4 problems found
A uniform elastic string lies on a smooth horizontal table. One end of the string is attached to a fixed peg, and the other end is pulled at constant speed \(u\). At time \(t=0\), the string is taut and its length is \(a\). Obtain an expression for the speed, at time \(t\), of the point on the string which is a distance \(x\) from the peg at time \(t\). An ant walks along the string starting at \(t=0\) at the peg. The ant walks at constant speed \(v\) along the string (so that its speed relative to the peg is the sum of \(v\) and the speed of the point on the string beneath the ant). At time \(t\), the ant is a distance \(x\) from the peg. Write down a first order differential equation for \(x\), and verify that \[ \frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,. \] Show that the time \(T\) taken for the ant to reach the end of the string is given by \[uT = a(\e^k-1)\,,\] where \(k=u/v\). On reaching the end of the string, the ant turns round and walks back to the peg. Find in terms of \(T\) and \(k\) the time taken for the journey back.
Solution: Points always maintain a constant fraction of the distance from the start, so the point distance \(x\) from the start at time \(t\) is moving with speed \(\frac{x}{a+ut} u\) The point is moving with speed \(v+\frac{x}{a+ut} u\) or in other words \begin{align*} && \frac{\d x}{\d t} &= v + \frac{x}{a+ut}u \\ \Rightarrow && \frac{\d x }{\d t} - \frac{u}{a+ut} x &= v \\ \Rightarrow && \frac{1}{a+ut} \frac{\d x}{\d t} - \frac{u}{(a+ut)^2} x &= \frac{1}{a+ut} v\\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{x}{a+ut} \right) &= \frac{v}{a+ut} \\ \Rightarrow && \frac{x}{a+ut} &=\frac{v}{u} \ln (a + ut) + C \\ t = 0, x = 0: && 0 &= \frac{v}{u} \ln a + C \\ \Rightarrow && x &= (a+ut) \frac{v}{u} \ln \left ( \frac{a+ut}{a} \right) \\ \\ \Rightarrow && 1 &= \frac{v}{u} \ln \left ( \frac{a+uT}{a} \right) \\ \Rightarrow && e^k &= 1 + \frac{uT}{a} \\ \Rightarrow && uT &= a(e^k-1) \end{align*} On the return journey, we have \begin{align*} && \frac{\d x}{\d t} &= \frac{x}{a+ut}u - v \\ \Rightarrow && \frac{\d x}{\d t} - \frac{u}{a+ut} x &= - v \\ \Rightarrow && \frac{\d }{\d x} \left ( \frac{x}{a+ut} \right) &= -\frac{v}{a+ut} \\ \Rightarrow &&f &= -\frac{v}{u} \ln(a+ut) + K \\ t = T, f = 1: && 1 &= -\frac{v}{u}\ln(a+uT) + K \\ \Rightarrow && f &= 1+\frac{v}{u}\ln \left ( \frac{a+uT}{a+ut} \right) \\ \Rightarrow && 0 &= 1+\frac{v}{u} \ln \left ( \frac{a+uT}{a+uT_2} \right) \\ \Rightarrow && e^k &= \frac{a+uT_2}{a+uT}\\ \Rightarrow && uT_2 &= (a+uT)e^k - a \\ \Rightarrow && T_2 - T &= \frac{1}{u} \left ( (a+uT)e^k - a - uT\right) \\ &&&= \frac{1}{u} \left ((a+a(e^k-1))e^k-a-a(e^k-1) \right) \\ &&&= \frac{1}{u} \left (ae^{2k} -ae^k \right) \\ &&&= \frac{ae^k}{u} \left ( e^k-1 \right) \\ &&&= Te^k \end{align*}
Let \(P,Q\) and \(R\) be functions of \(x\). Prove that, for any function \(y\) of \(x\), the function \[ Py''+Qy'+Ry \] can be written in the form \(\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),\) where \(p\) and \(q\) are functions of \(x\), if and only if \(P''-Q'+R=0.\) Solve the differential equation \[ (x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x}, \] given that when \(x=2,y=2\mathrm{e}^{2}\) and \(y'=0.\)
Solution: Suppose \(Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)\), then \begin{align*} Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\ &= py'' + p'y' + qy' + q' y \\ &= py'' + (p'+q)y' + q' y \end{align*} Therefore \(P = p, Q = p'+q, R = q'\), Therefore \(q = Q-P'\) and \(R = Q'-P''\) or \(P'' -Q'+R = 0\). \((\Rightarrow)\) Suppose it can be written in that form, then the logic we have applied shows that equation is true. \((\Leftarrow)\) Suppose \(P''-Q'+R = 0\), then letting \(p = P, q = Q-P'\) we find functions of the form which will be expressed correctly. Notice that if \(P = x-x^4, Q = (1-7x^3), R = -9x^2\) then: \begin{align*} P'' - Q' + R &= -12x^2+21x^2-9x^2 \\ &= 0 \end{align*} Therefore we can write our second order ODE as: \begin{align*} && (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y' +(1-7x^3-(1-4x^3))y \right) \\ &&&= \frac{\d}{\d x} \left ( (x-x^4) y' -3x^3y \right) \end{align*} Suppose \(z = (x-x^4)y' - 3x^2y\), then \(z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2\) when \(x = 2\). and we have: \begin{align*} && \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\ \Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\ &&&= x^3 e^x+c \\ \Rightarrow && -48e^2 &= e^2(8) + c \\ \Rightarrow && c &= -56e^2 \\ \Rightarrow && z &= e^x(x^3)-56e^2 \\ \end{align*} So our differential equation is: \begin{align*} && (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2 \\ \Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\ \Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\ \Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\ \underbrace{\Rightarrow}_{x=2} && (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\ \Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\ \Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)} \end{align*}
The matrix \(\mathbf{F}\) is defined by \[ \mathbf{F}=\mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}, \] where $\mathbf{A}=\begin{pmatrix}-3 & -1\\ 8 & 3 \end{pmatrix} \( and \) t \( is a variable scalar. Evaluate \)\mathbf{A}^{2}$, and show that \[ \mathbf{F}=\mathbf{I}\cosh t+\mathbf{A}\sinh t. \] Show also that \(\mathbf{F}^{-1}=\mathbf{I}\cosh t-\mathbf{A}\sinh t\), and that \(\dfrac{\mathrm{d}\mathbf{F}}{\mathrm{d}t}=\mathbf{FA}\). The vector $\mathbf{r}=\begin{pmatrix}x(t)\\ y(t) \end{pmatrix}$ satisfies the differential equation \[ \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}=\mathbf{0}, \] with \(x=\alpha\) and \(y=\beta\) at \(t=0.\) Solve this equation by means of a suitable matrix integrating factor, and hence show that \begin{alignat*}{1} x(t) & =\alpha\cosh t+(3\alpha+\beta)\sinh t\\ y(t) & =\beta\cosh t-(8\alpha+3\beta)\sinh t. \end{alignat*}
Solution: \begin{align*} \begin{pmatrix} -3 & -1 \\ 8 & 3 \end{pmatrix}^2 &= \begin{pmatrix} 9-8 & 3-3 \\ -24+24 & -8+9 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \mathbf{I} \end{align*} Therefore: \begin{align*} \mathbf{F} &= \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n} \\ &= \mathbf{I} + \sum_{n=1}^{\infty} \frac{1}{(2n)!}t^{2n} \mathbf{I} + \sum_{n=0}^{\infty} \frac{1}{(2n+1)!}t^{2n+1} \mathbf{A} \\ &= \cosh t \mathbf{I} + \sinh t \mathbf{A} \end{align*} Notice that \begin{align*} \mathbf{F} (\mathbf{I}\cosh t-\mathbf{A}\sinh t) &= (\mathbf{I}\cosh t+\mathbf{A}\sinh t)(\mathbf{I}\cosh t-\mathbf{A}\sinh t) \\ &= \mathbf{I}^2 \cosh^2 t+\mathbf{A}(\sinh t \cosh t - \cosh t \sinh t) - \mathbf{A}^2\sinh^2 t \\ &= \mathbf{I} \cosh^2 t - \mathbf{I} \sinh^2 t \\ &= \mathbf{I} \end{align*} Therefore \(\mathbf{F}^{-1} = \mathbf{I}\cosh t-\mathbf{A}\sinh t\) \begin{align*} \frac{\d \mathbf{F}}{\d t} &= \frac{\d }{\d t} \left [ \mathbf{I}+\sum_{n=1}^{\infty}\frac{1}{n!}t^{n}\mathbf{A}^{n}\right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^n \\ &= \left ( \sum_{n=1}^{\infty} \frac{1}{(n-1)!}t^{n-1} \mathbf{A}^{n-1} \right) \mathbf{A} \\ &= \mathbf{F}\mathbf{A} \end{align*} \begin{align*} && \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}+\mathbf{A}\mathbf{r}&=\mathbf{0} \\ \Rightarrow && \mathbf{F} \frac{\d \mathbf{r}}{\d t} + \mathbf{FAr} &= \mathbf{0} \\ && \frac{\d }{\d t} \left ( \mathbf{F} \mathbf{r}\right) &= 0 \\ \Rightarrow && \mathbf{Fr} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= \mathbf{F}^{-1}\mathbf{c} \\ &&&= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\mathbf{c} \\ t = 0: && \binom{\alpha}{\beta} &= \mathbf{c} \\ \Rightarrow && \mathbf{r} &= ( \mathbf{I}\cosh t-\mathbf{A}\sinh t)\binom{\alpha}{\beta} \\ &&&= \binom{\alpha \cosh t}{\beta \cosh t} - \binom{-3\alpha-\beta}{8\alpha + 3\beta}\sinh t \\ &&&= \binom{\alpha \cosh t + (3\alpha + \beta) \sinh t}{\beta \cosh t -(8\alpha + 3\beta)\sinh t} \end{align*} as required