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2016
P1 Q9
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DE
16 P1 Q9
2016 H2 Mathematics Paper 1 Question 9
Differential Equations (DEs)
Answers
(ib)
y
=
5
−
5
e
−
2
t
{y = 5 - 5 \mathrm{e}^{-2t}}
y
=
5
−
5
e
−
2
t
x
=
5
t
+
5
2
e
−
2
t
−
5
2
{x = 5t + \frac{5}{2}\mathrm{e}^{-2t} - \frac{5}{2}}
x
=
5
t
+
2
5
e
−
2
t
−
2
5
(ii)
x
=
5
t
2
−
10
t
+
20
sin
1
2
t
x = 5 t^2 - 10 t + 20 \sin \frac{1}{2} t
x
=
5
t
2
−
10
t
+
20
sin
2
1
t
(iii)
t
=
1.47
s
{t = 1.47 \textrm{ s}}
t
=
1.47
s
t
=
1.05
s
{t=1.05 \textrm{ s}}
t
=
1.05
s
Full solutions
(ia)
y
=
d
x
d
t
d
y
d
t
=
d
2
x
d
t
2
\begin{align*} y &= \frac{\mathrm{d}x}{\mathrm{d}t} \\ \frac{\mathrm{d}y}{\mathrm{d}t} &= \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} \end{align*}
y
d
t
d
y
=
d
t
d
x
=
d
t
2
d
2
x
Substituting
y
=
d
x
d
t
{\displaystyle y=\frac{\mathrm{d}x}{\mathrm{d}t}}
y
=
d
t
d
x
and
d
y
d
t
=
d
2
x
d
t
2
{\displaystyle \frac{\mathrm{d}y}{\mathrm{d}t}=\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}}
d
t
d
y
=
d
t
2
d
2
x
into the original differential equation,
d
y
d
t
+
2
y
=
10
d
y
d
t
=
10
−
2
y
■
\begin{gather*} \frac{\mathrm{d}y}{\mathrm{d}t} + 2 y = 10 \\ \frac{\mathrm{d}y}{\mathrm{d}t} = 10 - 2y \; \blacksquare \end{gather*}
d
t
d
y
+
2
y
=
10
d
t
d
y
=
10
−
2
y
■
(ib)
d
y
d
t
=
10
−
2
y
1
10
−
2
y
d
y
d
t
=
1
∫
1
10
−
2
y
d
y
=
∫
1
d
t
−
1
2
ln
∣
10
−
2
y
∣
=
t
+
c
ln
∣
10
−
2
y
∣
=
−
2
t
−
2
c
∣
10
−
2
y
∣
=
e
−
2
t
−
2
c
∣
10
−
2
y
∣
=
e
−
2
c
e
−
2
t
10
−
2
y
=
A
e
−
2
t
\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}t} &= 10 - 2y \\ \frac{1}{10 - 2 y} \frac{\mathrm{d}y}{\mathrm{d}t} &= 1 \\ \int \frac{1}{10 - 2 y} \;\mathrm{d}y &= \int 1 \; \mathrm{d}t \\ - \frac{1}{2} \ln \left| 10 - 2 y\right| &= t + c \\ \ln |10 - 2y | &= -2t-2c \\ |10 - 2y | &= \mathrm{e}^{-2t-2c} \\ |10 - 2y | &= \mathrm{e}^{-2c}\mathrm{e}^{-2t} \\ 10 - 2y &= A\mathrm{e}^{-2t} \\ \end{align*}
d
t
d
y
10
−
2
y
1
d
t
d
y
∫
10
−
2
y
1
d
y
−
2
1
ln
∣
10
−
2
y
∣
ln
∣10
−
2
y
∣
∣10
−
2
y
∣
∣10
−
2
y
∣
10
−
2
y
=
10
−
2
y
=
1
=
∫
1
d
t
=
t
+
c
=
−
2
t
−
2
c
=
e
−
2
t
−
2
c
=
e
−
2
c
e
−
2
t
=
A
e
−
2
t
When
t
=
0
,
y
=
d
x
d
t
=
0
,
{t=0, y=\frac{\mathrm{d}x}{\mathrm{d}t}=0,}
t
=
0
,
y
=
d
t
d
x
=
0
,
10
−
2
(
0
)
=
A
e
−
2
(
0
)
A
=
10
\begin{gather*} 10-2(0) = A\mathrm{e}^{-2(0)} \\ A = 10 \end{gather*}
10
−
2
(
0
)
=
A
e
−
2
(
0
)
A
=
10
10
−
2
y
=
10
e
−
2
t
y
=
5
−
5
e
−
2
t
■
\begin{gather*} 10-2y = 10\mathrm{e}^{-2t} \\ y = 5 - 5 \mathrm{e}^{-2t} \; \blacksquare \end{gather*}
10
−
2
y
=
10
e
−
2
t
y
=
5
−
5
e
−
2
t
■
d
x
d
t
=
5
−
5
e
−
2
t
x
=
∫
(
5
−
5
e
−
2
t
)
d
t
x
=
5
t
+
5
2
e
−
2
t
+
C
\begin{gather*} \frac{\mathrm{d}x}{\mathrm{d}t} = 5 - 5 \mathrm{e}^{-2t} \\ x = \int \left( 5 - 5 \mathrm{e}^{-2t} \right) \; \mathrm{d}t \\ x = 5t + \frac{5}{2}\mathrm{e}^{-2t} + C \\ \end{gather*}
d
t
d
x
=
5
−
5
e
−
2
t
x
=
∫
(
5
−
5
e
−
2
t
)
d
t
x
=
5
t
+
2
5
e
−
2
t
+
C
When
t
=
0
,
x
=
0
,
{t=0, x=0,}
t
=
0
,
x
=
0
,
0
=
5
(
0
)
+
5
2
e
−
2
(
0
)
+
C
C
=
−
5
2
\begin{gather*} 0 = 5(0) + \frac{5}{2}\mathrm{e}^{-2(0)} + C \\ C = -\frac{5}{2} \end{gather*}
0
=
5
(
0
)
+
2
5
e
−
2
(
0
)
+
C
C
=
−
2
5
x
=
5
t
+
5
2
e
−
2
t
−
5
2
■
x = 5t + \frac{5}{2}\mathrm{e}^{-2t} - \frac{5}{2} \; \blacksquare
x
=
5
t
+
2
5
e
−
2
t
−
2
5
■
(ii)
d
2
x
d
t
2
=
10
−
5
sin
1
2
t
d
x
d
t
=
10
t
+
10
cos
1
2
t
+
c
1
\begin{gather*} \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = 10 - 5 \sin \frac{1}{2} t \\ \frac{\mathrm{d}x}{\mathrm{d}t} = 10 t + 10 \cos \frac{1}{2} t + c_1 \\ \end{gather*}
d
t
2
d
2
x
=
10
−
5
sin
2
1
t
d
t
d
x
=
10
t
+
10
cos
2
1
t
+
c
1
When
t
=
0
,
{t=0, }
t
=
0
,
d
x
d
t
=
0
,
{\frac{\mathrm{d}x}{\mathrm{d}t}=0,}
d
t
d
x
=
0
,
0
=
10
(
0
)
+
10
cos
0
+
c
1
c
1
=
−
10
;
\begin{gather*} 0 = 10(0) + 10 \cos 0 + c_1 \\ c_1 = -10; \end{gather*}
0
=
10
(
0
)
+
10
cos
0
+
c
1
c
1
=
−
10
;
d
x
d
t
=
10
t
−
10
+
10
cos
1
2
t
x
=
5
t
2
−
10
t
+
20
sin
1
2
t
+
c
2
\begin{gather*} \frac{\mathrm{d}x}{\mathrm{d}t} = 10 t - 10 + 10 \cos \frac{1}{2} t \\ x = 5 t^2 - 10 t + 20 \sin \frac{1}{2} t + c_2 \\ \end{gather*}
d
t
d
x
=
10
t
−
10
+
10
cos
2
1
t
x
=
5
t
2
−
10
t
+
20
sin
2
1
t
+
c
2
When
t
=
0
,
{t=0, }
t
=
0
,
x
=
0
,
{x=0,}
x
=
0
,
0
=
5
(
0
)
2
−
10
(
0
)
+
20
sin
0
+
c
2
c
2
=
0
;
\begin{gather*} 0 = 5(0)^2 - 10(0) + 20 \sin 0 + c_2 \\ c_2 = 0; \end{gather*}
0
=
5
(
0
)
2
−
10
(
0
)
+
20
sin
0
+
c
2
c
2
=
0
;
x
=
5
t
2
−
10
t
+
20
sin
1
2
t
■
x = 5 t^2 - 10 t + 20 \sin \frac{1}{2} t \; \blacksquare
x
=
5
t
2
−
10
t
+
20
sin
2
1
t
■
(iii)
For the first model, when
x
=
5
,
{x=5,}
x
=
5
,
5
t
+
5
2
e
−
2
t
−
5
2
=
5
5t + \frac{5}{2}\mathrm{e}^{-2t} - \frac{5}{2} = 5
5
t
+
2
5
e
−
2
t
−
2
5
=
5
Using a GC,
t
=
1.47
s (2 dp)
■
t = 1.47 \textrm{ s} \textrm{ (2 dp)} \; \blacksquare
t
=
1.47
s
(2 dp)
■
For the second model, when
x
=
5
,
{x=5,}
x
=
5
,
5
t
2
−
10
t
+
20
sin
1
2
t
=
5
5 t^2 - 10 t + 20 \sin \frac{1}{2} t = 5
5
t
2
−
10
t
+
20
sin
2
1
t
=
5
Using a GC,
t
=
1.05
s (2 dp)
■
t = 1.05 \textrm{ s} \textrm{ (2 dp)} \; \blacksquare
t
=
1.05
s
(2 dp)
■
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