{2m} t \right) \]

Example

A mass weighing 2 lb stretches a spring 6 in. No damping.

\[ mu'' + \gamma u' + ku = 0 \]

Note: In this case, \( \gamma u' \) is crossed out as there is no damping.

Finding the Spring Constant (k)

Using Hooke's Law: \( F = kx \)

Where \( x \) is the deviation from natural length.

\[ 2 \text{ lb} = k \cdot (\frac{1}{2} \text{ ft}) \]

Note: 6 in = \( \frac{1}{2} \) ft.

\( k = 4 \text{ lb/ft} \)

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Finding the Mass (m)

\[ 2 \text{ lb} = m (32 \text{ ft/s}^2) \]\[ m = \frac{1}{16} \text{ lb} \cdot \text{s}^2/\text{ft} \]

The unit \( \text{lb} \cdot \text{s}^2/\text{ft} \) is also known as a slug.

Equation of Motion

Example (eg):

\[ mu'' + ku = 0 \]\[ \frac{1}{16} u'' + 4u = 0 \]\[ u'' + 64u = 0 \]

The general solution is:

\[ u = C_1 \cos 8t + C_2 \sin 8t \]

Initial Conditions (IC)

\( u(0) = 3 \text{ in} \)

Pull down 3 in from equilibrium.

\( u'(0) = 3 \text{ in/s} \)

Downward initial velocity.

Solving for constants:

\[ C_1 = \frac{1}{4}, \quad C_2 = \frac{1}{32} \]
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Phase-Amplitude Form Conversion

Starting with the expression for displacement:

\[ u = \left( \frac{1}{4} \right) \cos 8t + \left( \frac{1}{32} \right) \sin 8t \]

We rewrite this in the form:

\[ u = R \cos(\omega_0 t - \delta) \]
  • \( R \): amplitude
  • \( \omega_0 \): frequency
  • \( \delta \): phase angle

Expansion and Matching Coefficients

Using the cosine difference identity:

\[ u = R \cos \delta \cos \omega_0 t + R \sin \delta \sin \omega_0 t \]

By comparing coefficients with the original equation, we set:

\[ R \cos \delta = \frac{1}{4} \]
\[ R \sin \delta = \frac{1}{32} \]

Calculating Parameters

To find the phase angle \( \delta \):

\[ \tan \delta = \frac{1}{8} \implies \delta = \tan^{-1}\left( \frac{1}{8} \right) \]

To find the amplitude \( R \):

\[ R^2 = \left( \frac{1}{4} \right)^2 + \left( \frac{1}{32} \right)^2 = \frac{\sqrt{5}}{8} \]

The final rewritten form is:

\[ u = \frac{\sqrt{5}}{8} \cos\left( 8t - \tan^{-1}\left( \frac{1}{8} \right) \right) \]
Graph of a cosine wave starting at a peak on the y-axis and shifting right by a phase delta.
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Example: Underdamped Motion

Differential Equation

Given the second-order linear homogeneous equation with initial conditions:

\[ 4u'' + 4u' + 37u = 0, \quad u(0)=1, \, u'(0)=0 \]

Dividing by 4 to normalize:

\[ u'' + u' + 9.25u = 0 \]

Characteristic Equation

Solving for the roots \( r \):

\[ r = -\frac{1}{2} \pm 3i \]

This indicates an underdamped system.

General Solution

The solution for displacement \( u(t) \) is:

\[ u(t) = e^{-1/2 t} \left( \cos 3t + \frac{1}{6} \sin 3t \right) \]

Converting the trigonometric part to phase-amplitude form (see last example):

\[ u(t) = e^{-1/2 t} \frac{\sqrt{37}}{6} \left( \cos(3t - \tan^{-1}(1/6)) \right) \]

System Properties

  • Damped Frequency (Quasi-frequency): \( 3 \)
  • Quasi-period: \( \frac{2\pi}{3} \)
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Damped Harmonic Motion Analysis

The resulting displacement function for the system is given by the following expression:

\[ u(t) = \left[ \frac{\sqrt{37}}{6} e^{-1/2 t} \right] \left( \cos 3t - \tan^{-1}\left(\frac{1}{6}\right) \right) \]

Graphical Representation

The graph illustrates the behavior of the displacement \( u \) over time \( t \). It shows a decaying oscillation where the amplitude decreases exponentially due to the damping factor \( e^{-1/2 t} \). The initial displacement starts at a negative value and oscillates with a phase shift defined by \( \tan^{-1}(1/6) \).

A coordinate graph showing a damped oscillation u(t) vs t starting at -\frac{\sqrt{37}}{6} and decaying over time.

The amplitude is bounded by the envelopes \( \pm \frac{\sqrt{37}}{6} e^{-1/2 t} \), as indicated by the dashed lines on the plot. The phase shift \( \delta \) is marked on the time axis.