The differential equation
d 2 x d t 2 + 6 d x d t + 13 x = 8 e − 3 t t ⩾ 0 \begin{align*}
\frac{\mathrm{d}^2 x}{\mathrm{d}t^2} + 6\frac{\mathrm{d}x}{\mathrm{d}t} + 13x = 8\mathrm{e}^{-3t} \qquad t \geqslant 0
\end{align*} d t 2 d 2 x + 6 d t d x + 13 x = 8 e − 3 t t ⩾ 0 describes the motion of a particle along the x x x
(a) Determine the general solution of this differential equation.
(6)
Given that the motion of the particle satisfies x = 1 2 x = \frac{1}{2} x = 2 1 d x d t = 1 2 \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2} d t d x = 2 1 t = 0 t = 0 t = 0
(b) determine the particular solution for the motion of the particle.
(4)
On the graph of the particular solution found in part (b), the first turning point for t > 0 t > 0 t > 0 x = a x = a x = a
(c) Determine, to 3 significant figures, the value of a a a
[Solutions relying entirely on calculator technology are not acceptable.]
(4)
解答
(a)
To find the general solution of the differential equation d 2 x d t 2 + 6 d x d t + 13 x = 8 e − 3 t \frac{\mathrm{d}^2x}{\mathrm{d}t^2} + 6\frac{\mathrm{d}x}{\mathrm{d}t} + 13x = 8\mathrm{e}^{-3t} d t 2 d 2 x + 6 d t d x + 13 x = 8 e − 3 t
m 2 + 6 m + 13 = 0 ( m + 3 ) 2 + 4 = 0 m = − 3 ± 2 i \begin{align*}
m^2 + 6m + 13
= &\,0\\[4mm]
(m+3)^2 + 4
= &\,0\\[4mm]
m
= &\,-3 \pm 2\mathrm{i}
\end{align*} m 2 + 6 m + 13 = ( m + 3 ) 2 + 4 = m = 0 0 − 3 ± 2 i The CF is x c = e − 3 t ( A cos 2 t + B sin 2 t ) x_c = \mathrm{e}^{-3t}(A\cos 2t + B\sin 2t) x c = e − 3 t ( A cos 2 t + B sin 2 t )
For the particular integral (PI), let x p = λ e − 3 t x_p = \lambda \mathrm{e}^{-3t} x p = λ e − 3 t x p x_p x p
x p ′ = − 3 λ e − 3 t x p ′ ′ = 9 λ e − 3 t \begin{align*}
x_p'
= &\,-3\lambda \mathrm{e}^{-3t}\\[4mm]
x_p''
= &\,9\lambda \mathrm{e}^{-3t}
\end{align*} x p ′ = x p ′′ = − 3 λ e − 3 t 9 λ e − 3 t Substitute these into the differential equation:
9 λ e − 3 t + 6 ( − 3 λ e − 3 t ) + 13 ( λ e − 3 t ) = 8 e − 3 t ( 9 λ − 18 λ + 13 λ ) e − 3 t = 8 e − 3 t 4 λ = 8 λ = 2 \begin{align*}
9\lambda \mathrm{e}^{-3t} + 6(-3\lambda \mathrm{e}^{-3t}) + 13(\lambda \mathrm{e}^{-3t})
= &\,8\mathrm{e}^{-3t}\\[4mm]
(9\lambda - 18\lambda + 13\lambda)\mathrm{e}^{-3t}
= &\,8\mathrm{e}^{-3t}\\[4mm]
4\lambda
= &\,8\\[4mm]
\lambda
= &\,2
\end{align*} 9 λ e − 3 t + 6 ( − 3 λ e − 3 t ) + 13 ( λ e − 3 t ) = ( 9 λ − 18 λ + 13 λ ) e − 3 t = 4 λ = λ = 8 e − 3 t 8 e − 3 t 8 2 The PI is x p = 2 e − 3 t x_p = 2\mathrm{e}^{-3t} x p = 2 e − 3 t x = x c + x p x = x_c + x_p x = x c + x p
x = e − 3 t ( A cos 2 t + B sin 2 t ) + 2 e − 3 t \begin{align*}
x
= &\,\mathrm{e}^{-3t}(A\cos 2t + B\sin 2t) + 2\mathrm{e}^{-3t}
\end{align*} x = e − 3 t ( A cos 2 t + B sin 2 t ) + 2 e − 3 t
(b)
Given that at t = 0 t=0 t = 0 x = 1 2 x=\frac{1}{2} x = 2 1
1 2 = e 0 ( A cos 0 + B sin 0 ) + 2 e 0 1 2 = A + 2 A = − 3 2 \begin{align*}
\frac{1}{2}
= &\,\mathrm{e}^0(A\cos 0 + B\sin 0) + 2\mathrm{e}^0\\[4mm]
\frac{1}{2}
= &\,A + 2\\[4mm]
A
= &\,-\frac{3}{2}
\end{align*} 2 1 = 2 1 = A = e 0 ( A cos 0 + B sin 0 ) + 2 e 0 A + 2 − 2 3 Next, differentiate the general solution to use the second condition.
d x d t = − 3 e − 3 t ( A cos 2 t + B sin 2 t ) + e − 3 t ( − 2 A sin 2 t + 2 B cos 2 t ) − 6 e − 3 t \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}t}
= &\,-3\mathrm{e}^{-3t}(A\cos 2t + B\sin 2t) + \mathrm{e}^{-3t}(-2A\sin 2t + 2B\cos 2t) - 6\mathrm{e}^{-3t}
\end{align*} d t d x = − 3 e − 3 t ( A cos 2 t + B sin 2 t ) + e − 3 t ( − 2 A sin 2 t + 2 B cos 2 t ) − 6 e − 3 t Given that at t = 0 t=0 t = 0 d x d t = 1 2 \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2} d t d x = 2 1
1 2 = − 3 ( A cos 0 + B sin 0 ) + ( − 2 A sin 0 + 2 B cos 0 ) − 6 1 2 = − 3 A + 2 B − 6 \begin{align*}
\frac{1}{2}
= &\,-3(A\cos 0 + B\sin 0) + (-2A\sin 0 + 2B\cos 0) - 6\\[4mm]
\frac{1}{2}
= &\,-3A + 2B - 6
\end{align*} 2 1 = 2 1 = − 3 ( A cos 0 + B sin 0 ) + ( − 2 A sin 0 + 2 B cos 0 ) − 6 − 3 A + 2 B − 6 Substitute A = − 3 2 A = -\frac{3}{2} A = − 2 3
1 2 = − 3 ( − 3 2 ) + 2 B − 6 1 2 = 9 2 + 2 B − 6 1 2 = − 3 2 + 2 B 2 = 2 B B = 1 \begin{align*}
\frac{1}{2}
= &\,-3\left(-\frac{3}{2}\right) + 2B - 6\\[4mm]
\frac{1}{2}
= &\,\frac{9}{2} + 2B - 6\\[4mm]
\frac{1}{2}
= &\,-\frac{3}{2} + 2B\\[4mm]
2
= &\,2B\\[4mm]
B
= &\,1
\end{align*} 2 1 = 2 1 = 2 1 = 2 = B = − 3 ( − 2 3 ) + 2 B − 6 2 9 + 2 B − 6 − 2 3 + 2 B 2 B 1 The particular solution is:
x = e − 3 t ( − 3 2 cos 2 t + sin 2 t ) + 2 e − 3 t \begin{align*}
x
= &\,\mathrm{e}^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) + 2\mathrm{e}^{-3t}
\end{align*} x = e − 3 t ( − 2 3 cos 2 t + sin 2 t ) + 2 e − 3 t
(c)
For turning points, we set d x d t = 0 \frac{\mathrm{d}x}{\mathrm{d}t} = 0 d t d x = 0 d x d t \frac{\mathrm{d}x}{\mathrm{d}t} d t d x A A A B B B
d x d t = − 3 e − 3 t ( − 3 2 cos 2 t + sin 2 t ) + e − 3 t ( − 2 ( − 3 2 ) sin 2 t + 2 ( 1 ) cos 2 t ) − 6 e − 3 t = e − 3 t ( 9 2 cos 2 t − 3 sin 2 t + 3 sin 2 t + 2 cos 2 t − 6 ) = e − 3 t ( 13 2 cos 2 t − 6 ) \begin{align*}
\frac{\mathrm{d}x}{\mathrm{d}t}
= &\,-3\mathrm{e}^{-3t}\left(-\frac{3}{2}\cos 2t + \sin 2t\right) + \mathrm{e}^{-3t}\left(-2\left(-\frac{3}{2}\right)\sin 2t + 2(1)\cos 2t\right) - 6\mathrm{e}^{-3t}\\[4mm]
= &\,\mathrm{e}^{-3t}\left(\frac{9}{2}\cos 2t - 3\sin 2t + 3\sin 2t + 2\cos 2t - 6\right)\\[4mm]
= &\,\mathrm{e}^{-3t}\left(\frac{13}{2}\cos 2t - 6\right)
\end{align*} d t d x = = = − 3 e − 3 t ( − 2 3 cos 2 t + sin 2 t ) + e − 3 t ( − 2 ( − 2 3 ) sin 2 t + 2 ( 1 ) cos 2 t ) − 6 e − 3 t e − 3 t ( 2 9 cos 2 t − 3 sin 2 t + 3 sin 2 t + 2 cos 2 t − 6 ) e − 3 t ( 2 13 cos 2 t − 6 ) Set this to zero:
e − 3 t ( 13 2 cos 2 t − 6 ) = 0 13 2 cos 2 t = 6 cos 2 t = 12 13 \begin{align*}
\mathrm{e}^{-3t}\left(\frac{13}{2}\cos 2t - 6\right)
= &\,0\\[4mm]
\frac{13}{2}\cos 2t
= &\,6\\[4mm]
\cos 2t
= &\,\frac{12}{13}
\end{align*} e − 3 t ( 2 13 cos 2 t − 6 ) = 2 13 cos 2 t = cos 2 t = 0 6 13 12 The first turning point for t > 0 t > 0 t > 0 cos 2 t = 12 13 \cos 2t = \frac{12}{13} cos 2 t = 13 12 sin 2 t = 1 − ( 12 13 ) 2 = 5 13 \sin 2t = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} sin 2 t = 1 − ( 13 12 ) 2 = 13 5 t t t
t = 1 2 arccos ( 12 13 ) ≈ 0.197395... t = \frac{1}{2} \arccos\left(\frac{12}{13}\right) \approx 0.197395... t = 2 1 arccos ( 13 12 ) ≈ 0.197395... Now calculate the x x x a a a cos 2 t \cos 2t cos 2 t sin 2 t \sin 2t sin 2 t t t t
a = e − 3 t ( − 3 2 ( 12 13 ) + 5 13 ) + 2 e − 3 t = e − 3 t ( − 18 13 + 5 13 + 26 13 ) = e − 3 t ( 13 13 ) = e − 3 t \begin{align*}
a
= &\,\mathrm{e}^{-3t}\left(-\frac{3}{2}\left(\frac{12}{13}\right) + \frac{5}{13}\right) + 2\mathrm{e}^{-3t}\\[4mm]
= &\,\mathrm{e}^{-3t}\left(-\frac{18}{13} + \frac{5}{13} + \frac{26}{13}\right)\\[4mm]
= &\,\mathrm{e}^{-3t}\left(\frac{13}{13}\right)\\[4mm]
= &\,\mathrm{e}^{-3t}
\end{align*} a = = = = e − 3 t ( − 2 3 ( 13 12 ) + 13 5 ) + 2 e − 3 t e − 3 t ( − 13 18 + 13 5 + 13 26 ) e − 3 t ( 13 13 ) e − 3 t Substitute t ≈ 0.197395... t \approx 0.197395... t ≈ 0.197395...
a = e − 3 ( 0.197395... ) ≈ 0.553 (to 3 s.f.) \begin{align*}
a
= &\,\mathrm{e}^{-3(0.197395...)}\\[4mm]
\approx &\,0.553 \text{ (to 3 s.f.)}
\end{align*} a = ≈ e − 3 ( 0.197395... ) 0.553 (to 3 s.f.) 