Problems

2014 Paper 3 Q10

D: 1700.0 B: 1473.3

second order differential equation SHM elastic spring Hooke's law coupled oscillations normal modes simultaneous differential equations trigonometric

Two particles $X$ and $Y$, of equal mass $m$, lie on a smooth horizontal table and are connected by a light elastic spring of natural length $a$ and modulus of elasticity $\lambda$. Two more springs, identical to the first, connect $X$ to a point $P$ on the table and $Y$ to a point $Q$ on the table. The distance between $P$ and $Q$ is $3a$. Initially, the particles are held so that $XP=a$, $YQ= \frac12 a\,$, and $PXYQ$ is a straight line. The particles are then released. At time $t$, the particle $X$ is a distance $a+x$ from $P$ and the particle $Y$ is a distance $a+y$ from $Q$. Show that \[ m \frac{\.d ^2 x}{\.d t^2} = -\frac\lambda a (2x+y) \] and find a similar expression involving $\dfrac{\.d^2 y}{\.d t^2}$. Deduce that \[ x-y = A\cos \omega t +B \sin\omega t \] where $A$ and $B$ are constants to be determined and $ma\omega^2=\lambda$. Find a similar expression for $x+y$. Show that $Y$ will never return to its initial position.

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2009 Paper 3 Q2

D: 1700.0 B: 1484.0

second order differential equation power series series solution recurrence relation comparing coefficients trigonometric Maclaurin series

Let $\displaystyle y= \sum_{n=0}^\infty a_n x^n\,$, where the coefficients $a_n$ are independent of $x$ and are such that this series and all others in this question converge. Show that \[ \displaystyle y'= \sum_{n=1}^\infty na_n x^{n-1}\,, \] and write down a similar expression for $y''$. Write out explicitly each of the three series as far as the term containing $a_3$.
It is given that $y$ satisfies the differential equation \[ xy''-y'+4x^3y =0\,. \] By substituting the series of part (i) into the differential equation and comparing coefficients, show that $a_1=0$. Show that, for $n\ge4$, \[ a_n =- \frac{4}{n(n-2)}\, a_{n-4}\,, \] and that, if $a_0=1$ and $a_2=0$, then $ y=\cos (x^2)\,$. Find the corresponding result when $a_0=0$ and $a_2=1$.

Solution:

Let $\displaystyle y= \sum_{n=0}^\infty a_n x^n\,$ then \begin{align*} y' &= \frac{\d}{\d x} \l \sum_{n=0}^\infty a_n x^n \r \\ &= \sum_{n=0}^\infty \frac{\d}{\d x} \l a_n x^n \r \\ &= \sum_{n=0}^\infty n a_n x^{n-1} \\ &= \sum_{n=1}^\infty n a_n x^{n-1} \\ \\ y'' &= \frac{\d}{\d x} \l\sum_{n=1}^\infty n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty \frac{\d}{\d x} \l n a_n x^{n-1} \r \\ &= \sum_{n=1}^\infty n(n-1) a_n x^{n-2} \\ &= \sum_{n=2}^\infty n(n-1) a_n x^{n-2} \\ \\ y &= a_0 + a_1 x+ a_2x^2 + a_3x^3 + \cdots \\ y'&= a_1 + 2a_2x+3a_3x^2 + \cdots \\ y'' &= 2a_2 + 6a_3x + \cdots \end{align*}
\begin{align*} && 0 &= xy''-y'+4x^3y \\ &&&= x\sum_{n=2}^\infty n(n-1) a_n x^{n-2} - \sum_{n=1}^\infty n a_n x^{n-1} + 4x^3 \sum_{n=0}^\infty a_n x^n \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=0}^\infty 4a_n x^{n+3} \\ &&&= \sum_{n=2}^\infty n(n-1) a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^{n-1} + \sum_{n=4}^\infty 4a_{n-4} x^{n-1} \\ &&&= \sum_{n=4}^{\infty} \l n(n-1) a_n- n a_n +4a_{n-4} \r x^{n-1} + 2a_2x + 6a_3x^2-a_1-2a_2x-3a_3x^2 \\ &&&= \sum_{n=4}^{\infty} \l n(n-2) a_n +4a_{n-4} \r x^{n-1}+ 3a_3x^2-a_1 \\ \end{align*} Therefore since all coefficients are $0$, $a_1 = 0$, $a_3 = 0$ and $\displaystyle a_n = -\frac{4}{n(n-2)}a_{n-4}$. If $a_0 = 1, a_2 = 0$, and since $a_1 = 0, a_3 = 0$ the only values which will take non-zero value are $a_{4k}$. We can compute these values as: $a_{4k} = -\frac{4}{(4k)(4k-2)} a_{4k-4} = \frac{1}{2k(2k-1)}a_{4k-r}$ so $a_{4k} = \frac{(-1)^k}{(2k)!}$, which are precisely the coefficients in the expansion $\cos x^2$. If $a_0 = 0, a_2 = 1$ then since $a_1 = 0, a_3 = 0$ the only values which take non-zero values are $a_{4k+2}$ we can compute these values as: $a_{4k+2} = -\frac{4}{(4k+2)(4k)}a_{4k-2} = -\frac{1}{(2k+1)2k}a_{4k-2}$ so we can see that $a_{4k+2}= \frac{(-1)^k}{(2k+1)!}$ precisely the coefficients of $\sin x^2$

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2009 Paper 3 Q10

D: 1700.0 B: 1484.0

second order differential equation SHM elastic spring Hooke's law energy conservation particle on spring trigonometric initial value problem compression

A light spring is fixed at its lower end and its axis is vertical. When a certain particle $P$ rests on the top of the spring, the compression is $d$. When, instead, $P$ is dropped onto the top of the spring from a height $h$ above it, the compression at time $t$ after $P$ hits the top of the spring is $x$. Obtain a second-order differential equation relating $x$ and $t$ for $0\le t \le T$, where $T$ is the time at which $P$ first loses contact with the spring. Find the solution of this equation in the form \[ x= A + B\cos (\omega t) + C\sin(\omega t)\,, \] where the constants $A$, $B$, $C$ and $\omega$ are to be given in terms of $d$, $g$ and $h$ as appropriate. Show that \[ T = \sqrt{d/g\;} \left (2 \pi - 2 \arctan \sqrt{2h/d\;}\;\right)\,. \]

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1999 Paper 1 Q7

D: 1500.0 B: 1516.0

second order differential equation trigonometric inverse trigonometric chain rule polynomial solution verification triple angle formula sin 3θ identity

Show that $\sin(k\sin^{-1} x)$, where $k$ is a constant, satisfies the differential equation $$(1-x^{2})\frac {\d^2 y}{\d x^2} -x\frac{\d y}{\d x} +k^{2}y=0. \tag{*}$$ In the particular case when $k=3$, find the solution of equation $(*)$ of the form \[ y=Ax^{3}+Bx^{2}+Cx+D, \] that satisfies $y=0$ and $\displaystyle \frac{\d y}{\d x}=3$ at $x=0$. Use this result to express $\sin 3\theta$ in terms of powers of $\sin\theta$.

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1999 Paper 3 Q8

D: 1700.0 B: 1516.0

second order differential equation SHM piecewise function continuity conditions initial value problem infinite series integration trigonometric

The function $y(x)$ is defined for $x\ge0$ and satisfies the conditions \[ y=0 \mbox{ \ \ and \ \ } \frac{\d y}{\d x}=1 \mbox{ \ \ at $x=0$}. \] When $x$ is in the range $2(n-1)\pi< x <2n\pi$, where $n$ is a positive integer, $y(t)$ satisfies the differential equation $$ {\d^2y \over \d x^2} + n^2 y=0. $$ Both $y$ and $\displaystyle \frac{\d y}{\d x} $ are continuous at $x=2n\pi$ for $n=0,\; 1,\;2,\; \ldots\;$.

Find $y(x)$ for $0\le x \le 2\pi$.
Show that $y(x) = \frac12 \sin 2x $ for $2\pi\le x\le 4\pi$, and find $y(x)$ for all $x\ge0$.
Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$

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1996 Paper 2 Q8

D: 1600.0 B: 1485.5

second order differential equation functional equation complementary function particular integral trigonometric symmetry even and odd functions cos nθ

Suppose that \[{\rm f}''(x)+{\rm f}(-x)=x+3\cos 2x\] and ${\rm f}(0)=1$, ${\rm f}'(0)=-1$. If ${\rm g}(x)={\rm f}(x)+{\rm f}(-x)$, find ${\rm g}(0)$ and show that ${\rm g}'(0)=0$. Show that \[{\rm g}''(x)+{\rm g}(x)=6\cos 2x,\] and hence find ${\rm g}(x)$. Similarly, if ${\rm h}(x)={\rm f}(x)-{\rm f}(-x)$, find ${\rm h}(x)$ and show that \[{\rm f}(x)=2\cos x -\cos2x-x.\]

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1995 Paper 1 Q8

D: 1500.0 B: 1532.8

second order differential equation variable coefficients particular integral complementary function linear combination simultaneous equations superposition principle

Find functions $\mathrm{f,g}$ and $\mathrm{h}$ such that \[ \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}+\mathrm{f}(x)\frac{\mathrm{d}y}{\mathrm{d}x}+\mathrm{g}(x)y=\mathrm{h}(x)\tag{\ensuremath{*}} \] is satisfied by all three of the solutions $y=x,y=1$ and $y=x^{-1}$ for $0 < x < 1.$ If $\mathrm{f,g}$ and $\mathrm{h}$ are the functions you have found in the first paragraph, what condition must the real numbers $a,b$ and $c$ satisfy in order that \[ y=ax+b+\frac{c}{x} \] should be a solution of $(*)$?

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1988 Paper 3 Q7

D: 1700.0 B: 1554.3

second order differential equation Gaussian verification orthogonality Wronskian integration boundary conditions quantum harmonic oscillator

For $n=0,1,2,\ldots,$ the functions $y_{n}$ satisfy the differential equation \[ \frac{\mathrm{d}^{2}y_{n}}{\mathrm{d}x^{2}}-\omega^{2}x^{2}y_{n}=-(2n+1)\omega y_{n}, \] where $\omega$ is a positive constant, and $y_{n}\rightarrow0$ and $\mathrm{d}y_{n}/\mathrm{d}x\rightarrow0$ as $x\rightarrow+\infty$ and as $x\rightarrow-\infty.$ Verify that these conditions are satisfied, for $n=0$ and $n=1,$ by \[ y_{0}(x)=\mathrm{e}^{-\lambda x^{2}}\qquad\mbox{ and }\qquad y_{1}(x)=x\mathrm{e}^{-\lambda x^{2}} \] for some constant $\lambda,$ to be determined. Show that \[ \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right)=2(m-n)\omega y_{m}y_{n}, \] and deduce that, if $m\neq n,$ \[ \int_{-\infty}^{\infty}y_{m}(x)y_{n}(x)\,\mathrm{d}x=0. \]

Solution: \begin{align*} && y_0(x) &= e^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_0(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_0(x) &= \lim_{x \to \pm \infty} 2x\lambda e^{-\lambda x^2} \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= 4x^2 \lambda^2 e^{-\lambda x^2} + 2\lambda e^{-\lambda x^2} \\ \\ && y''_0 - \omega^2 x^2 y_0+(2\cdot 0 + 1) \omega y_0 &= e^{-\lambda x^2} \l 4x^2 \lambda^2 + 2 \lambda - \omega^2 x^2 + \omega\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore $y_0$ satisfies if $\lambda = \frac{\omega}{2}$ Similarly for $y_1$, \begin{align*} && y_1(x) &= xe^{-\lambda x^2} \\ && \lim_{x \to \pm \infty} y_1(x) &= 0 \Leftrightarrow \lambda > 0 \\ && \lim_{x \to \pm \infty} y'_1(x) &= \lim_{x \to \pm \infty} \l -2x^2 \lambda e^{-\lambda x^2} + e^{-\lambda x^2} \r \\ &&&= 0\Leftrightarrow \lambda > 0 \\ && y''_0(x) &= e^{-\lambda x^2} \l 4x^3 \lambda^2-4x\lambda - 2x\lambda \r \\ &&&= e^{-\lambda x^2} \l 4x^3 \lambda^2-6x\lambda \r \\ && y''_1 - \omega^2 x^2 y_1+(2\cdot 1 + 1) \omega y_1 &= e^{-\lambda x^2} \l 4x^3\lambda^2-6x\lambda-\omega^2x^3+3\omega x\r \\ &&&=0 \Leftrightarrow \lambda = \pm \frac{\omega}{2} \end{align*} Therefore $y_1$ satisfies if $\lambda = \frac{\omega}{2}$ \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) &= y'_my'_n+y_my''_n - y'_ny'_m-y_ny''_m \\ &= y_my''_n - y_ny''_m \\ &= y_m(\omega^2 x^2 y_n - (2n+1)\omega y_n) - y_n(\omega^2 x^2 y_m - (2m+1)\omega y_m) \\ &= y_my_n (2m-2n)\omega \\ &= 2(m-n) \omega y_my_n \end{align*} Therefore: \begin{align*} \int_{-\infty}^{\infty} y_m(x)y_n(x) \d x &= \int_{-\infty}^{\infty} \frac{1}{2(m-n)} \frac{\mathrm{d}}{\mathrm{d}x}\left(y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right) \d x \\ &= \frac{1}{2(m-n)} \left [ y_{m}\frac{\mathrm{d}y_{n}}{\mathrm{d}x}-y_{n}\frac{\mathrm{d}y_{m}}{\mathrm{d}x}\right]_{-\infty}^{\infty} \\ &\to 0 \end{align*} This condition is known as Orthogonality. In fact this question is talking about a Sturm-Liouville orthogonality condition, in particular for the quantum harmonic oscillator, and the eigenfunctions are related to Hermite polynomials.

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