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2024 Paper 3 Q6
D: 1500.0 B: 1500.0
differential equations systems of odes linear algebra eigenvalues invariant quantities necessary conditions exponential functions

  1. A particle moves in two-dimensional space. Its position is given by coordinates \((x, y)\) which satisfy \[\frac{\mathrm{d}x}{\mathrm{d}t} = -x + 3y + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x + y + u\] where \(t\) is the time and \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0)\).
    1. By considering \(\dfrac{\mathrm{d}x}{\mathrm{d}t} - \dfrac{\mathrm{d}y}{\mathrm{d}t}\), show that if the particle is at the origin \((0,0)\) at some time \(t > 0\), then it is necessary that \(x_0 = y_0\).
    2. Given that \(x_0 = y_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).
  2. A particle whose position in three-dimensional space is given by co-ordinates \((x, y, z)\) moves with time \(t\) such that \[\frac{\mathrm{d}x}{\mathrm{d}t} = 4y - 5z + u\] \[\frac{\mathrm{d}y}{\mathrm{d}t} = x - 2z + u\] \[\frac{\mathrm{d}z}{\mathrm{d}t} = x - 2y + u\] where \(u\) is a function of time. At time \(t = 0\), the particle has position \((x_0,\, y_0,\, z_0)\).
    1. Show that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(y_0\) is the mean of \(x_0\) and \(z_0\).
    2. Show further that, if the particle is at the origin \((0,0,0)\) at some time \(t > 0\), it is necessary that \(x_0 = y_0 = z_0\).
    3. Given that \(x_0 = y_0 = z_0\), find a constant value of \(u\) that ensures that the particle is at the origin at a time \(t = T\), where \(T > 0\).

Solution:

    1. \(\,\) \begin{align*} && \frac{\d x}{\d t} - \frac{\d y}{\d t} &= -2x + 2y \\ &&&= -2(x-y) \\ \Rightarrow && \dot{z} &= -2z \tag{\(z = x-y\)} \\ \Rightarrow && z &= Ae^{-2t} \\ z = 0, t : && A &= 0 \\ \Rightarrow && z &= 0 \quad \forall t \\ \Rightarrow && x &= y \quad \forall t \\ \Rightarrow && x_0 &= y_0 \end{align*}
    2. Since \(x = y\) for all \(t\) our equation can we written as \(\frac{\d x}{\d t} = 2x + u\). This has solution \(x = Ae^{2t} - \frac{u}{2}\) we also have \begin{align*} t = 0: && x_0 &= A - \frac{u}{2} \\ t = T: && 0 &= Ae^{2T} - \frac{u}{2} \\ \Rightarrow && x_0 &= \frac{u}{2} - e^{-2T}\frac{u}{2} \\ \Rightarrow && u &= \frac{2x_0}{1-e^{-2T}} \end{align*}
    1. Let \(w = y - \frac12(x+z)\) then \begin{align*} && \dot{w} &= \dot{y} - \tfrac12(\dot{x}+\dot{z}) \\ &&&= (x-2z+u) - \tfrac12(4y-5z+u+x-2y+u) \\ &&&= -y + \tfrac12(x+z) \\ &&&= -w \\ \Rightarrow && w &= Ae^{-t} \\ \text{at origin}: && w &= 0 \\ \Rightarrow && y &= \tfrac12(x+z) \end{align*}
    2. We now have \(\dot{x} = 2x+2z-5z+u = 2x-3z+u\) and \(\dot{z} = x - x- z +u = - z+u\) so in particular \((x - z)' = 2(x-z)\) or \(x-z = Ae^{2t}\) and since we hit the origin \(x = z\) for all \(t\) and so \(y = x = z\) for all \(t\).
    3. Notice we now have \(\dot{x} = -x + u\) or \(x = Ae^{-t} + u\) \begin{align*} t = 0: && x_0 &= A + u \\ t = T: && 0 &= Ae^{-T} + u \\ \Rightarrow && x_0 &= u-e^{T} u \\ \Rightarrow && u &= \frac{x_0}{1-e^{T}} \end{align*}

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2023 Paper 3 Q8
D: 1500.0 B: 1500.0
second order differential equations modulus function piecewise functions continuous differentiability substitution reflection symmetry exponential functions

If \[y = \begin{cases} \mathrm{k}_1(x) & x \leqslant b \\ \mathrm{k}_2(x) & x \geqslant b \end{cases}\] with \(\mathrm{k}_1(b) = \mathrm{k}_2(b)\), then \(y\) is said to be \emph{continuously differentiable} at \(x = b\) if \(\mathrm{k}_1'(b) = \mathrm{k}_2'(b)\).

  1. Let \(\mathrm{f}(x) = x\mathrm{e}^{-x}\). Verify that, for all real \(x\), \(y = \mathrm{f}(x)\) is a solution to the differential equation \[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\frac{\mathrm{d}y}{\mathrm{d}x} + y = 0\] and that \(y = 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) when \(x = 0\). Show that \(\mathrm{f}'(x) \geqslant 0\) for \(x \leqslant 1\).
  2. You are given the differential equation \[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right| + y = 0\] where \(y = 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) when \(x = 0\). Let \[y = \begin{cases} \mathrm{g}_1(x) & x \leqslant 1 \\ \mathrm{g}_2(x) & x \geqslant 1 \end{cases}\] be a solution of the differential equation which is continuously differentiable at \(x = 1\). Write down an expression for \(\mathrm{g}_1(x)\) and find an expression for \(\mathrm{g}_2(x)\).
  3. State the geometrical relationship between the curves \(y = \mathrm{g}_1(x)\) and \(y = \mathrm{g}_2(x)\).
  4. Prove that if \(y = \mathrm{k}(x)\) is a solution of the differential equation \[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + p\frac{\mathrm{d}y}{\mathrm{d}x} + qy = 0\] in the interval \(r \leqslant x \leqslant s\), where \(p\) and \(q\) are constants, then, in a suitable interval which you should state, \(y = \mathrm{k}(c - x)\) satisfies the differential equation \[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} - p\frac{\mathrm{d}y}{\mathrm{d}x} + qy = 0\,.\]
  5. You are given the differential equation \[\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 2\left|\frac{\mathrm{d}y}{\mathrm{d}x}\right| + 2y = 0\] where \(y = 0\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1\) when \(x = 0\). Let \(\mathrm{h}(x) = \mathrm{e}^{-x}\sin x\). Show that \(\mathrm{h}'\!\left(\frac{1}{4}\pi\right) = 0\). It is given that \(y = \mathrm{h}(x)\) satisfies the differential equation in the interval \(-\frac{3}{4}\pi \leqslant x \leqslant \frac{1}{4}\pi\) and that \(\mathrm{h}'(x) \geqslant 0\) in this interval. In a solution to the differential equation which is continuously differentiable at \((n + \frac{1}{4})\pi\) for all \(n \in \mathbb{Z}\), find \(y\) in terms of \(x\) in the intervals
    1. \(\frac{1}{4}\pi \leqslant x \leqslant \frac{5}{4}\pi\),
    2. \(\frac{5}{4}\pi \leqslant x \leqslant \frac{9}{4}\pi\).

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1994 Paper 1 Q2
D: 1484.0 B: 1468.1
differentiation exponential functions power functions logarithmic differentiation chain rule iterated exponentials

Given that \(a\) is constant, differentiate the following expressions with respect to \(x\):

  1. \(x^{a}\);
  2. \(a^{x}\);
  3. \(x^{x}\);
  4. \(x^{(x^{x})}\);
  5. \((x^{x})^{x}.\)

Solution: \begin{align*} && y &= x^a \\ && \frac{\d y}{\d x} &= \begin{cases} ax^{a-1} & a \neq 0 \\ 0 & a = 0 \end{cases} \\ \\ && y &= a^x \\ &&&= e^{(\ln a) \cdot x} \\ && \frac{\d y}{\d x} &= \ln a e^{(\ln a) x} \\ &&&= \ln a \cdot a^ x \\ \\ && y &= x^x \\ &&&= e^{x \ln x}\\ && \frac{\d y}{\d x} &= e^{x \ln x} \cdot \left ( \ln x + x \cdot \frac1x \right) \\ &&&= x^x \left (1 + \ln x \right) \\ \\ && y&= x^{(x^x)} \\ &&&= e^{x^ x \cdot \ln x} \\ && \frac{\d y}{\d x} &= e^{x^x \cdot \ln x} \left ( x^x \left (1 + \ln x \right) \cdot \ln x + x^x \cdot \frac1x\right) \\ &&&= x^{x^x} \left (x^x (1+ \ln x) \ln x +x^{x-1} \right) \\ &&&= x^{x^x+x-1} \left (1 + x \ln x + x (\ln x)^2 \right) \\ \\ && y &= (x^x)^x \\ &&&= x^{2x} \\ &&&= e^{2x \ln x} \\ && \frac{\d y}{\d x} &= e^{2 x \ln x} \left (2 \ln x + 2 \right) \\ &&&= 2(x^x)^x(1 + \ln x) \end{align*}

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1989 Paper 3 Q8
D: 1700.0 B: 1484.0
differential equation simultaneous differential equations second order elimination method exponential functions complementary function particular integral

Given that \[ \frac{\mathrm{d}x}{\mathrm{d}t}=4(x-y)\qquad\mbox{ and }\qquad\frac{\mathrm{d}y}{\mathrm{d}t}=x-12(\mathrm{e}^{2t}+\mathrm{e}^{-2t}), \] obtain a differential equation for \(x\) which does not contain \(y\). Hence, or otherwise, find \(x\) and \(y\) in terms of \(t\) given that \(x=y=0\) when \(t=0\).

Solution: \begin{align*} && \frac{\d x}{\d t} &= 4(x-y) \\ && \frac{\d y}{\d t} &= x - 12(e^{2t}+e^{-2t}) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} &= 4 \frac{\d x}{\d t}-4\frac{\d y}{\d t} \\ &&&= 4 \frac{\d x}{\d t}-4 \left ( x - 12(e^{2t}+e^{-2t}) \right) \\ \Rightarrow && \frac{\d^2 x}{\d t^2} - 4 \frac{\d x}{\d t}+4x &= 48 (e^{2t}+e^{-2t}) \end{align*} This differential equation has characteristic polynomial \(\lambda^2 - 4\lambda + 4 = (\lambda-2)^2\). Therefore we should expect a general solution of \((At+B)e^{2t}\). For particular integrals we should try \(ke^{-2t}\) and \(Ct^2 e^{2t}\). For the former, we have: \begin{align*} && 48 &= 4k+8k+k \\ \Rightarrow && k &= \frac{48}{13} \end{align*} For the latter we have: \begin{align*} &&4Ct^2e^{2t} -4C(2te^{2t}+2t^2e^{2t})+2C((1+2t)e^{2t}+2t^2e^{2t}) &= 48e^{2t} \\ \Rightarrow && 2C &= 48 \\ \Rightarrow && C &= 24 \end{align*} Therefore the solution should be: \begin{align*} x = (At+B)e^{2t} + \frac{48}{13}e^{-2t} + 24t^2 e^{2t} \\ x(0) = B + \frac{48}{13} \\ x'(0) = 2B+A-\frac{96}{13} \\ x =\frac{48}{13}((4t-1)e^{2t}+e^{-2t})+24t^2e^{2t} \\ y = x - \frac{1}{4} \frac{\d x}{\d t} \end{align*}

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