2014 H2 Mathematics Paper 1 Question 10

Differential Equations (DEs)

Answers

(i)

k=−15{k=- \frac{1}{5}}

(ii)

t=5ln⁡(5−2x+15+2x−1){t = \sqrt{5} \ln \left( \frac{ \sqrt{5}-2x+1 }{ \sqrt{5}+2x-1} \right)}
(iiia)
5ln⁡(25+125−1) min{\sqrt{5} \ln \left( \frac{ 2\sqrt{5}+1 }{ 2\sqrt{5}-1} \right) \textrm{ min}}
(iiib)
2.152 min{2.152 \textrm{ min}}

(iv)

x=(5+1)e−55t+1−52(1+e−55t){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)}}

Full solutions

(i)

When x=12,{x=\frac{1}{2}, } dxdt=−14{\frac{\mathrm{d}x}{\mathrm{d}t}=-\frac{1}{4} }
−14=k(1+12−(12)2)54k=−14k=−15  ■\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*}

(ii)

1+x−x2=−(x2−x−1)=−((x−12)2−54)=54−(x−12)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*}
dxdt=−15(1+x−x2)154−(x−12)2dxdt=−15∫154−(x−12)2  dx=∫−15  dt\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*}
Since 0≤x≤12,{0 \leq x \leq \frac{1}{2},}
1252ln⁡(52+(x−12)52−(x−12))=−15t+c15ln⁡(5+2x−15−2x+1)=−15t+cln⁡(5+2x−15−2x+1)=−55t+5c5+2x−15−2x+1=e−55t+5c5+2x−15−2x+1=Ae−55t\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*}
When t=0,{t=0, } x=12,{x=\frac{1}{2},}
5+2(12)−15−2(12)+1=Ae−55(0)A=55A=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+2x−15−2x+1=e−55t−55t=ln⁡(5+2x−15−2x+1)t=−55ln⁡(5+2x−15−2x+1)t=5ln⁡(5−2x+15+2x−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*}
(iiia)
When x=12(12)=14,{x=\frac{1}{2}\left(\frac{1}{2}\right)=\frac{1}{4},}
t=5ln⁡(5−2(14)+15+2(14)−1)=5ln⁡(5+125−12)=5ln⁡(25+125−1)  ■\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*}
(iiib)
When x=0,{x=0,}
t=5ln⁡(5−2(0)+15+2(0)−1)=5ln⁡(5+15−1)=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*}

(iv)

5+2x−15−2x+1=e−55t5+2x−1=(5−2x+1)e−55t2x(1+e−55t)=(5+1)e−55t+1−5x=(5+1)e−55t+1−52(1+e−55t)  ■\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*}