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1960 Paper 4 Q308
D: 1500.0 B: 1500.0

Solve completely the following differential equations:

  1. \(y' = y + e^{-x}\);
  2. \(x^3y'' + xy = 1\);
  3. \(y' = \frac{2x+3y}{3x+8y}\).

Solution:

  1. We can see the general solution of \(y'=y\) is \(Ae^x\), therefore we are looking for a particular integral of the form \(y = Ce^{-x}\), so \(-Ce^{-x} = 2e^{-x} \Rightarrow C = -2\). Therefore the general solution is \(y = Ae^x-2e^{-x}\)
  2. First consider \(x^2y'' + y = 0\). Consider solutions of the form \(y = x^t\), so \((t(t-1)+1)x^t = 0\), ie \(t^2 - t + 1 = 0 \Rightarrow t = \frac{1 \pm \sqrt{-3}}{2}\), ie the general solution is \(Ax^{\omega} + B x^{\omega^2}\). Now we seek a particular integral for \(x^2 y'' + y = \frac{1}{x}\). Trying \(y = \frac{C}{x}\) we find \(\frac{2C}{x} + \frac{C}{x} = \frac{1}{x} \Rightarrow C = \frac{1}{3}\).Therefore the general solution is \(y = Ae^x - 2e^{-x} + \frac{1}{3x}\)
  3. \begin{align*} && \frac{\d y}{\d x} &= \frac{2x+3y}{3x+8y} \\ y = vx: &&v + x\frac{\d v}{\d x} &= \frac{2+3v}{3+8v} \\ \Rightarrow && \frac{\d v}{\d x} &= \frac{2-8v^2}{x (3+8v)} \\ \Rightarrow && \int \frac{2}{x} \d x &= \int \frac{3+8v}{(1-2v)(1+2v)} \d v \\ &&&=\frac12 \int \left ( \frac{7}{1-2v} - \frac{1}{1+2v} \right) \d v \\ \Rightarrow && 8\ln |x| &= -7\ln|1-2v| - \ln|1+2v| + C \\ \Rightarrow && 8 \ln |x| &= - \ln |(1-2v)^7(1+2v)| +C \\ \Rightarrow && C &= \ln \left (x^8\left(1-2\frac{y}{x}\right)^7\left(1+2\frac{y}{x}\right) \right) \\ \Rightarrow && K &= (x-2y)^7(x+2y) \end{align*} Alternative method: Notice that \(\dot{y} = 2x+3y\) and \(\dot{x} = 3x+8y\), which is the equation \(\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\). This has eigenvalues \(\lambda = 7,-1\) and eigenvectors \(\binom{2}{1}\) and \(\binom{-2}{1}\), therefore \(\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{7t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} + Be^{-t}\begin{pmatrix} -2 \\ 1 \end{pmatrix}\). So we can write: \begin{align*} && x(t) &= 2Ae^{7t}-2Be^{-t} \\ && y(t) &= Ae^{7t}+Be^{-t} \\ \Rightarrow && 4Ae^{7t} &= x+2y \\ && 4Be^{-t} &= 2y-x \\ \Rightarrow && (2y-x)^7(x+2y) &= K \end{align*}

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1960 Paper 4 Q309
D: 1500.0 B: 1500.0

If \(y_m(x)\) is defined as a function of \(x\) by the equation $$y_m(x) = (-1)^m e^{x^2} \frac{d^m}{dx^m} e^{-x^2},$$ show that \(y_m\) is a polynomial in \(x\) and that $$y_{n+1}(x) = -\frac{d}{dx} y_n(x) + 2xy_n(x).$$ Deduce, by induction on \(n\) or otherwise, that $$\frac{d^2}{dx^2} y_n(x) - 2x \frac{d}{dx} y_n(x) + 2ny_n(x) = 0.$$

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1960 Paper 4 Q310
D: 1500.0 B: 1500.0

Let $$f_m(x) = \frac{x}{2} \left[ \sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \ldots + (-1)^{m+1} \frac{\sin mx}{m} \right].$$ By considering \(df_m(x)/dx\), or otherwise, show that $$(-1)^m f_m(x) > 0$$ for \(0 < x < \pi/(2m+1)\). Show also that $$(-1)^m f_m\left(\frac{\pi}{m+\frac{1}{3}}\right) < 0.$$

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1959 Paper 1 Q101
D: 1500.0 B: 1500.0

Discover all the real roots of each of the equations

  1. [(i)] \((x-1)^3 + (x-2)^3 = 0\),
  2. [(ii)] \((x-1)^4 + (x-2)^4 = 1\),
  3. [(iii)] \((x-1)^4 + (x-2)^4 = 0\),
indicating your methods.

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1959 Paper 1 Q102
D: 1500.0 B: 1500.0

(i) If \(k = 9^9\), use the information given in four-figure tables to prove that \(9^k\) is a number of more than 368,000,000 figures. (ii) Prove that, if \(m\), \(n\), \(p\) are positive integers such that $$(m^n)^p = m^{(n,p)},$$ then the only possibilities are that either \(m = 1\) or \(p = 1\) or \(n = p = 2\).

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1959 Paper 1 Q103
D: 1500.0 B: 1500.0

Obtain conditions on the positive integer \(n\) and the constants \(a\), \(b\) in order that the \(n+1\) equations for \(x_0\), \(\ldots\), \(x_n\) $$x_k - x_{k-1} + x_{k-2} = 0 \quad (k = 2, 3, \ldots, n), \quad x_0 = a, \quad x_n = b,$$ shall have (i) exactly one solution, (ii) no solution, (iii) more than one solution.

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1959 Paper 1 Q104
D: 1500.0 B: 1500.0

Prove that, if \(h(x)\) is the H.C.F. of two polynomials \(p(x)\), \(q(x)\), then polynomials \(A(x)\), \(B(x)\) exist such that $$A(x)p(x) + B(x)q(x) \equiv h(x).$$ Obtain an identity of this form when $$p(x) = x^{10} - 1, \quad q(x) = x^6 - 1.$$

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1959 Paper 1 Q105
D: 1500.0 B: 1500.0

Prove that, if \(n > 1\), $$\sum_{r=1}^n \left(1 + \frac{1}{r}\right) > n(n+1)^{1/n}.$$ If you appeal to any general inequality, prove it. Prove that $$n\{(n+1)^{1/n} - 1\} < \sum_{r=1}^n \frac{1}{r} < n\left\{1 + \frac{1}{n+1} - \frac{1}{(n+1)^{1/n}}\right\}.$$

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1959 Paper 1 Q106
D: 1500.0 B: 1500.0

On the sides of a triangle \(Z_1Z_2Z_3\) are constructed isosceles triangles \(Z_2Z_3W_1\), \(Z_3Z_1W_2\), \(Z_1Z_2W_3\) lying outside the triangle \(Z_1Z_2Z_3\). The angles at \(W_1\), \(W_2\), \(W_3\) are all \(\frac{2\pi}{13}\). By assuming complex numbers \(z_1\), \(z_2\), \(z_3\) to \(Z_1\), \(Z_2\), \(Z_3\) and calculating the numbers representing \(W_1\), \(W_2\), \(W_3\), or otherwise, prove that \(W_1W_2W_3\) is equilateral.

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1959 Paper 1 Q107
D: 1500.0 B: 1500.0

A regular polygon \(\Pi\) of \(n\) sides is given. A variable regular polygon of \(n\) sides is inscribed in \(\Pi\), having one vertex on each side of \(\Pi\). Prove that the sides of the variable polygon envelop parabolas. When \(n = 4\), identify the foci and latera recta of the parabolas.

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