Chapter 8 Challenge Problems Solutions

SOLUTION The rate of change of \(Q\) with respect to \(t\) is \(\textderiv{Q}{t}.\) We are given that this rate is directly proportional to \(Q^3\) and inversely proportional to \(t^2.\) So letting \(k\) be a constant, we write \[\deriv{Q}{t} = \frac{k Q^3}{t^2} \pd\] To solve for \(k\) we substitute the given \(Q = 4,\) \(t = 2,\) and \(\textderiv{Q}{t} = 8 \col\) \[8 = \frac{k (4)^3}{(2)^2} \implies k = \frac{1}{2} \pd\] So the differential equation becomes \[\deriv{Q}{t} = \frac{Q^3}{2t^2} \pd\] Performing Separation of Variables, we see \[ \ba \int \frac{\dd Q}{Q^3} &= \int \frac{\dd t}{2t^2} \nl -\frac{1}{2 Q^2} &= -\frac{1}{2t} + C \pd \ea \] Substituting the initial condition \(Q(2) = 4\) gives \[-\frac{1}{2(4)^2} = -\frac{1}{2(2)} + C \implies C = \frac{7}{32} \pd\] We therefore find \[ \ba -\frac{1}{2 Q^2} &= -\frac{1}{2t} + \frac{7}{32} \nl \frac{1}{Q^2} &= \frac{1}{t} - \frac{7}{16} = \frac{16 - 7t}{16t} \nl Q^2 &= \frac{16t}{16 - 7t} \pd \ea \] We are given the initial condition \(Q(2) = 4,\) meaning \(Q\) must take on positive values. So let's choose the positive square root: \[Q = \sqrt{\frac{16t}{16 - 7t}} = \boxed{4 \sqrt{\frac{t}{16 - 7t}}}\]

SOLUTION

1. Taking \(a = x\) and \(b = h\) shows \[f(x + h) = f(x) f(h) \pd\] By the limit definition of a derivative (Section 2.1), \[ \ba f'(x) &= \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \nl &= \lim_{h \to 0} \frac{f(x) f(h) - f(x)}{h} \nl &= f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \pd \ea \] Since \(f(0) = 1,\) note that \[\lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = f'(0) \pd\] (Alternatively, you could use L'Hôpital's Rule to resolve the limit.) But \(f'(0) = k,\) so we have \[f'(x) = k f(x) \cma\] as requested.

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2. The differential equation \(f'(x) = k f(x)\) can be rewritten as \[ \deriv{y}{x} = k y \cma \] where \(y = f(x).\) Using Separation of Variables, we see \[ \ba \int \frac{\dd y}{y} &= \int k \di x \nl \ln \abs y &= kx + C_1 \nl \abs y &= e^{C_1} e^{kx} \nl y &= C e^{kx} \cma \ea \] where \(C = \pm e^{C_1}.\) Substituting \(y(0) = 1\) shows \[1 = C e^{0} \implies C = 1 \pd\] Thus, the identity of \(f\) is \[f(x) = \boxed{e^{kx}}\]

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3. Exponential functions in the form \(f(x) = e^{kx},\) \(k \ne 0,\) satisfy the Multiplicative Cauchy functional equation.

SOLUTION The slope field is horizontal when \(\textderiv{y}{t} = 0 \col\) \[\frac{\par{4y^3 + 8y^2 - y - 2} e^{y + 8} \sqrt{y + 5} \, \sin t}{t^4 + 2} = 0 \pd\] We seek the values of \(y\) that fix the derivative to \(0;\) the only way these arise is when the \(y\)-dependent factors equal \(0 \col\) \[\par{4y^3 + 8y^2 - y - 2} e^{y + 8} \sqrt{y + 5} = 0 \pd\] Because \(e^{y + 8} \ne 0\) for all \(y,\) dividing it on both sides gives \[ \par{4y^3 + 8y^2 - y - 2} \sqrt{y + 5} = 0 \pd \] This gives two possibilities: \[4y^3 + 8y^2 - y - 2 = 0 \or \sqrt{y + 5} = 0 \pd\] In the second case, the solution \(y = -5\) is trivial. In the first case, factoring the left side (using factoring by grouping) gives \[ \ba 4y^2 (y + 2) - (y + 2) &= 0 \nl (4y^2 - 1) (y + 2) &= 0 \nl \implies y &= -2 \cma -\tfrac{1}{2} \cma \tfrac{1}{2} \pd \ea \] Thus, we have equilibrium solutions of \[y = \boxed{-5} \cma \boxed{-2} \cma \boxed{-\tfrac{1}{2}} \cma \boxed{\tfrac{1}{2}}\]

SOLUTION

1. We have the initial condition \(I(0) = 0.\) Let's rewrite the given differential equation as \[\deriv{I}{t} = 4 - 2I \pd\] The first step of Euler's Method gives \[I_1 = 0 + 0.25 \parbr{4 - 2(0)} = 1 \un A \pd\] Likewise, subsequent approximations are as follows: \[ \ba I_2 &= 1 + 0.25 \parbr{4 - 2(1)} = 1.5 \un A \cma \nl I_3 &= 1.5 + 0.25 \parbr{4 - 2(1.5)} = 1.75 \un A \cma \nl I(1) \approx I_4 &= 1.75 + 0.25 \parbr{4 - 2(1.75)} = \boxed{1.875 \un A} \ea \]

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2. We see \(\textderiv{I}{t} \to 0\) as \(I \to 2,\) meaning \(I\) cannot increase past \(2 \un A.\) Thus, the current after a very long time is \(\boxed{2 \un A}.\)

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3. Differentiating \(I,\) we attain \[I'(t) = 4e^{-2t} \pd\] We then see \[20 - \underbrace{10 \parbr{2\par{1 - e^{-2t}}}}_{10 I} - \underbrace{5 \par{4e^{-2t}}}_{5 \, \textderiv{I}{t}} \equalsCheck 0 \pd\] In addition, the initial condition \(I(0) = 0\) is met: \[I(0) = 2 \par{1 - e^0} \equalsCheck 0 \pd\] Hence, \(I(t) = 2 \par{1 - e^{-2t}}\) is a particular solution because it satisfies the differential equation with \(I(0) = 0.\) The solution makes physical sense because the inductor gradually allows more current to flow in the circuit. Initially, the inductor opposes most of the current from flowing, but it eventually allows the current to grow to its maximum of \(2 \un A.\) With this behavior, inductors are used as shut-off devices—for example, in surge protectors—to prevent electronic damage by a current overload.

SOLUTION We divide the given differential equation by \(m\) to get \begin{equation} \deriv{v}{t} + \frac{b}{m} v = g \pd \label{eq:drag-linear} \end{equation} We then choose an integrating factor to be \[\mu(t) = e^{\int (b/m) \di t} = e^{bt/m} \pd\] Multiplying \(\eqRefer{eq:drag-linear}\) by \(\mu(t) = e^{bt/m}\) yields \[e^{bt/m} \deriv{v}{t} + \frac{b}{m} e^{bt/m} v = g e^{bt/m} \pd\] The left side is the Product Rule expansion of \(\par{ve^{bt/m}}',\) so we rewrite the equation as \begin{equation*} \par{ve^{bt/m}}' = g e^{bt/m} \pd \end{equation*} Integrating both sides yields \begin{equation*} ve^{bt/m} = \frac{mg}{b} e^{bt/m} + C \cma \end{equation*} where \(C\) is a constant. Substituting \(v(0) = 0\) shows \[0= \frac{mg}{b} + C \implies C = - \frac{mg}{b} \pd\] The solution is therefore \[ \ba ve^{bt/m} &= \frac{mg}{b} e^{bt/m} - \frac{mg}{b} \nl \implies v(t) &= \frac{mg}{b} \par{1 - e^{-bt/m}} \cma \ea \] as requested. Additionally, the terminal velocity is \[ \ba v_T &= \lim_{t \to \infty} v(t) \nl &= \lim_{t \to \infty} \frac{mg}{b} \par{1 - e^{-bt/m}} \nl &= \frac{mg}{b} \pd \ea \]

SOLUTION Note the initial conditions \(s(0) = 0\) (the nest's position) and \(v(0) = 1/2.\)

Velocity Function Recall that acceleration is the time derivative of velocity—that is, \(a = \textderiv{v}{t}.\) Given \(a^2 = 1 - 4v^2,\) we have the following differential equation: \[ \deriv{v}{t} = \pm \sqrt{1 - 4v^2} \pd \] Because \(v(0) = 1/2\) ensures the right side equals \(0,\) \(\textderiv{v}{t} = \pm \sqrt{1 - 4v^2}\) agree at \(t = 0.\) It turns out that choosing either branch yields the same volution, as we will see later. Performing Separation of Variables gives \[\int \frac{1}{\sqrt{1 - 4v^2}} \di v = \int \dd t \pd\] To evaluate the integral on the left, we substitute \(u = 2v.\) Then \(\dd u = 2 \di v\) and so \[ \ba \int \frac{1}{\sqrt{1 - 4v^2}} \di v &= \int \frac{1/2}{\sqrt{1 - u^2}} \di u \nl &= \tfrac{1}{2} \asin u \nl &= \tfrac{1}{2} \asin 2v \cma \ea \] where we omitted the constant of integration. We therefore have \[\tfrac{1}{2} \asin 2v = t + C \pd\] Substituting the initial condition \(v(0) = 1/2,\) we find \[\tfrac{1}{2} \asin 1 = 0 + C \implies C = \frac{\pi}{4} \pd\] Hence, we have \[ \ba \tfrac{1}{2} \asin 2v &= t + \frac{\pi}{4} \nl \asin 2v &= 2t + \frac{\pi}{2} \nl 2v &= \sin \par{2t + \frac{\pi}{2}} \nl v(t) &= \boxed{\tfrac{1}{2} \sin \par{2t + \frac{\pi}{2}}} \ea \] Note that \(v(t)\) is even. If the negative branch of the square root were selected in the differential equation, then the solution would be \[v(t) = \tfrac{1}{2} \sin \par{-2t + \frac{\pi}{2}} \cma\] which is equivalent to our answer.

Acceleration Function The acceleration function is given by \(a(t) = \textderiv{v}{t},\) namely, \[ \ba a(t) &= \deriv{}{t} \parbr{\tfrac{1}{2} \sin \par{2t + \frac{\pi}{2}}} \nl &= \boxed{\cos \par{2t + \frac{\pi}{2}}} \ea \]

Position Function The position function is given by \(s(t) = \int v(t) \di t,\) that is, \[s(t) = \int \tfrac{1}{2} \sin \par{2t + \frac{\pi}{2}} \di t \pd\] Substituting \(u = 2t + \pi/2,\) we have \(\dd u = 2 \di t\) and therefore \[ \ba \int \tfrac{1}{2} \sin \par{2t + \frac{\pi}{2}} \di t &= \int \tfrac{1}{4} \sin u \di u \nl &= - \tfrac{1}{4} \cos u + D \nl &= -\tfrac{1}{4} \cos \par{2t + \frac{\pi}{2}} + D \pd \ea \] Accordingly, the position function is \[s(t) = -\tfrac{1}{4} \cos \par{2t + \frac{\pi}{2}} + D \pd\] Since \(s(0) = 0,\) we find \[0 = -\tfrac{1}{4} \cos \par{0 + \frac{\pi}{2}} + D \implies D = 0 \pd\] Thus, the position function is \[s(t) = \boxed{-\tfrac{1}{4} \cos \par{2t + \frac{\pi}{2}}}\]

SOLUTION In the first iteration of Euler's Method, the student finds \(y(k + h) \approx Q,\) so \[ y(k + h) \approx Q + h \deriv{y}{x} \intEval_{x = k \cma \, y = Q} = Q \pd \] We then see \[ \ba Q + \frac{k^3 - 4k}{Q^2 + 1} &= Q \nl \frac{k^3 - 4k}{Q^2 + 1} &= 0 \nl k^3 - 4k &= 0 \nl \implies k &= -2 \cma 0 \cma 2 \pd \ea \] Because \(k \geq 0,\) we reject \(k = -2.\) The solutions \(k = 0\) and \(k = 2\) are left open. On the second iteration of Euler's Method, the student again finds \(y(k + 2h) \approx Q,\) or \[ y(k + 2h) \approx Q + h \deriv{y}{x} \intEval_{x = k + h \cma \, y = Q} = Q \pd \] This gives \[ \ba Q + \frac{(k + h)^3 - 4(k + h)}{Q^2 + 1} &= Q \nl (k + h)^3 - 4(k + h) &= 0 \nl (k + h) [(k + h)^2 - 4] &= 0 \cma \ea \] which is satisfied by \(h = -k,\) \(k + h = 2,\) or \(k + h = -2.\) But we need \(h \gt 0\) and \(k \geq 0,\) so we immediately reject \(h = -k.\) For \(k = 0,\) we have \[0 + h = 2 \implies h = 2 \or 0 + h = -2 \implies h = -2 \pd\] For \(k = 2,\) we have \[2 + h = 2 \implies h = 0 \or 2 + h = -2 \implies h = -4 \pd \] To ensure \(k \geq 0\) and \(h \gt 0,\) we must have \(\boxed{k = 0}\) and \(\boxed{h = 2}.\)

SOLUTION Differentiating the given solution gives \[ \ba y' &= K_1 r_1 e^{r_1 x} + K_2 r_2 e^{r_2 x} \cma \nl y'' &= K_1 r_1^2 e^{r_1 x} + K_2 r_2^2 e^{r_2 x}\pd \ea \] Substituting these expressions into the differential equation gives \[ a \par{K_1 r_1^2 e^{r_1 x} + K_2 r_2^2 e^{r_2 x}} + b \par{K_1 r_1 e^{r_1 x} + K_2 r_2 e^{r_2 x}} + c \par{K_1 e^{r_1x} + K_2 e^{r_2 x}} = 0 \pd \] Grouping terms with \(K_1 e^{r_1 x}\) and \(K_2 e^{r_2 x},\) we have \[ \ba K_1 e^{r_1 x} \par{a r_1^2 + br_1 + c} + K_2 e^{r_2 x} \par{a r_2^2 + br_2 + c} &= 0 \nl K_1 e^{r_1 x} \par{a r_1^2 + br_1 + c} &= -K_2 e^{r_2 x} \par{a r_2^2 + br_2 + c} \pd \ea \] But because \(K_1\) and \(K_2\) are arbitrary coefficients, the only way the equation can hold true is if each expression equals \(0\)—that is, \[ \ba a r_1^2 + br_1 + c &= 0 \cma \nl a r_2^2 + br_2 + c &= 0 \pd \ea \] Thus, \(r_1\) and \(r_2\) are the zeros to the quadratic equation \[a r^2 + br + c = 0 \cma\] so \[r_{1, 2} = \boxed{\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}\]

SOLUTION Combining the fractions on the right gives \[ \ba \sqrt{\deriv{y}{x}} &= \frac{4y}{e^{2x^2}} + \frac{2}{e^{2x^2}} \nl &= \frac{4y + 2}{e^{2x^2}} \nl &= (4y + 2) e^{-2x^2} \pd \ea \] Then squaring both sides and using Separation of Variables, we have \[ \ba \deriv{y}{x} &= (4y + 2)^2 e^{-4x^2} \nl \int \frac{\dd y}{(4y + 2)^2} &= \int e^{-4x^2} \di x \nl -\frac{1}{4 (4y + 2)} &= \int e^{-4x^2} \di x \pd \ea \] On the right, substituting \(u = 2x\) gives \(\dd u = 2 \di x.\) We then see \[ \ba \int e^{-4x^2} \di x &= \tfrac{1}{2} \int e^{-u^2} \di u \nl &= \frac{\sqrt \pi}{4} \erf u + C_1 \nl &= \frac{\sqrt \pi}{4} \erf (2x) + C_1 \pd \ea \] Thus, our solution to the differential equation is \[ -\frac{1}{4 (4y + 2)} = \frac{\sqrt \pi}{4} \erf (2x) + C_1 \pd \] Then we solve for \(y,\) as follows: \[ \ba \frac{1}{4y + 2} &= - \sqrt{\pi} \erf(2x) - 4 C_1 \nl 4y + 2 &= - \frac{1}{\sqrt{\pi} \erf(2x) - 4 C_1} \nl y &= -\frac{1}{2} - \frac{1}{4 \sqrt{\pi} \erf(2x) - 16 C_1} \ea \] By defining a new constant \(C = -16 C_1,\) we have \[\boxed{y = -\frac{1}{2} - \frac{1}{4 \sqrt{\pi} \erf(2x) + C}}\] Because \(\sqrt{\textderiv{y}{x}} \geq 0,\) we require \[(4y + 2) e^{-2x^2} \geq 0 \implies 4y + 2 \geq 0 \implies y \geq -\tfrac{1}{2} \pd\]

SOLUTION

1. We aim to show that \[\lim_{t \to \infty} x_T(t) = \lim_{t \to \infty} C e^{-\lambda t} \cos \par{\alpha t + \beta} = 0 \pd\] Because cosine's range is \([-1, 1],\) we have \(-1 \leq \cos \par{\alpha t + \beta} \leq 1.\) In addition, \(-\abs C \leq C \leq \abs C,\) so \[-\abs C e^{-\lambda t} \leq C e^{-\lambda t} \cos \par{\alpha t + \beta} \leq \abs C e^{-\lambda t} \pd\] Since \(\lambda \gt 0,\) we see \[\lim_{t \to \infty} \par{-\abs C e^{-\lambda t}} = \lim_{t \to \infty} \par{\abs C e^{-\lambda t}} = 0 \pd\] Hence, by the Squeeze Theorem, \(\lim_{t \to \infty} x_T(t) = 0.\) Accordingly, the transient solution dies out over time, and the system's behavior is governed by \(x_S(t)\) as \(t \to \infty.\)

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2. Differentiating \(x_S(t) = A \cos \omega t + B \sin \omega t\) shows \[ \ba \deriv{x_S}{t} &= -A \omega \sin \omega t + B \omega \cos \omega t \cma \nl \derivOrder{x_S}{t}{2} &= -A \omega^2 \cos \omega t - B \omega^2 \sin \omega t \pd \ea \] Substituting these expressions into the differential equation yields \[ \small \ba m \par{-A \omega^2 \cos \omega t - B \omega^2 \sin \omega t} + c \par{-A \omega \sin \omega t + B \omega \cos \omega t} + k \par{A \cos \omega t + B \sin \omega t} &= F_0 \cos \omega t \nl \par{-mA \omega^2 + Bc \, \omega + kA} \cos \omega t + \par{-m B \omega^2 - A c \, \omega + kB} \sin \omega t &= F_0 \cos \omega t \pd \ea \] On both sides, the coefficients of the cosines must match, so \begin{equation} -mA \omega^2 + Bc \, \omega + kA = F_0 \pd \label{eq:response-cos} \end{equation} In addition, the coefficients of the sines must match, meaning \begin{equation} -m B \omega^2 - A c \, \omega + kB = 0 \pd \label{eq:response-sin} \end{equation} From \(\eqref{eq:response-sin},\) solving for \(B\) gives \[B = A \, \frac{c \, \omega}{k - m \omega^2} \pd\] Substituting this result into \(\eqref{eq:response-cos}\) shows \[ \ba A \par{k - m \omega^2} + \par{A \, \frac{c \, \omega}{k - m \omega^2}} (c \, \omega) &= F_0 \nl \implies A &= \frac{F_0}{k - m \omega^2 + \frac{(c \, \omega)^2}{k - m \omega^2}} \nl &= \boxed{\frac{F_0 \par{k - m \omega^2}}{\par{k - m \omega^2}^2 + \par{c \, \omega}^2}} \ea \] Then \[B = \boxed{\frac{F_0 (c \, \omega)}{\par{k - m \omega^2}^2 + (c \, \omega)^2}}\] The steady-state solution is therefore \[x_S(t) = \frac{F_0 \par{k - m \omega^2}}{\par{k - m \omega^2}^2 + \par{c \, \omega}^2} \cos \omega t + \frac{F_0 c \, \omega}{\par{k - m \omega^2}^2 + (c \, \omega)^2} \sin \omega t \pd \]

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3. We have \[ \ba X &= \sqrt{A^2 + B^2} \nl &= \sqrt{\par{\frac{F_0 \par{k - m \omega^2}}{\par{k - m \omega^2}^2 + \par{c \, \omega}^2}}^2 + \par{\frac{F_0 (c \, \omega)}{\par{k - m \omega^2}^2 + (c \, \omega)^2}}^2} \nl &= \sqrt{\frac{\par{F_0}^2 \parbr{\par{k - m \omega^2}^2 + (c \, \omega)^2}}{\parbr{\par{k - m \omega^2}^2 + \par{c \, \omega}^2}^2}} \nl &= \sqrt{\frac{\par{F_0}^2}{\par{k - m \omega^2}^2 + \par{c \, \omega}^2}} \nl &= \boxed{\frac{F_0}{\sqrt{\par{k - m \omega^2}^2 + \par{c \, \omega}^2}}} \ea \]