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12 P2 Q1
2012 H2 Mathematics Paper 2 Question 1
Differential Equations (DEs)
Answers
(a)
y
=
8
x
2
−
3
4
x
4
+
c
x
+
d
{y = 8 x^2 - \frac{3}{4} x^4+cx+d}
y
=
8
x
2
−
4
3
x
4
+
c
x
+
d
(b)
t
=
1
24
ln
(
4
+
3
u
7
(
4
−
3
u
)
)
{t = \frac{1}{24} \ln \left( \frac{4+3u}{7(4-3u)} \right) }
t
=
24
1
ln
(
7
(
4
−
3
u
)
4
+
3
u
)
Full solutions
(a)
d
2
y
d
x
2
=
16
−
9
x
2
d
y
d
x
=
16
x
−
3
x
3
+
c
y
=
8
x
2
−
3
4
x
4
+
c
x
+
d
■
\begin{gather*} \frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} = 16 - 9 x^2 \\ \frac{\mathrm{d}y}{\mathrm{d}x} = 16 x - 3 x^3 + c \\ y = 8 x^2 - \frac{3}{4} x^4 + cx + d \; \blacksquare \end{gather*}
d
x
2
d
2
y
=
16
−
9
x
2
d
x
d
y
=
16
x
−
3
x
3
+
c
y
=
8
x
2
−
4
3
x
4
+
c
x
+
d
■
(b)
d
u
d
t
=
16
−
9
u
2
∫
1
16
−
9
u
2
d
u
=
∫
1
d
t
1
9
∫
1
16
9
−
u
2
d
u
=
∫
1
d
t
1
9
⋅
1
2
(
4
3
)
ln
∣
4
3
+
u
4
3
−
u
∣
=
t
+
C
1
24
ln
∣
4
+
3
u
4
−
3
u
∣
=
t
+
C
ln
∣
4
+
3
u
4
−
3
u
∣
=
24
t
+
24
C
∣
4
+
3
u
4
−
3
u
∣
=
e
24
t
+
24
C
\begin{gather*} \frac{\mathrm{d}u}{\mathrm{d}t} = 16 - 9 u^2 \\ \int \frac{1}{16 - 9 u^2} \; \mathrm{d}u = \int 1 \; \mathrm{d}t \\ \frac{1}{9} \int \frac{1}{\frac{16}{9} - u^2} \; \mathrm{d}u = \int 1 \; \mathrm{d}t \\ \frac{1}{9} \cdot \frac{1}{2\left(\frac{4}{3}\right)}\ln \left| \frac{\frac{4}{3}+u}{\frac{4}{3}-u} \right| = t + C \\ \frac{1}{24} \ln \left| \frac{4 + 3 u}{4 - 3 u} \right| = t + C \\ \ln \left| \frac{4+3u}{4-3u} \right| = 24t + 24C \\ \left| \frac{4+3u}{4-3u} \right| = \mathrm{e}^{24t + 24C} \\ \end{gather*}
d
t
d
u
=
16
−
9
u
2
∫
16
−
9
u
2
1
d
u
=
∫
1
d
t
9
1
∫
9
16
−
u
2
1
d
u
=
∫
1
d
t
9
1
⋅
2
(
3
4
)
1
ln
3
4
−
u
3
4
+
u
=
t
+
C
24
1
ln
4
−
3
u
4
+
3
u
=
t
+
C
ln
4
−
3
u
4
+
3
u
=
24
t
+
24
C
4
−
3
u
4
+
3
u
=
e
24
t
+
24
C
4
+
3
u
4
−
3
u
=
±
e
24
C
e
24
t
=
A
e
24
t
\begin{align*} \frac{4+3u}{4-3u} &= \pm \mathrm{e}^{24C} \mathrm{e}^{24t} \\ &= A \mathrm{e}^{24t} \end{align*}
4
−
3
u
4
+
3
u
=
±
e
24
C
e
24
t
=
A
e
24
t
When
t
=
0
,
{t=0, }
t
=
0
,
u
=
1
{u=1}
u
=
1
4
+
3
(
1
)
4
−
3
(
1
)
=
A
e
24
(
0
)
A
=
7
\begin{gather*} \frac{4+3(1)}{4-3(1)} = A \mathrm{e}^{24(0)} \\ A = 7 \end{gather*}
4
−
3
(
1
)
4
+
3
(
1
)
=
A
e
24
(
0
)
A
=
7
4
+
3
u
4
−
3
u
=
7
e
24
t
e
24
t
=
4
+
3
u
7
(
4
−
3
u
)
24
t
=
ln
(
4
+
3
u
7
(
4
−
3
u
)
)
t
=
1
24
ln
(
4
+
3
u
7
(
4
−
3
u
)
)
■
\begin{gather*} \frac{4+3u}{4-3u} = 7 \mathrm{e}^{24t} \\ \mathrm{e}^{24t} = \frac{4+3u}{7(4-3u)} \\ 24 t = \ln \left( \frac{4+3u}{7(4-3u)} \right) \\ t = \frac{1}{24} \ln \left( \frac{4+3u}{7(4-3u)} \right) \; \blacksquare \end{gather*}
4
−
3
u
4
+
3
u
=
7
e
24
t
e
24
t
=
7
(
4
−
3
u
)
4
+
3
u
24
t
=
ln
(
7
(
4
−
3
u
)
4
+
3
u
)
t
=
24
1
ln
(
7
(
4
−
3
u
)
4
+
3
u
)
■
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