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2019 Paper 3 Q1
D: 1500.0
B: 1500.0
differential equation
second order
coupled differential equations
curve sketching
parametric
auxiliary equation
repeated roots
particle path
The coordinates of a particle at time \(t\) are \(x\) and \(y\). For \(t \geq 0\), they satisfy the pair of coupled differential equations \[ \begin{cases} \dot{x} &= -x -ky \\ \dot{y} &= x - y \end{cases}\] where \(k\) is a constant. When \(t = 0\), \(x = 1\) and \(y = 0\).
- Let \(k = 1\). Find \(x\) and \(y\) in terms of \(t\) and sketch \(y\) as a function of \(t\). Sketch the path of the particle in the \(x\)-\(y\) plane, giving the coordinates of the point at which \(y\) is greatest and the coordinates of the point at which \(x\) is least.
- Instead, let \(k = 0\). Find \(x\) and \(y\) in terms of \(t\) and sketch the path of the particle in the \(x\)-\(y\) plane.
Solution:
- Let \(k = 1\), then
\begin{align*}
\dot{x} &= - x - y \\
\dot{y} &= x - y \\
\dot{x}-\dot{y} &= -2x \\
\ddot{x} &= -\dot{x}-\dot{y} \\
&= -\dot{x} - (\dot{x}+2x) \\
&= -2\dot{x}- 2x \\
\dot{x}+\dot{y} &= -2y \\
\ddot{y} &= \dot{x}-\dot{y} \\
&= -2y-2\dot{y}
\end{align*}
So we have an auxiliary equation for \(x\) and \(y\) which is \(\lambda^2 + 2 \lambda+2 = 0 \Rightarrow \lambda = -1 \pm i\).
Therefore \(x = Ae^{-t} \cos t + B e^{-t} \sin t, y = Ce^{-t}\cos t + De^{-t} \sin t\). We also must have that, \(A = 1, C = 0\), so \(x = e^{-t} \cos t + Be^{-t} \sin t\) and \(y = De^{-t} \sin x\).
\begin{align*}
\dot{y} &= -De^{-t} \sin t +De^{-t} \cos t \\
&= e^{-t} \cos x + Be^{-t} \sin t- De^{-t} \sin t \\
\end{align*}
therefore \(B = 0, D = 1\) and \(x = e^{-t} \cos t, y = e^{-t} \sin t\)
\begin{align*} y &= e^{-t} \sin t \\ \dot{y} &= -e^{-t} \sin t + e^{-t} \cos t \\ \dot{x} &= e^{-t} \cos t -e^{-t} \sin t \end{align*}
- \begin{align*}
\dot{x} = -x \\
\dot{y} = x-y
\end{align*}
So \(x = e^{-t}\). \(\dot{y} + y = e^{-t}\) so \(y = (t+B)e^{-t}\) and so \(y =te^{-t}\).
