Math 308 Week in Review 7

Directions: Full worked solutions for Week in Review 7 (Differential Equations). Click Solution under any problem to reveal the worked answer. Use the Contents panel or the Prev / Up / Next buttons to move between sections, and Search to find terms on this page.

Reference

Step & Impulse Function Reference

Step and impulse function Laplace transform pairs
$f(t)$	$\mathcal{L}\{f(t)\}$
$u_c(t)$	$\dfrac{e^{-cs}}{s}$
$u_c(t)\,g(t-c)$	$e^{-cs}\,G(s)$, $G(s)=\mathcal{L}\{g(t)\}$
$u_c(t)\,f(t)$	$e^{-cs}\,\mathcal{L}\{f(t+c)\}$
$\delta(t-c)$	$e^{-cs}$
$f(t)\,\delta(t-c)$	$e^{-cs}\,f(c)$

Section 6.3: Step Functions

1.

This question explores the unit step function and its basic properties.

Convert $f(t) = 1 + u_2(t) - 2\,u_5(t)$ to a piecewise function. Sketch the graph of $f$ for $t \geq 0$ and state its range.
Express the following piecewise function using unit step functions: \[ h(t) = \begin{cases} -1, & 0 \leq t < 3 \\ 2, & 3 \leq t < 7 \\ 0, & t \geq 7 \end{cases} \]
The pulse (or gate) function is defined by $p_{a,b}(t) = u_a(t) - u_b(t)$ for $0 \leq a < b$. Sketch $p_{a,b}(t)$, describe it as a piecewise function, and compute $\mathcal{L}\{p_{a,b}(t)\}$.

SolutionSolution

(a) Evaluate on each interval. For $0 \leq t < 2$: $u_2 = 0,\; u_5 = 0$, so $f = 1$. For $2 \leq t < 5$: $u_2 = 1,\; u_5 = 0$, so $f = 2$. For $t \geq 5$: $u_2 = u_5 = 1$, so $f = 0$.

\[ f(t) = \begin{cases} 1, & 0 \leq t < 2 \\ 2, & 2 \leq t < 5 \\ 0, & t \geq 5 \end{cases} \]

The graph is a staircase: height 1 on $[0,2)$, rising to 2 on $[2,5)$, dropping to 0 at $t = 5$. Range: $\{0,1,2\}$.

Graph of $f(t) = 1 + u_2(t) - 2\,u_5(t)$

(b) At $t=3$ the value jumps from $-1$ to $2$ (change $+3$); at $t=7$ it drops to $0$ (change $-2$). Starting value is $-1$:

\[ h(t) = -1 + 3\,u_3(t) - 2\,u_7(t) \]

Verify: $[0,3)$: $-1$. $[3,7)$: $-1+3=2$. $[7,\infty)$: $-1+3-2=0$. ✓

(c) Since $u_a(t) - u_b(t) = 1$ only when $a \leq t < b$:

\[ p_{a,b}(t) = \begin{cases} 0, & t < a \\ 1, & a \leq t < b \\ 0, & t \geq b \end{cases} \]

The graph is a rectangular pulse of height 1 on $[a,b)$. By linearity of $\mathcal{L}$:

Graph of $p_{a,b}(t) = u_a(t) - u_b(t)$

\[ \boxed{ \mathcal{L}\{p_{a,b}(t)\} = \dfrac{e^{-as} - e^{-bs}}{s} } \]

2.

Find $\mathcal{L}\{f(t)\}$.

(a) $\displaystyle f(t) = u_2(t)\,(t-2)^3$
(b) $\displaystyle f(t) = u_1(t)\,e^{-(t-1)}\cos\!\bigl(2(t-1)\bigr)$
(c) $\displaystyle f(t) = u_3(t)\,\sin\!\bigl(\pi(t-3)\bigr)$
(d) $\displaystyle f(t) = u_3(t)\,(2t - 6)$

SolutionSolution

Each function is in the form $u_c(t)\,g(t-c)$. Apply $\mathcal{L}\{u_c(t)\,g(t-c)\} = e^{-cs}\,G(s)$ where $G(s) = \mathcal{L}\{g(t)\}$.

(a) $c = 2$, $g(t) = t^3$, $G(s) = 6/s^4$:

\[ \mathcal{L}\{u_2(t)(t-2)^3\} = \dfrac{6\,e^{-2s}}{s^4} \]

(b) $c = 1$, $g(t) = e^{-t}\cos(2t)$, $G(s) = \dfrac{s+1}{(s+1)^2+4}$:

\[ \mathcal{L}\bigl\{u_1(t)\,e^{-(t-1)}\cos(2(t-1))\bigr\} = \dfrac{e^{-s}(s+1)}{(s+1)^2+4} \]

(c) $c = 3$, $g(t) = \sin(\pi t)$, $G(s) = \dfrac{\pi}{s^2+\pi^2}$:

\[ \mathcal{L}\bigl\{u_3(t)\sin(\pi(t-3))\bigr\} = \dfrac{\pi\,e^{-3s}}{s^2+\pi^2} \]

(d) Factor first: $2t - 6 = 2(t-3)$. So $c = 3$, $g(t) = 2t$, $G(s) = 2/s^2$:

\[ \mathcal{L}\{u_3(t)(2t-6)\} = \dfrac{2\,e^{-3s}}{s^2} \]

\[\boxed{(a)\;\dfrac{6e^{-2s}}{s^4};\quad (b)\;\dfrac{e^{-s}(s+1)}{(s+1)^2+4};\quad (c)\;\dfrac{\pi e^{-3s}}{s^2+\pi^2};\quad (d)\;\dfrac{2e^{-3s}}{s^2}}\]

3.

For each piecewise function, (i) rewrite it using unit step functions and (ii) compute $\mathcal{L}\{g(t)\}$.

\[ g(t) = \begin{cases} t, & 0 \leq t < 2 \\ 4 - t, & t \geq 2 \end{cases} \]
\[ g(t) = \begin{cases} 0, & 0 \leq t < 1 \\ 2(t-1), & 1 \leq t < 3 \\ 4, & t \geq 3 \end{cases} \]
\[ g(t) = \begin{cases} 1, & 0 \leq t < 2 \\ t, & 2 \leq t < 4 \\ 3, & t \geq 4 \end{cases} \]

SolutionSolution

(a) Check at $t=2$: left $= 2$, right $= 4-2 = 2$. Continuous. The change at $t=2$ is $(4-t)-t = -2(t-2)$:

\[ g(t) = t - 2\,u_2(t)(t-2) \]

Verify: $[0,2)$: $t$. $[2,\infty)$: $t - 2(t-2) = 4-t$. ✓

\[ \mathcal{L}\{g\} = \dfrac{1}{s^2} - \dfrac{2e^{-2s}}{s^2} = \dfrac{1-2e^{-2s}}{s^2} \]

(b) Check: at $t=1$: $0 = 2(0)$. ✓ At $t=3$: $2(2) = 4$. ✓ Continuous. At $t=1$ the ramp $2(t-1)$ switches on. At $t=3$ the change is $4 - 2(t-1) = -2(t-3)$:

\[ g(t) = 2\,u_1(t)(t-1) - 2\,u_3(t)(t-3) \]

\[ \mathcal{L}\{g\} = \dfrac{2e^{-s}}{s^2} - \dfrac{2e^{-3s}}{s^2} = \dfrac{2(e^{-s}-e^{-3s})}{s^2} \]

(c) Check: at $t=2$: left $=1$, right $=2$. Jump $+1$. At $t=4$: left $=4$, right $=3$. Jump $-1$. Discontinuous.

The change at $t=2$ is $t - 1 = (t-2)+1$. The change at $t=4$ is $3 - t = -(t-3)$. So:

\[ g(t) = 1 + (t-1)\,u_2(t) - (t-3)\,u_4(t) \]

Verify: $[0,2)$: $1$. $[2,4)$: $1+(t-1)=t$. $[4,\infty)$: $1+(t-1)-(t-3) = 3$. ✓

Use $\mathcal{L}\{u_c(t)f(t)\} = e^{-cs}\mathcal{L}\{f(t+c)\}$:

\[ \mathcal{L}\{(t-1)\,u_2(t)\} = e^{-2s}\,\mathcal{L}\{t+1\} = e^{-2s}\!\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right) \]

\[ \mathcal{L}\{(t-3)\,u_4(t)\} = e^{-4s}\,\mathcal{L}\{t+1\} = e^{-4s}\!\left(\dfrac{1}{s^2}+\dfrac{1}{s}\right) \]

\[\boxed{\mathcal{L}\{g\} = \dfrac{1}{s} + \dfrac{s+1}{s^2}\!\left(e^{-2s} - e^{-4s}\right)}\]

4.

Consider the continuous piecewise linear function

\[ g(t) = \begin{cases} 2t, & 0 \leq t < 1 \\ 2, & t \geq 1. \end{cases} \]

Use the definition $\displaystyle\mathcal{L}\{g(t)\} = \int_0^{\infty} e^{-st}\,g(t)\,dt$ to compute $\mathcal{L}\{g(t)\}$ by splitting the integral at $t = 1$.
Write $g(t)$ using unit step functions, then apply the Second Translation Theorem to find $\mathcal{L}\{g(t)\}$. Verify that your answer agrees with part (a).

SolutionSolution

(a) Definition method. Split at $t = 1$:

\[\mathcal{L}\{g\} = \int_0^1 2t\,e^{-st}\,dt + \int_1^\infty 2\,e^{-st}\,dt\]

First integral — integrate by parts with $u = 2t$, $dv = e^{-st}\,dt$:

\[\int_0^1 2t\,e^{-st}\,dt = \left[-\dfrac{2t\,e^{-st}}{s}\right]_0^1 + \dfrac{2}{s}\int_0^1 e^{-st}\,dt = -\dfrac{2e^{-s}}{s} + \dfrac{2}{s}\left[-\dfrac{e^{-st}}{s}\right]_0^1 = -\dfrac{2e^{-s}}{s} + \dfrac{2}{s^2}(1-e^{-s})\]

Second integral:

\[\int_1^\infty 2\,e^{-st}\,dt = \left[-\dfrac{2e^{-st}}{s}\right]_1^\infty = \dfrac{2e^{-s}}{s}\]

Adding: the $-2e^{-s}/s$ and $+2e^{-s}/s$ terms cancel exactly:

\[\mathcal{L}\{g\} = \dfrac{2}{s^2}(1-e^{-s}) = \dfrac{2(1-e^{-s})}{s^2}\]

(b) Step function method. Check continuity at $t=1$: $2(1) = 2$. ✓ The change at $t=1$ is $2 - 2t = -2(t-1)$:

\[g(t) = 2t - 2\,u_1(t)(t-1)\]

Applying the Second Translation Theorem to $u_1(t)(t-1)$ (here $c=1$, $g(t)=t$, $G(s)=1/s^2$):

\[\mathcal{L}\{g\} = \dfrac{2}{s^2} - \dfrac{2e^{-s}}{s^2} = \dfrac{2(1-e^{-s})}{s^2}\]

Both methods agree. ✓

\[\boxed{\mathcal{L}\{g(t)\} = \dfrac{2(1-e^{-s})}{s^2}}\]

5.

Find $f(t) = \mathcal{L}^{-1}\{F(s)\}$. Express each answer involving the unit step function and, where possible, also as a piecewise function.

(a) $\displaystyle F(s) = \dfrac{e^{-2s}}{s^2 + 4s + 13}$
(b) $\displaystyle F(s) = \dfrac{e^{-s}(2s + 1)}{s^2 + 2s + 5}$
(c) $\displaystyle F(s) = \dfrac{3\,e^{-4s}}{s(s^2 + 9)}$
(d) $\displaystyle F(s) = \dfrac{e^{-3s}(s + 5)}{(s+1)(s+3)}$

SolutionSolution

Use: if $\mathcal{L}^{-1}\{G(s)\} = g(t)$, then $\mathcal{L}^{-1}\{e^{-cs}G(s)\} = u_c(t)\,g(t-c)$.

(a) Complete the square: $s^2+4s+13 = (s+2)^2+9$. The $G(s)$ factor is $\dfrac{1}{(s+2)^2+9}$:

\[\dfrac{1}{(s+2)^2+9} = \dfrac{1}{3}\cdot\dfrac{3}{(s+2)^2+9} \;\Rightarrow\; g(t) = \tfrac{1}{3}\,e^{-2t}\sin(3t)\]

\[f(t) = \tfrac{1}{3}\,u_2(t)\,e^{-2(t-2)}\sin(3(t-2))\]

(b) Complete the square: $s^2+2s+5=(s+1)^2+4$. Rewrite $2s+1 = 2(s+1)-1$:

\[\dfrac{2(s+1)-1}{(s+1)^2+4} = 2\cdot\dfrac{s+1}{(s+1)^2+4} - \dfrac{1}{2}\cdot\dfrac{2}{(s+1)^2+4}\]

\[g(t) = 2e^{-t}\cos(2t) - \tfrac{1}{2}e^{-t}\sin(2t)\]

Applying the shift $c = 1$ (replace $t$ by $t-1$):

\[f(t) = u_1(t)\,e^{-(t-1)}\!\left[2\cos(2(t-1)) - \tfrac{1}{2}\sin(2(t-1))\right]\]

(c) Partial fractions on $\dfrac{3}{s(s^2+9)} = \dfrac{A}{s}+\dfrac{Bs+C}{s^2+9}$:

$s=0$: $A = 1/3$. Coeff. $s^2$: $0 = A+B \Rightarrow B = -1/3$. Coeff. $s^1$: $C = 0$.

\[g(t) = \mathcal{L}^{-1}\!\left\{\dfrac{1/3}{s} - \dfrac{s/3}{s^2+9}\right\} = \tfrac{1}{3} - \tfrac{1}{3}\cos(3t)\]

\[f(t) = \tfrac{1}{3}\,u_4(t)\!\left[1-\cos(3(t-4))\right]\]

(d) Partial fractions on $\dfrac{s+5}{(s+1)(s+3)} = \dfrac{A}{s+1}+\dfrac{B}{s+3}$:

$s=-1$: $A = 2$. $s=-3$: $B = -1$. So $g(t) = 2e^{-t} - e^{-3t}$.

\[f(t) = u_3(t)\!\left[2e^{-(t-3)} - e^{-3(t-3)}\right]\]

Section 6.4: Discontinuous Forcing Functions

6.

Use the Laplace transform to solve the initial value problem. A constant force turns on at $t = 2$ and turns off at $t = 5$.

\[ y'' + 4y = u_2(t) - u_5(t), \quad y(0) = 0, \quad y'(0) = 1. \]

Take the Laplace transform of both sides and solve for $Y(s)$.
Find $y(t)$.
Write $y(t)$ as an explicit piecewise function on the intervals $[0, 2)$, $[2, 5)$, and $[5, \infty)$.
What do you observe about $y(t)$ for $t \geq 5$ compared with $y(t)$ for $0 \leq t < 2$? Explain this in terms of superposition.

SolutionSolution

(a) Take $\mathcal{L}$ with $y(0)=0$, $y'(0)=1$:

\[(s^2+4)Y - 1 = \dfrac{e^{-2s}-e^{-5s}}{s} \;\Rightarrow\; Y(s) = \dfrac{1}{s^2+4} + \dfrac{e^{-2s}-e^{-5s}}{s(s^2+4)}\]

(b) Let $h(t) = \mathcal{L}^{-1}\!\left\{\dfrac{1}{s(s^2+4)}\right\}$. Partial fractions: $\dfrac{1}{s(s^2+4)} = \dfrac{1/4}{s} - \dfrac{s/4}{s^2+4}$, giving $h(t) = \tfrac{1}{4}(1-\cos 2t)$.

\[y(t) = \tfrac{1}{2}\sin(2t) + u_2(t)\,h(t-2) - u_5(t)\,h(t-5)\]

(c) Expanding $h(t-2) = \tfrac{1}{4}[1-\cos(2(t-2))]$ and $h(t-5) = \tfrac{1}{4}[1-\cos(2(t-5))]$:

\[y(t) = \begin{cases} \dfrac{1}{2}\sin(2t), & 0 \leq t < 2 \\[8pt] \dfrac{1}{2}\sin(2t) + \dfrac{1}{4}\bigl[1-\cos(2(t-2))\bigr], & 2 \leq t < 5 \\[8pt] \dfrac{1}{2}\sin(2t) + \dfrac{1}{4}\bigl[\cos(2(t-5))-\cos(2(t-2))\bigr], & t \geq 5 \end{cases}\]

(d) For $t \geq 5$ the two constant $\tfrac{1}{4}$ terms cancel, so the sustained offset disappears and the system returns to pure sinusoidal free oscillation. By superposition: the combined effect of switching the force on at $t=2$ and off at $t=5$ is the sum of a step-on response and a delayed step-off response; when both are active they cancel the constant offset, leaving only oscillation with accumulated phase shifts.

7.

Solve the initial value problem, where the forcing function is piecewise constant.

\[ y'' + 3y' + 2y = g(t), \quad y(0) = 0, \quad y'(0) = 0, \] \[ \text{where}\quad g(t) = \begin{cases} 4, & 0 \leq t < 2 \\ 0, & t \geq 2. \end{cases} \]

Rewrite $g(t)$ using unit step functions.
Take the Laplace transform and solve for $Y(s)$.
Find $h(t) = \mathcal{L}^{-1}\{H(s)\}$.
Write $y(t)$ as an explicit piecewise function.

SolutionSolution

(a) $g(t) = 4\bigl(1 - u_2(t)\bigr)$.

(b) Take $\mathcal{L}$ with zero ICs:

\[(s^2+3s+2)Y = \dfrac{4(1-e^{-2s})}{s} \;\Rightarrow\; Y(s) = H(s)(1-e^{-2s}), \quad H(s) = \dfrac{4}{s(s+1)(s+2)}\]

(c) Partial fractions on $H(s)$:

\[\dfrac{4}{s(s+1)(s+2)} = \dfrac{A}{s}+\dfrac{B}{s+1}+\dfrac{C}{s+2}\]

$s=0$: $A=2$. $\;s=-1$: $4=B(-1)(1)\Rightarrow B=-4$. $\;s=-2$: $4=C(-2)(-1)\Rightarrow C=2$.

\[h(t) = 2 - 4e^{-t} + 2e^{-2t}\]

Check: $h(0) = 0$ and $h'(0) = 0+4-4 = 0$. ✓

(d) $y(t) = h(t) - u_2(t)\,h(t-2)$, where $h(t-2) = 2 - 4e^{-(t-2)} + 2e^{-2(t-2)}$:

\[\boxed{y(t) = \begin{cases} 2 - 4e^{-t} + 2e^{-2t}, & 0 \leq t < 2 \\[6pt] 2 - 4e^{-t} + 2e^{-2t} - 2 + 4e^{-(t-2)} - 2e^{-2(t-2)}, & t \geq 2 \end{cases}}\]

8.

A 1 kg mass hangs on a spring. When the mass is attached and the system is at rest, the spring stretches 0.5 m. A dashpot provides a resistive force of 4 N when the mass moves at 1 m/s. Starting from rest at equilibrium, a constant external downward force of 10 N is applied for the first 3 seconds and then abruptly removed. Use $g = 10$ m/s$^2$.

Determine the spring constant $k$ and the damping coefficient $\gamma$.
Write the external force $F_{\text{ext}}(t)$ using unit step functions. Then write the full IVP for displacement $u(t)$ (positive downward) in standard form.
Classify the damping (underdamped, critically damped, or overdamped).
Take the Laplace transform and write $U(s)$ explicitly.

SolutionSolution

(a) At static equilibrium $k\,\Delta\ell = mg$:

\[k = \dfrac{(1)(10)}{0.5} = 20\;\text{N/m}, \qquad \gamma = 4\;\text{N\textperiodcentered s/m}\]

(b) $F_{\text{ext}}(t) = 10\bigl(1 - u_3(t)\bigr)$. Dividing by $m=1$:

\[u'' + 4u' + 20u = 10\bigl(1 - u_3(t)\bigr), \quad u(0) = 0,\quad u'(0) = 0\]

(c) Discriminant: $\gamma^2 - 4mk = 16 - 80 = -64 < 0$. Underdamped.

Characteristic roots: $r = -2 \pm 4i$, so $\omega_d = 4$ rad/s.

(d) Take $\mathcal{L}$ with zero ICs and factor the denominator:

\[(s^2+4s+20)U = \dfrac{10(1-e^{-3s})}{s}\]

\[\boxed{U(s) = \dfrac{10(1-e^{-3s})}{s\bigl[(s+2)^2+16\bigr]}}\]

Section 6.5: Impulse Functions

9.

Use the sifting property to compute each Laplace transform.

(a) $\displaystyle f(t) = 5\,\delta(t - 3)$
(b) $\displaystyle f(t) = t^2 e^{t}\,\delta(t - 2)$
(c) $\displaystyle f(t) = \sin(2t)\,\delta(t - 3)$
(d) $\displaystyle f(t) = (3t - 1)\,\delta(t - 1) + \cos(t)\,\delta(t - 3)$

SolutionSolution

Apply $\mathcal{L}\{f(t)\,\delta(t-c)\} = e^{-cs}f(c)$: evaluate the coefficient at $t=c$ and multiply by $e^{-cs}$.

(a) $f(t) = 5$ evaluated at $t=3$: $5$. $\mathcal{L} = 5e^{-3s}$.

(b) $f(t) = t^2 e^t$ evaluated at $t=2$: $4e^2$. $\mathcal{L} = 4e^2\cdot e^{-2s}$.

(c) $f(t) = \sin(2t)$ evaluated at $t=3$: $\sin 6$. $\mathcal{L} = e^{-3s}\sin 6$.

(d) Two terms. $(3t-1)$ at $t=1$: $2$, giving $2e^{-s}$. $\;\cos t$ at $t=3$: $\cos 3$, giving $e^{-3s}\cos 3$.

\[\boxed{(a)\;5e^{-3s};\quad (b)\;4e^{2}\,e^{-2s};\quad (c)\;e^{-3s}\sin 6;\quad (d)\;2e^{-s}+e^{-3s}\cos 3}\]

10.

Use the Laplace transform to solve the initial value problem.

\[ y'' + 4y' + 13y = \delta(t - 2), \quad y(0) = 1, \quad y'(0) = 0. \]

Take the Laplace transform and solve for $Y(s)$.
Find $y(t)$.
Write $y(t)$ as an explicit piecewise function on $[0, 2)$ and $[2, \infty)$.
Sketch a rough graph of $y(t)$. Describe what happens to the amplitude of oscillation just after $t = 2$.

SolutionSolution

(a) Take $\mathcal{L}$ with $y(0)=1$, $y'(0)=0$:

\[(s^2Y-s) + 4(sY-1) + 13Y = e^{-2s}\]

\[Y(s^2+4s+13) = s+4+e^{-2s} \;\Rightarrow\; Y(s) = \dfrac{s+4}{(s+2)^2+9} + \dfrac{e^{-2s}}{(s+2)^2+9}\]

(b) Rewrite $s+4 = (s+2)+2$:

\[\dfrac{(s+2)+2}{(s+2)^2+9} = \dfrac{s+2}{(s+2)^2+9} + \dfrac{2}{3}\cdot\dfrac{3}{(s+2)^2+9}\]

\[y(t) = e^{-2t}\cos(3t) + \tfrac{2}{3}\,e^{-2t}\sin(3t) + \tfrac{1}{3}\,u_2(t)\,e^{-2(t-2)}\sin(3(t-2))\]

(c)

\[y(t) = \begin{cases} e^{-2t}\cos(3t) + \dfrac{2}{3}\,e^{-2t}\sin(3t), & 0 \leq t < 2 \\[10pt] e^{-2t}\cos(3t) + \dfrac{2}{3}\,e^{-2t}\sin(3t) + \dfrac{1}{3}\,e^{-2(t-2)}\sin(3(t-2)), & t \geq 2 \end{cases}\]

(d) For $0 \leq t < 2$: the system undergoes decaying underdamped oscillation driven by the initial condition, with amplitude envelope $e^{-2t}\sqrt{1 + 4/9}$. At $t=2$ the impulse injects a new decaying oscillatory term $\tfrac{1}{3}e^{-2(t-2)}\sin(3(t-2))$. Just after $t=2$ the combined amplitude increases; both terms decay to zero as $t \to \infty$.

Rough sketch of $y(t)$; dashed line marks impulse at $t=2$

11.

This IVP has two different types of forcing: an instantaneous impulse and a step (sustained) input. Solve using the Laplace transform.

\[ y'' + 4y = \delta(t - 1) + u_3(t), \quad y(0) = 0, \quad y'(0) = 0. \]

Take the Laplace transform and solve for $Y(s)$.
Find $y(t)$.
Write $y(t)$ as an explicit piecewise function on the three intervals $[0, 1)$, $[1, 3)$, and $[3, \infty)$.
Interpret each piece physically: what does the system do before the impulse, between the impulse and the step, and after the step turns on?

SolutionSolution

(a) Take $\mathcal{L}$ with zero ICs:

\[(s^2+4)Y = e^{-s} + \dfrac{e^{-3s}}{s} \;\Rightarrow\; Y(s) = \dfrac{e^{-s}}{s^2+4} + \dfrac{e^{-3s}}{s(s^2+4)}\]

(b) Invert each term. $\mathcal{L}^{-1}\!\left\{1/(s^2+4)\right\} = \tfrac{1}{2}\sin(2t)$. Partial fractions: $\dfrac{1}{s(s^2+4)} = \dfrac{1/4}{s} - \dfrac{s/4}{s^2+4}$, so $\mathcal{L}^{-1}\!\left\{1/(s(s^2+4))\right\} = \tfrac{1}{4}(1-\cos 2t)$.

\[y(t) = \tfrac{1}{2}\,u_1(t)\sin(2(t-1)) + \tfrac{1}{4}\,u_3(t)\bigl[1-\cos(2(t-3))\bigr]\]

(c)

\[y(t) = \begin{cases} 0, & 0 \leq t < 1 \\[4pt] \dfrac{1}{2}\sin(2(t-1)), & 1 \leq t < 3 \\[6pt] \dfrac{1}{2}\sin(2(t-1)) + \dfrac{1}{4}\bigl[1-\cos(2(t-3))\bigr], & t \geq 3 \end{cases}\]

(d) Physical interpretation.

$0 \leq t < 1$: The system is completely at rest; no forcing has acted.
$1 \leq t < 3$: The impulse at $t=1$ instantaneously imparts velocity, exciting undamped free oscillation $\tfrac{1}{2}\sin(2(t-1))$ at natural frequency $\omega = 2$ rad/s.
$t \geq 3$: The step input adds the particular solution $\tfrac{1}{4}(1-\cos(2(t-3)))$. The motion becomes a superposition of two sinusoidal components at the same frequency, oscillating about the shifted equilibrium $y = \tfrac{1}{4}$. With no damping, the oscillations persist indefinitely.

12.

Solve the initial value problem.

\[ y'' + 6y' + 9y = \delta(t - 2), \quad y(0) = 2, \quad y'(0) = 1. \]

Take the Laplace transform and solve for $Y(s)$.
Find $y(t)$.
Verify your answer satisfies the initial conditions. Describe the behavior of $y(t)$ as $t \to \infty$.

SolutionSolution

(a) Take $\mathcal{L}$ with $y(0)=2$, $y'(0)=1$:

\[(s^2Y-2s-1)+6(sY-2)+9Y = e^{-2s}\]

\[(s+3)^2 Y = 2s+13+e^{-2s} \;\Rightarrow\; Y(s) = \dfrac{2s+13}{(s+3)^2} + \dfrac{e^{-2s}}{(s+3)^2}\]

(b) Decompose the IC term: $2s+13 = 2(s+3)+7$:

\[\dfrac{2(s+3)+7}{(s+3)^2} = \dfrac{2}{s+3}+\dfrac{7}{(s+3)^2} \;\Rightarrow\; \mathcal{L}^{-1} = (2+7t)e^{-3t}\]

\[\mathcal{L}^{-1}\!\left\{\dfrac{e^{-2s}}{(s+3)^2}\right\} = u_2(t)(t-2)e^{-3(t-2)}\]

\[\boxed{y(t) = (2+7t)e^{-3t} + u_2(t)(t-2)e^{-3(t-2)}}\]

(c) Verify ICs. At $t=0$: $y(0) = (2+0)\cdot 1 = 2$ ✓. Differentiating the IC term: $\dfrac{d}{dt}(2+7t)e^{-3t} = (7-3(2+7t))e^{-3t} = (1-21t)e^{-3t}$, so $y'(0)=1$ ✓.

Long-term behavior. Both $(2+7t)e^{-3t}\to 0$ and $(t-2)e^{-3(t-2)}\to 0$ as $t\to\infty$ since exponential decay dominates polynomial growth. Thus $y(t)\to 0$.

13.

A 1 kg mass is suspended from a spring. When the mass hangs at rest, the spring stretches 0.4 m. A dashpot exerts a resistive force of 6 N when the mass moves at 1 m/s. The mass is set in motion from equilibrium with an initial downward velocity of 2 m/s. At time $t = \pi$ seconds, a hammer strikes the mass sharply in the upward direction, delivering an instantaneous impulse of 3 N·s. Use $g = 10$ m/s$^2$ and take displacement positive downward.

Find the spring constant $k$ and the damping coefficient $\gamma$. Compute the discriminant $\gamma^2 - 4mk$ and classify the damping. Find the damped natural frequency $\omega_d$.
Write the IVP governing $u(t)$.
Take the Laplace transform of the IVP and solve for $U(s)$.
Find $u(t)$ and write the solution as a piecewise function.
Describe the motion: what happens for $0 \leq t < \pi$? What changes at $t = \pi$ when the hammer strikes? What is the long-term behavior?

SolutionSolution

(a) $k = mg/\Delta\ell = (1)(10)/0.4 = 25$ N/m. $\;\gamma = 6$ N·s/m.

Discriminant: $36 - 4(1)(25) = 36 - 100 = -64 < 0$. Underdamped.

Characteristic roots: $r = (-6\pm\sqrt{-64})/2 = -3\pm 4i$, so $\omega_d = 4$ rad/s.

(b) The hammer is upward (negative in downward-positive convention), magnitude 3 N·s:

\[u'' + 6u' + 25u = -3\,\delta(t-\pi), \quad u(0) = 0,\quad u'(0) = 2\]

(c) Take $\mathcal{L}$ with $u(0)=0$, $u'(0)=2$:

\[(s^2U-2)+6sU+25U = -3e^{-\pi s} \;\Rightarrow\; U\bigl[(s+3)^2+16\bigr] = 2-3e^{-\pi s}\]

\[\boxed{U(s) = \dfrac{2}{(s+3)^2+16} - \dfrac{3\,e^{-\pi s}}{(s+3)^2+16}}\]

(d) Invert each term:

\[\mathcal{L}^{-1}\!\left\{\dfrac{2}{(s+3)^2+16}\right\} = \tfrac{1}{2}\,e^{-3t}\sin(4t)\]

\[\mathcal{L}^{-1}\!\left\{\dfrac{3\,e^{-\pi s}}{(s+3)^2+16}\right\} = \tfrac{3}{4}\,u_\pi(t)\,e^{-3(t-\pi)}\sin(4(t-\pi))\]

Use $\sin(4(t-\pi)) = \sin(4t-4\pi) = \sin(4t)$ since $4\pi$ is an integer multiple of $2\pi$:

\[u(t) = \tfrac{1}{2}\,e^{-3t}\sin(4t) - \tfrac{3}{4}\,u_\pi(t)\,e^{-3(t-\pi)}\sin(4t)\]

\[\boxed{u(t) = \begin{cases} \dfrac{1}{2}\,e^{-3t}\sin(4t), & 0 \leq t < \pi \\[10pt] \sin(4t)\!\left[\dfrac{1}{2}\,e^{-3t} - \dfrac{3}{4}\,e^{-3(t-\pi)}\right], & t \geq \pi \end{cases}}\]

(e) Description of motion.

$0 \leq t < \pi$: Underdamped oscillation driven by the initial downward velocity of 2 m/s, with frequency $\omega_d = 4$ rad/s and amplitude decaying at rate $e^{-3t}$.
At $t = \pi$: By $t=\pi$ the original oscillation has decayed to a negligibly small amplitude (the factor $e^{-3\pi}$ is extremely small). The upward impulse adds a new term with initial amplitude $\tfrac{3}{4}$, which is much larger. The motion is effectively re-excited at a larger amplitude in the upward direction, causing the displacement to reverse sign just after the strike.
Long-term: Both exponentials $e^{-3t}$ and $e^{-3(t-\pi)}$ decay to zero, so $u(t)\to 0$ as $t\to\infty$.

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