Quick Recall
Quick check โ pick the correct option in each question before moving on.
Separating variables: $\dfrac{\text{d}y}{y} = \text{d}x$, so $\ln y = x + C$ and $y = Ae^{x}$, where $A = e^C$.
Separating variables: $\dfrac{\text{d}N}{N} = 0.4\,\text{d}t$, so $\ln N = 0.4t + C$ and $N = Ae^{0.4t}$.
The general solution represents a whole family of curves โ one for each value of the constant $C$. The initial condition selects the single curve that matches the real situation.
The Wall
A biologist models a bacteria colony with $\dfrac{\text{d}N}{\text{d}t} = 0.4N$. From W9-L2 she finds the general solution $N = Ae^{0.4t}$. The colony starts with 50 bacteria at $t = 0$. She wants to predict the population at $t = 5$.
But the general solution gives a whole family of curves โ one for each value of $A$. Without knowing $A$, she cannot make a prediction. The general solution alone is not enough; she needs a way to select the one curve that matches the real starting condition.
The Need
We need a method to determine the constant $C$ using an initial condition (a known value of $y$ at a known value of $x$). This pins down one specific curve from the family โ the particular solution that matches the real situation.
๐ก The Bridge (click to reveal)
Substitute the initial condition $N = 50$ at $t = 0$ into the general solution $N = Ae^{0.4t}$:
$$50 = Ae^{0.4 \cdot 0} = Ae^{0} = A \cdot 1 = A$$
So $A = 50$, and the particular solution is $N = 50e^{0.4t}$. Now the biologist can predict $N(5) = 50e^{2} \approx 369$ bacteria.
This is today's lesson: substitute the initial condition into the general solution to find $C$, then write the particular solution.
Last lesson (W9-L2) we found the general solution of a separable DE โ a family of curves with an arbitrary constant $C$. Today we learn to find the particular solution: the single member of the family that matches a real-world starting condition.
If the general solution is $y = G(x) + C$ (or $y = AG(x)$), and we know that $y = y_0$ when $x = x_0$, substitute these values to solve for $C$ (or $A$):
$$\boxed{\;\text{Substitute } x = x_0,\ y = y_0 \text{ into the general solution, then solve for } C.\;}$$The result is the particular solution โ one specific curve from the family.
An initial condition is not just a number โ it has a physical meaning:
$\bullet$ Population $N(0) = N_0$: the starting population at $t = 0$.
$\bullet$ Cooling $\theta(0) = \theta_0$: the initial temperature of the liquid.
$\bullet$ Motion $v(0) = v_0$: the starting velocity of the object.
$\bullet$ Displacement $s(0) = s_0$: the starting position.
The constant $C$ (or $A$) in the general solution is determined by this physical initial state.
For a first-order separable DE, each initial condition $y(x_0) = y_0$ selects exactly one member of the family. There is a one-to-one correspondence:
$$\boxed{\;\text{One initial condition } (x_0, y_0) \;\longleftrightarrow\; \text{one value of } C \;\longleftrightarrow\; \text{one particular solution.}\;}$$Different initial conditions select different curves โ no two initial conditions give the same particular solution (in the region where the solution exists).
Work in a group of four. Each stage of this solution belongs to one role โ Substituter, Solver, Builder, Checker. Only the owner of a stage may click, and the group must agree on the answer first, because each stage is locked until the one above it is right. The task: the general solution of the differential equation is $y = Ae^{x^{2}}$, and $y = 5$ when $x = 0$. Find the particular solution.
Role rule: the owner explains the reasoning out loud before anyone clicks โ the rest of the group may only ask โwhy?โ.
On your own. The general solution of $\dfrac{\text{d}y}{\text{d}x} = 0.5y$ is $y = Ae^{0.5x}$. The dotted curve is the particular solution that passes through $(0, 4)$. Adjust $A$ until your solid curve sits exactly on the dotted target, then answer the two follow-up questions.
Dotted amber = the particular solution through $(0,4)$. Solid emerald = your current value of $A$.
Given that $y = 1$ when $x = 0$, solve the differential equation $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x(3y^2 + 10y + 3)$, obtaining an expression for $y$ in terms of $x$.
๐ Reveal Solution
Factorise: $3y^2 + 10y + 3 = (3y + 1)(y + 3)$.
Separate: $\dfrac{\text{d}y}{(3y+1)(y+3)} = 4x\,\text{d}x$.
Write $\dfrac{1}{(3y+1)(y+3)} = \dfrac{A}{3y+1} + \dfrac{B}{y+3}$, then multiply through by the full denominator:
$$1 \equiv A(y+3) + B(3y+1)$$
Substitute $y = -\tfrac{1}{3}$: $\;1 = A\!\left(\tfrac{8}{3}\right) \Rightarrow A = \tfrac{3}{8}$.
Substitute $y = -3$: $\;1 = B(-8) \Rightarrow B = -\tfrac{1}{8}$.
So $\dfrac{1}{(3y+1)(y+3)} = \dfrac{3/8}{3y+1} - \dfrac{1/8}{y+3}$.
$\displaystyle\int \!\left(\dfrac{3/8}{3y+1} - \dfrac{1/8}{y+3}\right)\text{d}y = \int 4x\,\text{d}x$
$\dfrac{3}{8} \cdot \dfrac{1}{3}\ln|3y+1| - \dfrac{1}{8}\ln|y+3| = 2x^2 + C$
$\dfrac{1}{8}\ln|3y+1| - \dfrac{1}{8}\ln|y+3| = 2x^2 + C$
$\dfrac{1}{8}\ln\!\left|\dfrac{3y+1}{y+3}\right| = 2x^2 + C$
Multiply by 8: $\ln\!\left|\dfrac{3y+1}{y+3}\right| = 16x^2 + 8C$.
Exponentiate: $\dfrac{3y+1}{y+3} = Ae^{16x^2}$ where $A = e^{8C}$.
Substitute: $\dfrac{3(1)+1}{1+3} = Ae^{0} \implies \dfrac{4}{4} = A \implies A = 1$.
So $\dfrac{3y+1}{y+3} = e^{16x^2}$.
$3y + 1 = (y+3)e^{16x^2}$
$3y + 1 = ye^{16x^2} + 3e^{16x^2}$
$3y - ye^{16x^2} = 3e^{16x^2} - 1$
$y\!\left(3 - e^{16x^2}\right) = 3e^{16x^2} - 1$
$\boxed{y = \dfrac{3e^{16x^2} - 1}{3 - e^{16x^2}}}$
Verification: at $x = 0$, $y = \dfrac{3 - 1}{3 - 1} = \dfrac{2}{2} = 1$. ✓
The temperature of a quantity of liquid at time $t$ is $\theta$. The liquid is cooling in atmosphere whose temperature is constant and equal to $A$. The rate of decrease of $\theta$ is proportional to the temperature difference $(\theta - A)$. Thus $\theta$ and $t$ satisfy the differential equation $\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - A)$, where $k$ is a positive constant.
(i) Find, in any form, the solution of this differential equation, given that $\theta = 4A$ when $t = 0$.
(ii) Given also that $\theta = 3A$ when $t = 1$, show that $k = \ln\frac{3}{2}$.
(iii) Find $\theta$ in terms of $A$ when $t = 2$, expressing your answer in its simplest form.
๐ Reveal Solution
Separate: $\dfrac{\text{d}\theta}{\theta - A} = -k\,\text{d}t$.
Integrate: $\ln|\theta - A| = -kt + C$.
Exponentiate: $\theta - A = Be^{-kt}$ where $B = e^C$.
Apply $\theta = 4A$ at $t = 0$: $4A - A = Be^{0} = B$, so $B = 3A$.
$\boxed{\theta = A + 3Ae^{-kt} = A\!\left(1 + 3e^{-kt}\right)}$
Substitute $\theta = 3A$, $t = 1$ into $\theta = A(1 + 3e^{-kt})$:
$3A = A(1 + 3e^{-k})$
$3 = 1 + 3e^{-k}$
$2 = 3e^{-k}$
$e^{-k} = \dfrac{2}{3}$
$-k = \ln\!\left(\dfrac{2}{3}\right) = -\ln\!\left(\dfrac{3}{2}\right)$
$\boxed{k = \ln\frac{3}{2}}$ โ (as required)
From part (i): $\theta = A(1 + 3e^{-2k})$.
Since $e^{-k} = \dfrac{2}{3}$, we have $e^{-2k} = \left(\dfrac{2}{3}\right)^{2} = \dfrac{4}{9}$.
$\theta = A\!\left(1 + 3 \cdot \dfrac{4}{9}\right) = A\!\left(1 + \dfrac{4}{3}\right) = A \cdot \dfrac{7}{3}$.
$\boxed{\theta = \dfrac{7A}{3}}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
The variables $x$ and $t$ are related by the differential equation $e^{2t}\dfrac{\mathrm{d}x}{\mathrm{d}t} = \cos^2 x$, where $t \geq 0$. When $t = 0$, $x = 0$.
(i) Solve the differential equation, obtaining an expression for $x$ in terms of $t$.
(ii) State what happens to the value of $x$ when $t$ becomes very large.
(iii) Explain why $x$ increases as $t$ increases.
Cambridge International AS & A Level Mathematics 9709 Paper 32 Q7 June 2010
๐ก Hint
For (i): rewrite as $\dfrac{\text{d}x}{\text{d}t} = \cos^2 x \cdot e^{-2t}$, then separate: $\dfrac{\text{d}x}{\cos^2 x} = e^{-2t}\,\text{d}t$. Recall $\dfrac{1}{\cos^2 x} = \sec^2 x$ and $\int \sec^2 x\,\text{d}x = \tan x$.
For (ii): take $t \to \infty$ in your solution โ what does $e^{-2t}$ approach?
For (iii): look at the sign of $\dfrac{\text{d}x}{\text{d}t} = \cos^2 x \cdot e^{-2t}$.
๐ Reveal Solution
Rewrite: $\dfrac{\text{d}x}{\text{d}t} = \cos^2 x \cdot e^{-2t}$.
Separate: $\sec^2 x\,\text{d}x = e^{-2t}\,\text{d}t$.
Integrate: $\tan x = -\dfrac{1}{2}e^{-2t} + C$.
Apply $x = 0$ at $t = 0$: $\tan 0 = -\dfrac{1}{2}e^{0} + C \implies 0 = -\dfrac{1}{2} + C \implies C = \dfrac{1}{2}$.
So $\tan x = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t} = \dfrac{1 - e^{-2t}}{2}$.
$\boxed{x = \arctan\!\left(\dfrac{1 - e^{-2t}}{2}\right)}$
As $t \to \infty$, $e^{-2t} \to 0$, so $\tan x \to \dfrac{1 - 0}{2} = \dfrac{1}{2}$.
Therefore $x \to \arctan\!\left(\dfrac{1}{2}\right)$.
$\boxed{x \to \arctan\!\left(\tfrac{1}{2}\right) \approx 0.464 \text{ rad} \approx 26.6^{\circ}}$
From the DE: $\dfrac{\text{d}x}{\text{d}t} = \cos^2 x \cdot e^{-2t}$.
Since $\cos^2 x \geq 0$ for all $x$ and $e^{-2t} > 0$ for all $t \geq 0$, we have $\dfrac{\text{d}x}{\text{d}t} \geq 0$ (and $> 0$ except where $\cos x = 0$).
For the relevant range $0 \leq x < \arctan(1/2) < \pi/2$, we have $\cos x \neq 0$, so $\dfrac{\text{d}x}{\text{d}t} > 0$.
Therefore $x$ is an increasing function of $t$.
An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time $t$ hours after filling begins, the volume of liquid is $V \, \text{m}^3$ and the depth of liquid is $h \, \text{m}$. It is given that $V = \frac{4}{3}h^3$. The liquid is poured in at a rate of $20 \, \text{m}^3$ per hour, but owing to leakage, liquid is lost at a rate proportional to $h^2$. When $h = 1$, $\frac{dh}{dt} = 4.95$.
(i) Show that $h$ satisfies the differential equation $\frac{dh}{dt} = \frac{5}{h^2} - \frac{1}{20}$.
(ii) Verify that $\frac{20h^2}{100 - h^2} \equiv -20 + \frac{2000}{(10 - h)(10 + h)}$.
(iii) Hence solve the differential equation in part (i), obtaining an expression for $t$ in terms of $h$.
Cambridge International AS & A Level Mathematics 9709 Paper 3 Q8 November 2008
๐ก Hint 1
For (i): use the chain rule $\dfrac{\text{d}V}{\text{d}t} = \dfrac{\text{d}V}{\text{d}h} \cdot \dfrac{\text{d}h}{\text{d}t}$. The net rate of change of volume is $20 - kh^2$ (inflow minus leakage).
For (ii): combine the RHS over a common denominator $(10-h)(10+h) = 100 - h^2$ and simplify.
๐ก Hint 2
For (iii): separate as $\dfrac{20h^2}{100 - h^2}\,\text{d}h = \text{d}t$, then use the identity from (ii) and partial fractions on $\dfrac{1}{(10-h)(10+h)}$.
Apply the initial condition $h = 0$ at $t = 0$ (tank initially empty) to find $C$.
๐ Reveal Solution
Net rate of volume change: $\dfrac{\text{d}V}{\text{d}t} = 20 - kh^2$ (inflow 20, leakage $kh^2$).
Chain rule: $V = \tfrac{4}{3}h^3 \Rightarrow \dfrac{\text{d}V}{\text{d}h} = 4h^2$, so $\dfrac{\text{d}V}{\text{d}t} = 4h^2 \dfrac{\text{d}h}{\text{d}t}$.
Therefore $4h^2 \dfrac{\text{d}h}{\text{d}t} = 20 - kh^2$, giving $\dfrac{\text{d}h}{\text{d}t} = \dfrac{5}{h^2} - \dfrac{k}{4}$.
Use $h = 1$, $\dfrac{\text{d}h}{\text{d}t} = 4.95$: $4.95 = 5 - \dfrac{k}{4} \implies \dfrac{k}{4} = 0.05 \implies k = 0.2 = \dfrac{1}{5}$.
So $\dfrac{k}{4} = \dfrac{1}{20}$, hence $\dfrac{\text{d}h}{\text{d}t} = \dfrac{5}{h^2} - \dfrac{1}{20}$. ✓
RHS $= -20 + \dfrac{2000}{(10-h)(10+h)} = \dfrac{-20(10-h)(10+h) + 2000}{(10-h)(10+h)}$.
Since $(10-h)(10+h) = 100 - h^2$:
RHS $= \dfrac{-20(100 - h^2) + 2000}{100 - h^2} = \dfrac{-2000 + 20h^2 + 2000}{100 - h^2} = \dfrac{20h^2}{100 - h^2} =$ LHS. ✓
Rewrite the DE: $\dfrac{\text{d}h}{\text{d}t} = \dfrac{100 - h^2}{20h^2}$.
Separate: $\dfrac{20h^2}{100 - h^2}\,\text{d}h = \text{d}t$.
Using part (ii): $\left(-20 + \dfrac{2000}{(10-h)(10+h)}\right)\text{d}h = \text{d}t$.
Partial fractions: $\dfrac{2000}{(10-h)(10+h)} = \dfrac{100}{10-h} + \dfrac{100}{10+h}$ (since $\dfrac{1}{(10-h)(10+h)} = \dfrac{1/20}{10-h} + \dfrac{1/20}{10+h}$).
Integrate: $\int\!\left(-20 + \dfrac{100}{10-h} + \dfrac{100}{10+h}\right)\text{d}h = \int \text{d}t$.
$-20h - 100\ln|10-h| + 100\ln|10+h| = t + C$.
$-20h + 100\ln\!\left|\dfrac{10+h}{10-h}\right| = t + C$.
Apply $h = 0$ at $t = 0$: $0 + 100\ln\!\left|\dfrac{10}{10}\right| = 0 + C \implies 100\ln 1 = C \implies C = 0$.
$\boxed{t = 100\ln\!\left(\dfrac{10+h}{10-h}\right) - 20h}$
Naturalists are managing a wildlife reserve to increase the number of plants of a rare species. The number of plants at time $t$ years is denoted by $N$, where $N$ is treated as a continuous variable.
(i) It is given that the rate of increase of $N$ with respect to $t$ is proportional to $(N - 150)$. Write down a differential equation relating $N$, $t$ and a constant of proportionality.
(ii) Initially, when $t = 0$, the number of plants was 650. It was noted that, at a time when there were 900 plants, the number of plants was increasing at a rate of 60 per year. Express $N$ in terms of $t$.
(iii) The naturalists had a target of increasing the number of plants from 650 to 2000 within 15 years. Will this target be met?
Cambridge International AS & A Level Mathematics 9709 Paper 33 Q10 November 2015
๐ก Hint
For (i): โrate of increase of $N$ is proportional to $(N-150)$โ means $\dfrac{\text{d}N}{\text{d}t} = k(N - 150)$.
For (ii): use โ$\dfrac{\text{d}N}{\text{d}t} = 60$ when $N = 900$โ to find $k$. Then separate, integrate, and apply $N = 650$ at $t = 0$ to find the constant.
For (iii): substitute $t = 15$ into your particular solution and check if $N \geq 2000$.
๐ Reveal Solution
$\boxed{\dfrac{\text{d}N}{\text{d}t} = k(N - 150)}$ where $k$ is the constant of proportionality.
Use $\dfrac{\text{d}N}{\text{d}t} = 60$ when $N = 900$: $60 = k(900 - 150) = 750k \implies k = \dfrac{60}{750} = \dfrac{2}{25}$.
Separate: $\dfrac{\text{d}N}{N - 150} = \dfrac{2}{25}\,\text{d}t$.
Integrate: $\ln|N - 150| = \dfrac{2}{25}t + C$.
Exponentiate: $N - 150 = Ae^{(2/25)t}$ where $A = e^C$.
Apply $N = 650$ at $t = 0$: $650 - 150 = Ae^{0} = A \implies A = 500$.
$\boxed{N = 150 + 500e^{(2/25)t}}$
Substitute $t = 15$: $N = 150 + 500e^{(2/25)(15)} = 150 + 500e^{30/25} = 150 + 500e^{6/5}$.
$e^{6/5} \approx 3.3201$, so $N \approx 150 + 500(3.3201) = 150 + 1660.07 \approx 1810$.
Since $1810 < 2000$, the target will not be met.
Alternative: find the time to reach 2000: $2000 = 150 + 500e^{(2/25)t} \implies 3.7 = e^{(2/25)t} \implies t = \dfrac{25}{2}\ln(3.7) \approx 16.35$ years. Since $16.35 > 15$, the target is not met.
$\boxed{\text{No, the target will not be met. } N(15) \approx 1810 < 2000.}$
The number of birds of a certain species in a forested region is recorded over several years. At time $t$ years, the number of birds is $N$, where $N$ is treated as a continuous variable. The variation in the number of birds is modelled by $\dfrac{\mathrm{d}N}{\mathrm{d}t} = \dfrac{N(1800 - N)}{3600}$. It is given that $N = 300$ when $t = 0$.
(i) Find an expression for $N$ in terms of $t$.
(ii) According to the model, how many birds will there be after a long time?
๐ก Hint 1
This is a logistic differential equation. Separate as $\dfrac{3600}{N(1800 - N)}\,\text{d}N = \text{d}t$.
Use partial fractions: $\dfrac{3600}{N(1800 - N)} = \dfrac{A}{N} + \dfrac{B}{1800 - N}$. Find $A$ and $B$.
After integrating, exponentiate and apply $N = 300$ at $t = 0$.
๐ก Hint 2
For the partial fractions: $3600 = A(1800 - N) + BN$. Set $N = 0$: $A = 2$. Set $N = 1800$: $B = 2$.
So $\dfrac{3600}{N(1800 - N)} = \dfrac{2}{N} + \dfrac{2}{1800 - N}$.
Integrating gives $2\ln|N| - 2\ln|1800 - N| = t + C$, i.e. $\ln\!\left|\dfrac{N}{1800 - N}\right| = \dfrac{t}{2} + C'$.
๐ Reveal Solution
Separate: $\dfrac{3600}{N(1800 - N)}\,\text{d}N = \text{d}t$.
Partial fractions: $\dfrac{3600}{N(1800 - N)} = \dfrac{2}{N} + \dfrac{2}{1800 - N}$ (since $A(1800-N) + BN = 3600$ gives $A = B = 2$).
Integrate: $\int\!\left(\dfrac{2}{N} + \dfrac{2}{1800 - N}\right)\text{d}N = \int \text{d}t$.
$2\ln|N| - 2\ln|1800 - N| = t + C$.
$\ln\!\left|\dfrac{N}{1800 - N}\right| = \dfrac{t}{2} + \dfrac{C}{2}$. Let $K = \dfrac{C}{2}$: $\dfrac{N}{1800 - N} = Ae^{t/2}$ where $A = e^K$.
Apply $N = 300$ at $t = 0$: $\dfrac{300}{1800 - 300} = Ae^{0} \implies \dfrac{300}{1500} = A \implies A = \dfrac{1}{5}$.
So $\dfrac{N}{1800 - N} = \dfrac{1}{5}e^{t/2}$.
Solve for $N$: $5N = (1800 - N)e^{t/2} \implies 5N = 1800e^{t/2} - Ne^{t/2} \implies N(5 + e^{t/2}) = 1800e^{t/2}$.
$N = \dfrac{1800e^{t/2}}{5 + e^{t/2}}$. Dividing numerator and denominator by $e^{t/2}$:
$\boxed{N = \dfrac{1800}{1 + 5e^{-t/2}}}$
Verification: at $t = 0$, $N = \dfrac{1800}{1 + 5} = \dfrac{1800}{6} = 300$. ✓
As $t \to \infty$, $e^{-t/2} \to 0$, so $N \to \dfrac{1800}{1 + 0} = 1800$.
The population approaches the carrying capacity of 1800 birds.
$\boxed{N \to 1800 \text{ birds}}$
This is the value at which $\dfrac{\text{d}N}{\text{d}t} = \dfrac{N(1800-N)}{3600} = 0$ โ the population stops growing when it reaches 1800.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides next lesson's review.