Consider the differential equatiml+s -fz8.k th5wabnx /4i3j mon $\frac{\mathrm{d} y}{\mathrm{~d} x}-e^{x}=1$
Given that y(0)=1 , use Euler's method with step length h=0.25 to find an approximation for y(1) . Give your answer correct to two decimal places.

≈   

  Solve the differential equ ,zd5hh,)jp gcation

$\left(1+x^{2}\right) \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x y^{2}$

for y , which satisfies the initial condition $y(0)=-\frac{1}{2} $ y =  (代数式) 

  Consider the differen.c jf*9tqg 2:qcb f;pak 2ip8vtial equation $ \frac{\mathrm{d} y}{\mathrm{~d} x}+\left(\frac{6 x}{3 x^{2}-2}\right) y=4 x$ , given that y=4 when x=0 .
1. Show that $3 x^{2}-2$ is an integrating factor for this differential equation.  (代数式) 
2. Hence solve this differential equation. Give the answer in the form y=f(x) .  (代数式) 

  There is a rumour spreading about the questions that will appear in an upcomidtdg45 awgr4)zxe/sl- 6 es0c ng chemistry exam in a class with a large number of students. Let x be the proportion of students who have heard sd z6sg4 ta5xr-4g c0d/)ee wlthe rumor and let t be the time in hours, after 10.00 a.m.

The situation can be modelled by the differential equation $ \frac{\mathrm{d} x}{\mathrm{~d} t}=k x(1-x)$ where k is a constant.
1. Use partial fractions to solve this differential equation and hence show that $\frac{x}{1-x}=A e^{k t} $, where A is a constant  (代数式) 
2. At 10.00 $\mathrm{a}$ .$ \mathrm{m}$ . one tenth of the students know about the rumour. Find the value of A   
3. At 12.00 p.m., the proportion of students who knew about the rumor is 0.55 . Find the value of k ≈   
4. Hence, find the proportion of students who knew about the rumour at 1.00 p.m.≈   

Solve the differential equatt p q2,/uhprxobc6jl)5 l;9iiion

$\sqrt{1-x^{2}} \frac{\mathrm{d} y}{\mathrm{~d} x}=\sqrt{1-y^{2}}$

for y , which satisfies the initial condition $y(0)=\frac{1}{2} $

1. Show that $y=\frac{1}{2 x^{2}} \int f(x) \mathrm{d} x$ is a solution of the differential equation

$2 x^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}+4 x y=f(x)$ .

2. Hence solve $2 x^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}+4 x y=\frac{1}{x}$, x>0 , given that y=2 when x=1 .

  Consider the differential equat pi6y)-ma prno6-*+yic8 koyk ion $\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{y}{x}=\frac{1}{2}$ , where x>0 .
1. Given that y(1)=2 , use Euler's method with step length h=0.5 to find an approximation for y(3) . Give your answer correct to two significant figures.≈   
2. Solve the equation $ \frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{y}{x}=\frac{1}{2}$ for y(1)=2 .  (代数式) 
3. Find the percentage error when y(3) is approximated by the final rounded value found in part (a). Give your answer correct to two significant figures.≈    %

Consider the different9 up c.pp6 8jx-1 i+ahj/e xddf*9suyzial equation

$x^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}+6 x^{2}=y^{2}$

for x>0 and y>3 x . It is given that y=4 when x=1 .
1. Use Euler's method, with a step length of 0.08 , to find an approximate value for y when x=1.4 .
2. Use the substitution y=v x to show that $x \frac{\mathrm{d} v}{\mathrm{~d} x}=v^{2}-v-6 $.
3. By solving the differential equation, show that $y=\frac{18 x+2 x^{6}}{6-x^{5}}$ .
4. 1. Find the actual value of y when x=1.4 .
2. Using the graph of $ y=\frac{18 x+2 x^{6}}{6-x^{5}} $, suggest a reason why the approximation given by Euler's method in part (a) is not a good estimate to the actual value of y at x=1.4

Consider the differential equatl;dq * ds+;nft/gbc/q ion $x \frac{\mathrm{d} y}{\mathrm{~d} x}+y=x^{p+1} $ where $x \in \mathbb{R}$, $x \neq 0 $ and p is a positive integer, p>0 .
1. Solve the differential equation given that y=1 when x=1 . Give your answer in the form y=f(x) .
2. 1. Show that the x -coordinate(s) of the points on the curve y=f(x) where $\frac{\mathrm{d} y}{\mathrm{~d} x}=0 $ satisfy the equation $x^{p+2}=1$ .
2. Deduce the set of values for p such that there are two points on the curve y=f(x) where $ \frac{\mathrm{d} y}{\mathrm{~d} x}=0 $. Give a reason for your answer.

  Consider the differential equatfom6, ;sfs:b)1ydfbxhf g1 9 cion $\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{2 x^{3}}{1+x^{4}} y=4 x^{3} $ where y=1 when x=0 .
1. Show that $\sqrt{1+x^{4}} $ is an integrating factor for this differential equation.  (代数式) 
2. Solve the differential equation giving your answer in the form y=f(x) .  (代数式) 

1. Consider the differential equatioevwo43 ,s+aqh n

$\frac{\mathrm{d} y}{\mathrm{~d} x}=f\left(\frac{y}{x}\right), \quad x\lt 0$ .

Use the substitution$ v=\frac{y}{x}$ to show that the general solution of this differential equation is

$\int \frac{\mathrm{d} v}{f(v)-v}=\ln x+C$

2. Hence, or otherwise, solve the differential equation

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x^{2}+5 x y+y^{2}}{x^{2}}, \quad x\lt 0$,

given that y=2 when x=1 . Give your answer in the form y=g(x) .

Consider the differential equationzr /72oo kn)dr $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x}{3 y^{2}}+x y $, where y=1 when x=0 .
1. Show that $ z=y^{3}$ transforms the differential equation into $ \frac{\mathrm{d} z}{\mathrm{~d} x}-3 x z=x $.
2. By solving this differential equation in z , obtain an expression for y in terms of x .

The curves y=f(x) and y=g(x) usv j/)hj m/t1 both pass through the point (1,0) and are defined by the diffe )/vt usjmj/1hrential equations $ \frac{\mathrm{d} y}{\mathrm{~d} x}=2 x-y^{2}$ and $\frac{\mathrm{d} y}{\mathrm{~d} x}=3 y-\frac{x}{2} $ respectively.
1. Show that the tangent to the curve y=f(x) at the point (1,0) is normal to the curve y=g(x) at the point (1,0) .
2. Find g(x) .
3. Use Euler's method with steps of 0.2 to estimate f(2) to 5 decimal places.

Consider the differen.h3:+k4iqjd h ih6g oitial equation $ \frac{\mathrm{d} y}{\mathrm{~d} x}-(\tan x) y=1$ , where $x \neq \frac{(2 n+1) \pi}{2} $, for any integer n .
1. Given that y(0)=1 , use Euler's method with step length h=0.2 to find an approximation for y(1) . Give your answer correct to two decimal places.
2. Solve the equation $ \frac{\mathrm{d} y}{\mathrm{~d} x}-(\tan x) y=1 $. Give your answer in the form y=f(x) .
3. Find the percentage error when y(1) is approximated by the final rounded value found in part (a). Give your answer correct to two significant figures.
4. Show that the x -coordinate(s) of the points on the curve y=f(x) where $ \frac{\mathrm{d} y}{\mathrm{~d} x}=0 $ are of the form $x=\frac{1}{2}(4 \pi n-\pi)$ , where $ n \in \mathbb{Z}$ .

The acceleration, $a \mathrm{~ms}^{-2} $ of a particle moving in a vertical trajectory at time t seconds, $ t \geq 0$ , is given by $a(t)=-(3+v) $ where v is the particle's velocity in $\mathrm{ms}^{-1} $. At t=0 , the particle is at a fixed origin O and has an initial velocity of $v_{0} \mathrm{~ms}^{-1}$ .
1. By solving an appropriate differential equation, show that the particle's velocity is given by v$(t)=\left(v_{0}+3\right) e^{-t}-3 $.

The particle initially moves upwards until it reaches its maximum height from O , and then returns to O .
Let s metres represent the particle's displacement from O , and $s_{\max }$ the maximum displacement from O .
2. 1. Show that the time T taken for the particle to reach$ s_{\max } $ satisfies the equation $e^{-T}=\frac{3}{v_{0}+3} $.
2. Hence, solve for T in terms of $ v_{0}$ .
3. By solving an appropriate differential equation and using the results from part (b) (i) and (ii), find an expression for $s_{\text {max }} $ in terms of $ v_{0}$ .

Let v(T-k) represent the particle's velocity k seconds before it reaches $ s_{\max }$ , where

$v(T-k)=\left(v_{0}+3\right) e^{-(T-k)}-3$

3. By using the result from part (b) (i), show that v(T-k)=3 e^{k}-3 .

Similarly, let v(T+k) represent the particle's velocity k seconds after it reaches $s_{\text {max }}$
4. Deduce a similar expression for v(T+k) in terms of k .
5. Hence, show that $ v(T-k)+v(T+k) \geq 0 $.

A video streaming service company are monitoring their.m-kfyc l(ru r;j x /b41b7ddf market share in a region in which they have recently commencj/7rmbk; c1f r-d. ( ybdxluf4ed operations.
The number of households, N , they predict will subscribe to the streaming service can be modelled by the logistic differential equation

$\frac{\mathrm{d} N}{\mathrm{~d} t}=\frac{3 k N(L-N)}{2 L}$

where t is time measured in years and k, L are positive constants.
The constant L represents the total number of households in the region who could possibly subscribe to the streaming service.
1. Show that $\frac{\mathrm{d}^{2} N}{\mathrm{~d} t^{2}}=\left(\frac{3 k}{2 L}\right)^{2}(N)(L-N)(L-2 N)$ .
2. Hence show that the number of households subscribing to the streaming service is predicted to increase at its maximum rate when $N=\frac{L}{2}$ .
3. Hence determine the maximum value of $\frac{\mathrm{d} N}{\mathrm{~d} t}$ in terms of k and L .

Let N_{0} be the number of households who have subscribed to the streaming service at the time the company start monitoring their market share.
4. By solving the logistic differential equation, show that its solution can be expressed in the form

k$ t=\left(\frac{2}{3}\right) \ln \left(\frac{N\left(L-N_{0}\right)}{N_{0}(L-N)}\right)$

After 12 years, the number of subscribed households is predicted to be 4$ N_{0} $. It is known that L=5$ N_{0} $.
5. Find the value of k for this model.

A tank has been prepared in order.gz) j ):f5*fun hom.vvmgw-j to mix a color for a fabric dyeing process. The tank initially contains water. A color concentr):zg.nj5ojf- . wfvgm)hvm *u ate is a premix of color powder and a small amount of water The color concentrate begins to flow into the tank. The color solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let x grams represent the amount of the color powder in the tank and let t minutes represent the time since the color concentrate began flowing into the tank.
The rate of change of the amount of color powder in the tank, $\frac{\mathrm{d} x}{\mathrm{~d} t}$ , is described by the differential equation

$\frac{\mathrm{d} x}{\mathrm{~d} t}=4 e^{-\frac{t}{5}}-\frac{x}{t+3}$

1. Show that t+3 is an integrating factor for this differential equation.
2. Hence, by solving this differential equation, show that $ x(t)=\frac{160-20 e^{-\frac{t}{5}}(t+8)}{t+3}$ .
3. Sketch the graph of x versus t for $0 \leq t \leq 50 $ and hence find the maximum amount of color powder in the tank and the value of t at which this occurs.
4. Find the value of t at which the amount of color powder in the tank is decreasing most rapidly.

The rate of change of the amount of color powder leaving the tank is equal to
5. Find the amount of color powder that left the tank during the first 50 minutes.

The population P of fish in a lake after t weeks od.hg q e1tn vrc;(hih* k8uban+2+s.can be modelled by the differential equa( ona*.e+;hgbck+i dn18 . qsvurh th2tion.

$\frac{\mathrm{d} P}{\mathrm{~d} t}=k \sqrt{P}, \quad k, t>0$

1. Show that the population of fish is given by

$P(t)=\left(\frac{k t}{2}+\sqrt{P_{0}}\right)^{2}, \quad t>0$

where $ P_{0}$ is the initial fish population.
It is known that the initial fish population was 3000 , and that 24 weeks later the population had doubled in size.
2. Find the value of k to three significant figures.
3. Estimate the number of fish after 30 weeks to the nearest integer.

After a careful adjustment it is found that the model that best describes the fish population is given by

$\frac{\mathrm{d} P_{2}}{\mathrm{~d} t}=(1.89+3 \cos (0.2 \pi t)) \sqrt{P_{2}}$

where t is the time measured in weeks, $t \geq 0 $.
4. Verify that $ P_{2}=\left(\frac{1.89 t}{2}+\frac{30 \sin (0.2 \pi t)}{4 \pi}+\sqrt{3000}\right)^{2} $ is the solution of this new differential equation.
5. Sketch the graph of P_{2}(t) and the graph of P(t) found in parts (a) and (b) on the same axes, for $ 0 \leq t \leq 50 $.
6. Use $ P_{2}(t) $ to estimate the number of whole weeks it takes for the population to reach 5000 fish.