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2014
P1 Q10
Topical
DE
14 P1 Q10
2014 H2 Mathematics Paper 1 Question 10
Differential Equations (DEs)
Answers
(i)
k
=
−
1
5
{k=- \frac{1}{5}}
k
=
−
5
1
(ii)
t
=
5
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
{t = \sqrt{5} \ln \left( \frac{ \sqrt{5}-2x+1 }{ \sqrt{5}+2x-1} \right)}
t
=
5
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
(iiia)
5
ln
(
2
5
+
1
2
5
−
1
)
min
{\sqrt{5} \ln \left( \frac{ 2\sqrt{5}+1 }{ 2\sqrt{5}-1} \right) \textrm{ min}}
5
ln
(
2
5
−
1
2
5
+
1
)
min
(iiib)
2.152
min
{2.152 \textrm{ min}}
2.152
min
(iv)
x
=
(
5
+
1
)
e
−
5
5
t
+
1
−
5
2
(
1
+
e
−
5
5
t
)
{x = \frac{\left( \sqrt{5}+1 \right) \mathrm{e}^{-\frac{\sqrt{5}}{5}t} + 1 - \sqrt{5}}{2\left(1 + \mathrm{e}^{-\frac{\sqrt{5}}{5}t}\right)}}
x
=
2
(
1
+
e
−
5
5
t
)
(
5
+
1
)
e
−
5
5
t
+
1
−
5
Full solutions
(i)
When
x
=
1
2
,
{x=\frac{1}{2}, }
x
=
2
1
,
d
x
d
t
=
−
1
4
{\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{1}{4} }
d
t
d
x
=
−
4
1
−
1
4
=
k
(
1
+
1
2
−
(
1
2
)
2
)
5
4
k
=
−
1
4
k
=
−
1
5
■
\begin{gather*} {\textstyle - \frac{1}{4}} = k \left( 1 + {\textstyle \frac{1}{2} - \left( \frac{1}{2} \right)^2} \right) \\ \frac{5}{4} k = - \frac{1}{4} \\ k = - \frac{1}{5} \; \blacksquare \end{gather*}
−
4
1
=
k
(
1
+
2
1
−
(
2
1
)
2
)
4
5
k
=
−
4
1
k
=
−
5
1
■
(ii)
1
+
x
−
x
2
=
−
(
x
2
−
x
−
1
)
=
−
(
(
x
−
1
2
)
2
−
5
4
)
=
5
4
−
(
x
−
1
2
)
2
\begin{align*} & 1 + x - x^2 \\ &= -(x^2 - x - 1) \\ &= -\left(\left(x - \frac{1}{2}\right)^2 - \frac{5}{4}\right) \\ & = \frac{5}{4} - \left( x-\frac{1}{2} \right)^2 \end{align*}
1
+
x
−
x
2
=
−
(
x
2
−
x
−
1
)
=
−
(
(
x
−
2
1
)
2
−
4
5
)
=
4
5
−
(
x
−
2
1
)
2
d
x
d
t
=
−
1
5
(
1
+
x
−
x
2
)
1
5
4
−
(
x
−
1
2
)
2
d
x
d
t
=
−
1
5
∫
1
5
4
−
(
x
−
1
2
)
2
d
x
=
∫
−
1
5
d
t
\begin{gather*} \frac{\mathrm{d}x}{\mathrm{d}t} = - \frac{1}{5} (1 + x - x^2) \\ \frac{1}{\frac{5}{4} - \left( x-\frac{1}{2} \right)^2} \frac{\mathrm{d}x}{\mathrm{d}t} = - \frac{1}{5} \\ \int \frac{1}{\frac{5}{4} - \left( x-\frac{1}{2} \right)^2} \; \mathrm{d}x = \int - \frac{1}{5} \; \mathrm{d}t \\ \end{gather*}
d
t
d
x
=
−
5
1
(
1
+
x
−
x
2
)
4
5
−
(
x
−
2
1
)
2
1
d
t
d
x
=
−
5
1
∫
4
5
−
(
x
−
2
1
)
2
1
d
x
=
∫
−
5
1
d
t
Since
0
≤
x
≤
1
2
,
{0 \leq x \leq \frac{1}{2},}
0
≤
x
≤
2
1
,
1
2
5
2
ln
(
5
2
+
(
x
−
1
2
)
5
2
−
(
x
−
1
2
)
)
=
−
1
5
t
+
c
1
5
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
=
−
1
5
t
+
c
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
=
−
5
5
t
+
5
c
5
+
2
x
−
1
5
−
2
x
+
1
=
e
−
5
5
t
+
5
c
5
+
2
x
−
1
5
−
2
x
+
1
=
A
e
−
5
5
t
\begin{gather*} \frac{1}{2\frac{\sqrt{5}}{2}} \ln \left( \frac{ \frac{\sqrt{5}}{2}+\left(x-\frac{1}{2}\right)}{ \frac{\sqrt{5}}{2}-\left(x-\frac{1}{2}\right)} \right) = - \frac{1}{5}t + c \\ \frac{1}{\sqrt{5}} \ln \left( \frac{ \sqrt{5}+2x-1}{ \sqrt{5}-2x+1} \right) = - \frac{1}{5}t + c \\ \ln \left( \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} \right) = -\frac{\sqrt{5}}{5}t + \sqrt{5}c \\ \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} = \mathrm{e}^{-\frac{\sqrt{5}}{5}t+\sqrt{5}c} \\ \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} = A\mathrm{e}^{-\frac{\sqrt{5}}{5}t} \\ \end{gather*}
2
2
5
1
ln
(
2
5
−
(
x
−
2
1
)
2
5
+
(
x
−
2
1
)
)
=
−
5
1
t
+
c
5
1
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
=
−
5
1
t
+
c
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
=
−
5
5
t
+
5
c
5
−
2
x
+
1
5
+
2
x
−
1
=
e
−
5
5
t
+
5
c
5
−
2
x
+
1
5
+
2
x
−
1
=
A
e
−
5
5
t
When
t
=
0
,
{t=0, }
t
=
0
,
x
=
1
2
,
{x=\frac{1}{2},}
x
=
2
1
,
5
+
2
(
1
2
)
−
1
5
−
2
(
1
2
)
+
1
=
A
e
−
5
5
(
0
)
A
=
5
5
A
=
1
\begin{gather*} \frac{\sqrt{5}+2\left(\frac{1}{2}\right)-1}{\sqrt{5}-2\left(\frac{1}{2}\right)+1} = A \mathrm{e}^{-\frac{\sqrt{5}}{5}(0)} \\ A = \frac{\sqrt{5}}{\sqrt{5}} \\ A = 1 \\ \end{gather*}
5
−
2
(
2
1
)
+
1
5
+
2
(
2
1
)
−
1
=
A
e
−
5
5
(
0
)
A
=
5
5
A
=
1
5
+
2
x
−
1
5
−
2
x
+
1
=
e
−
5
5
t
−
5
5
t
=
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
t
=
−
5
5
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
t
=
5
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
■
\begin{gather*} \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} = \mathrm{e}^{-\frac{\sqrt{5}}{5}t} \\ -\frac{\sqrt{5}}{5}t = \ln \left( \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} \right) \\ t = -\frac{5}{\sqrt{5}}\ln \left( \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} \right) \\ t = \sqrt{5} \ln \left( \frac{ \sqrt{5}-2x+1 }{ \sqrt{5}+2x-1} \right) \; \blacksquare \end{gather*}
5
−
2
x
+
1
5
+
2
x
−
1
=
e
−
5
5
t
−
5
5
t
=
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
t
=
−
5
5
ln
(
5
−
2
x
+
1
5
+
2
x
−
1
)
t
=
5
ln
(
5
+
2
x
−
1
5
−
2
x
+
1
)
■
(iiia)
When
x
=
1
2
(
1
2
)
=
1
4
,
{x=\frac{1}{2}\left(\frac{1}{2}\right)=\frac{1}{4},}
x
=
2
1
(
2
1
)
=
4
1
,
t
=
5
ln
(
5
−
2
(
1
4
)
+
1
5
+
2
(
1
4
)
−
1
)
=
5
ln
(
5
+
1
2
5
−
1
2
)
=
5
ln
(
2
5
+
1
2
5
−
1
)
■
\begin{align*} t &= \sqrt{5} \ln \left( \frac{ \sqrt{5}-2\left(\frac{1}{4}\right)+1 }{ \sqrt{5}+2\left(\frac{1}{4}\right)-1} \right) \\ &= \sqrt{5} \ln \left( \frac{ \sqrt{5}+\frac{1}{2} }{ \sqrt{5}- \frac{1}{2}} \right) \\ &= \sqrt{5} \ln \left( \frac{ 2\sqrt{5}+1 }{ 2\sqrt{5}-1} \right) \; \blacksquare \end{align*}
t
=
5
ln
(
5
+
2
(
4
1
)
−
1
5
−
2
(
4
1
)
+
1
)
=
5
ln
(
5
−
2
1
5
+
2
1
)
=
5
ln
(
2
5
−
1
2
5
+
1
)
■
(iiib)
When
x
=
0
,
{x=0,}
x
=
0
,
t
=
5
ln
(
5
−
2
(
0
)
+
1
5
+
2
(
0
)
−
1
)
=
5
ln
(
5
+
1
5
−
1
)
=
2.152
min (3 dp)
■
\begin{align*} t &= \sqrt{5} \ln \left( \frac{ \sqrt{5}-2(0)+1 }{ \sqrt{5}+2(0)-1} \right) \\ &= \sqrt{5} \ln \left( \frac{ \sqrt{5}+1 }{ \sqrt{5}-1} \right) \\ &= 2.152 \textrm{ min} \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}
t
=
5
ln
(
5
+
2
(
0
)
−
1
5
−
2
(
0
)
+
1
)
=
5
ln
(
5
−
1
5
+
1
)
=
2.152
min
(3 dp)
■
(iv)
5
+
2
x
−
1
5
−
2
x
+
1
=
e
−
5
5
t
5
+
2
x
−
1
=
(
5
−
2
x
+
1
)
e
−
5
5
t
2
x
(
1
+
e
−
5
5
t
)
=
(
5
+
1
)
e
−
5
5
t
+
1
−
5
x
=
(
5
+
1
)
e
−
5
5
t
+
1
−
5
2
(
1
+
e
−
5
5
t
)
■
\begin{gather*} \frac{ \sqrt{5}+2x-1 }{ \sqrt{5}-2x+1} = \mathrm{e}^{-\frac{\sqrt{5}}{5}t} \\ \sqrt{5}+2x-1 = \left( \sqrt{5}-2x+1 \right) \mathrm{e}^{-\frac{\sqrt{5}}{5}t} \\ 2x\left( 1 + \mathrm{e}^{-\frac{\sqrt{5}}{5}t} \right) = \left( \sqrt{5}+1 \right) \mathrm{e}^{-\frac{\sqrt{5}}{5}t} + 1 - \sqrt{5} \\ x = \frac{\left( \sqrt{5}+1 \right) \mathrm{e}^{-\frac{\sqrt{5}}{5}t} + 1 - \sqrt{5}}{2\left(1 + \mathrm{e}^{-\frac{\sqrt{5}}{5}t}\right)} \; \blacksquare \end{gather*}
5
−
2
x
+
1
5
+
2
x
−
1
=
e
−
5
5
t
5
+
2
x
−
1
=
(
5
−
2
x
+
1
)
e
−
5
5
t
2
x
(
1
+
e
−
5
5
t
)
=
(
5
+
1
)
e
−
5
5
t
+
1
−
5
x
=
2
(
1
+
e
−
5
5
t
)
(
5
+
1
)
e
−
5
5
t
+
1
−
5
■
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