Math 308 Week in Review 9 — Solutions

Step 1: Find the spring constant. At equilibrium the spring force balances gravity: $kL = mg$.

Step 2: Set up and solve the IVP. With $m = 1\,\text{kg}$, $\gamma = 0$ (no damping), $k = 100\,\text{N/m}$:

(Initial displacement is $+3\,\text{cm} = 0.03\,\text{m}$ downward; initial velocity is $-40\,\text{cm/s} = -0.40\,\text{m/s}$ upward.)

The angular frequency is $\omega_0 = \sqrt{k/m} = \sqrt{100} = 10\,\text{rad/s}$. The general solution is:

Apply initial conditions: $y(0) = A = 0.03$ and $y'(0) = 10B = -0.40$, so $B = -0.04$.

Step 3: Compute amplitude and period.

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Step 1: Extract parameters. The weight $W = mg = 20\,\text{N}$. Using $g = 10\,\text{m/s}^2$ gives $m = 2\,\text{kg}$.

The damping coefficient: $\gamma = |F_{\text{damp}}|/\text{speed} = 6/2 = 3\,\text{N}\cdot\text{s/m}$.

The external force is $F(t) = 4\,u_3(t)\,\text{N}$ (a constant $4\,\text{N}$ downward force that switches on at $t = 3$).

Step 2: Write the IVP.

Step 3: Classify damping. Compute the discriminant:

Since $\Delta < 0$, the system is underdamped.

Step 4: Find the critical damping value. Set $\Delta = 0$:

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The natural frequency is $\omega_0 = \sqrt{k/m} = \sqrt{4} = 2\,\text{rad/s}$.

Case 1: $\omega \neq 2$. Try $y_p = A\cos(\omega t)$. Then $y_p'' = -A\omega^2\cos(\omega t)$, so:

General solution: $y = c_1\cos(2t) + c_2\sin(2t) + \dfrac{3}{4-\omega^2}\cos(\omega t)$.

Apply $y(0) = 0$: $\;c_1 + \dfrac{3}{4-\omega^2} = 0 \;\Rightarrow\; c_1 = -\dfrac{3}{4-\omega^2}$.

Apply $y'(0) = 0$: $\;2c_2 = 0 \;\Rightarrow\; c_2 = 0$.

Long-term behavior ($\omega \neq 2$): The solution is a superposition of two frequencies ($\omega$ and $\omega_0 = 2$). Since there is no damping, the motion remains bounded forever — a perpetual oscillation. When $\omega$ is close to (but not equal to) $2$, the solution exhibits a beat pattern: slow amplitude modulation at the beat frequency $|\omega - 2|/2$ with carrier frequency $(\omega + 2)/2$.

Case 2: $\omega = 2$ (resonance). Since the forcing frequency matches $\omega_0 = 2$, multiply the trial solution by $t$. Try $y_p = At\cos(2t) + Bt\sin(2t)$.

Substituting into $y_p'' + 4y_p$: the $t$-terms cancel (as expected) and we get:

Matching coefficients: $4B = 3 \Rightarrow B = \tfrac{3}{4}$ and $-4A = 0 \Rightarrow A = 0$.

So $y_p = \tfrac{3}{4}\,t\sin(2t)$. General solution: $y = c_1\cos(2t) + c_2\sin(2t) + \tfrac{3}{4}\,t\sin(2t)$.

Apply $y(0) = 0$: $c_1 = 0$. Apply $y'(0) = 0$: $2c_2 = 0 \Rightarrow c_2 = 0$.

Long-term behavior ($\omega = 2$): The amplitude envelope is $\tfrac{3}{4}t$, which grows linearly without bound. The solution oscillates with ever-increasing amplitude; the system experiences resonance.

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Method 1: Repeated Integration by Parts. Set $u = t^2$, $dv = e^{-st}\,dt$, so $du = 2t\,dt$, $v = -\dfrac{e^{-st}}{s}$:

The boundary term vanishes for $s > 0$. Apply IBP again on $\displaystyle\int_0^\infty t\,e^{-st}\,dt$ with $u = t$, $dv = e^{-st}\,dt$:

Combining: $\mathcal{L}\{t^2\} = 0 + \dfrac{2}{s}\cdot\dfrac{1}{s^2} = \dfrac{2}{s^3}$. $\checkmark$

Method 2: Tabular Integration. Differentiate $t^2$ repeatedly down the left column and integrate $e^{-st}$ repeatedly down the right column. Read off diagonal products with alternating signs.

Multiply each D-entry by the I-entry one row below, attach the sign from its row, and sum:

As $t\to\infty$ every term vanishes ($s>0$). At $t = 0$ only the last term survives: $-\dfrac{2}{s^3}$. Therefore:

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Part 1 — piecewise to step function. Multiply each formula by the window of step functions over which it is active: $u_0 - u_2$ for $[0,2)$, $\;u_2 - u_5$ for $[2,5)$, and $u_5$ for $[5,\infty)$. Since $u_0(t)=1$ on $[0,\infty)$, write:

Expand and collect by step function:

Check: For $0 \leq t < 2$: $f = t^2$. For $2 \leq t < 5$: $f = t^2 + (4-t^2) = 4$. For $t \geq 5$: $f = t^2 + (4-t^2) + (\sin t - 4) = \sin t$. ✓

Part 2 — step function to piecewise. $u_2(t)$ activates at $t = 2$; $u_5(t)$ activates at $t = 5$.

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Step 1: Take $\mathcal{L}$ of both sides. With $Y = \mathcal{L}\{y\}$:

Step 2: Solve for $Y(s)$ and decompose. Write the numerator as $3s + 13 = 3(s+2) + 7$:

Step 3: Invert. Using $\mathcal{L}^{-1}\!\left\{\dfrac{1}{s+a}\right\} = e^{-at}$ and $\mathcal{L}^{-1}\!\left\{\dfrac{1}{(s+a)^2}\right\} = te^{-at}$:

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Step 1: Take $\mathcal{L}$. Since $y(0) = y'(0) = 0$ and $\mathcal{L}\{u_\pi(t)\} = e^{-\pi s}/s$:

Step 2: Partial fractions on $\dfrac{1}{s(s^2+9)}$.

Multiply through: $1 = A(s^2+9) + (Bs+C)s$. Setting $s=0$: $A = \tfrac{1}{9}$. Comparing $s^2$: $0 = A + B \Rightarrow B = -\tfrac{1}{9}$. Comparing $s^1$: $C = 0$.

Step 3: Invert. Let $f(t) = \mathcal{L}^{-1}\!\left\{\dfrac{1}{9s} - \dfrac{s}{9(s^2+9)}\right\} = \dfrac{1}{9}\bigl(1 - \cos(3t)\bigr)$. Then by the shift theorem:

Simplification: $\cos(3(t-\pi)) = \cos(3t - 3\pi) = \cos(3t)\cos(3\pi) = -\cos(3t)$, so $1 - \cos(3(t-\pi)) = 1 + \cos(3t)$.

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Direct convolution. With $f(\tau) = e^{2\tau}$ and $g(t-\tau) = e^{-(t-\tau)}$:

Confirmation via partial fractions.

At $s=2$: $A = \tfrac{1}{3}$. At $s=-1$: $B = \tfrac{1}{(-3)} = -\tfrac{1}{3}$.

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Step 1: Take $\mathcal{L}$. Note $y(0) = 0$ and $y'(0) = 1$:

Step 2: Invert term by term.

$\mathcal{L}^{-1}\!\left\{\dfrac{1}{(s+3)^2}\right\} = t\,e^{-3t}$.

$\mathcal{L}^{-1}\!\left\{\dfrac{e^{-2s}}{(s+3)^2}\right\} = u_2(t)\,(t-2)\,e^{-3(t-2)}$   (shift theorem with $c = 2$).

Behavior: For $0 \leq t < 2$ the system is driven only by the initial velocity, decaying as $te^{-3t}$. At $t = 2$ the impulse instantaneously imparts additional energy, exciting a second decaying term $(t-2)e^{-3(t-2)}$ that activates at the moment of the impulse.

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Method 1: Partial Fractions + Shift Theorem.

Step 1: Partial fractions on $F(s) = \dfrac{2}{(s+2)(s+5)}$.

At $s = -2$: $A = 2/3$. At $s = -5$: $B = 2/(-3) = -2/3$.

Step 2: Invert $F(s)$.

Step 3: Apply the shift theorem. Since $H(s) = e^{-3s}\,F(s)$:

Method 2: Convolution + Shift Theorem.

Factor $H(s)$ as a product of three transforms:

Step 1: Use the convolution theorem on $F(s)$. Write $F(s) = 2 \cdot \dfrac{1}{s+2} \cdot \dfrac{1}{s+5}$, so by the convolution theorem:

Step 2: Evaluate the convolution integral. Factor out $e^{-5t}$:

This matches the partial-fraction inversion exactly. $\checkmark$

Step 3: Apply the shift theorem. Since $H(s) = e^{-3s}\,F(s)$:

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Series 1: $a_n = \dfrac{(-1)^n(x+2)^n}{n\cdot 3^n}$.

Converges when $\dfrac{|x+2|}{3} < 1$, i.e., $|x+2| < 3$. So $R = 3$ and the interval of absolute convergence is $(-5,\,1)$.

Endpoint check (requested for Series 1 only):

• $x = 1$ (so $x+2 = 3$): $\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^n\cdot 3^n}{n\cdot 3^n} = \sum_{n=1}^\infty \dfrac{(-1)^n}{n}$ — the alternating harmonic series, which converges (conditionally).
• $x = -5$ (so $x+2 = -3$): $\displaystyle\sum_{n=1}^\infty \dfrac{(-1)^n(-3)^n}{n\cdot 3^n} = \sum_{n=1}^\infty \dfrac{1}{n}$ — the harmonic series, which diverges.

Series 2: $a_n = \dfrac{n^2+1}{4^n}(x-1)^n$.

Converges when $|x-1| < 4$, so $R = 4$ and the interval of absolute convergence is $(-3,\,5)$.

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Standard form: $y'' + P(x)y' + Q(x)y = 0$ with $P(x) = -x$ and $Q(x) = 5$. Both are polynomials, so $x = 0$ is an ordinary point. Substitute $y = \displaystyle\sum_{n=0}^\infty a_n x^n$:

Substituting into $y'' - xy' + 5y = 0$:

By the identity theorem, each coefficient must vanish:

Note: At $n = 5$ the numerator is zero, so $a_7 = 0$, and consequently $a_9 = a_{11} = \cdots = 0$. The solution $y_2$ (determined by $a_1$) is therefore a polynomial of degree 5 (terminating series).

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Ordinary point check: $P(x) = 2x$ and $Q(x) = -6$ are both polynomials, so $x = 0$ is an ordinary point. Since there are no singular points, the series converge for all $x$ ($R = \infty$).

Substitute $y = \displaystyle\sum_{n=0}^\infty a_n x^n$:

Recurrence relation:

Compute coefficients (even and odd chains are independent):

Even-indexed (from $a_0$):

Odd-indexed (from $a_1$):

Since $a_5 = 0$, all subsequent odd-indexed terms vanish: $a_5 = a_7 = \cdots = 0$. So $y_2$ terminates after $x^3$.

Basis solutions (setting $(a_0,a_1) = (1,0)$ and $(0,1)$):

Linear independence: $W(y_1,y_2)(0) = y_1(0)\,y_2'(0) - y_1'(0)\,y_2(0) = (1)(1)-(0)(0) = 1 \neq 0$. ✓

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Initial values from the initial conditions. Since $y(x) = \displaystyle\sum_{n=0}^\infty a_n(x-1)^n$:

Apply the recurrence.

The nonzero coefficients are $a_0 = 1$, $a_2 = -3$, $a_3 = -3$, $a_4 = 1$  (note $a_1 = 0$).

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