The Scenario
A biologist is studying a bacteria colony. She finds that the population $N$ (in hundreds) grows at a rate proportional to the current population. The differential equation is:
$$\dfrac{\text{d}N}{\text{d}t} = 0.5\,N$$
She wants to find $N$ as a function of $t$ — that is, predict the population at any future time. But there is a problem: $N$ appears on both sides. It is inside the derivative $\dfrac{\text{d}N}{\text{d}t}$ on the left, and it appears as a factor on the right. Last lesson (W9-L1) we learned to set up this equation. Now we need to solve it.
The key idea: if we can rearrange the equation so that all $N$'s go to the left and all $t$'s go to the right, we can integrate each side separately.
The Question
The differential equation $\dfrac{\text{d}N}{\text{d}t} = 0.5\,N$ has $N$ on both sides. We cannot integrate it in its current form because the left side involves both $N$ and $t$.
How can we rearrange $\dfrac{\text{d}N}{\text{d}t} = 0.5\,N$ so that everything involving $N$ is on the left and everything involving $t$ is on the right? And once separated, how do we find $N(t)$?
💡 The Bridge (click to reveal)
Treat $\dfrac{\text{d}N}{\text{d}t}$ as a fraction (informally): multiply both sides by $\text{d}t$ and divide both sides by $N$:
$$\frac{\text{d}N}{N} = 0.5\,\text{d}t$$
Now the left side has only $N$ and the right side has only $t$. Integrate both sides:
$$\int \frac{1}{N}\,\text{d}N = \int 0.5\,\text{d}t \implies \ln|N| = 0.5\,t + C$$
Exponentiate to solve for $N$: $N = e^{0.5t + C} = e^C \cdot e^{0.5t} = Ae^{0.5t}$, where $A = e^C$.
This is the separation of variables method — today's lesson formalises it for any DE of the form $\dfrac{\text{d}y}{\text{d}x} = f(x)\,g(y)$.
Last lesson (W9-L1) we learned to set up differential equations from real-world scenarios. Today we learn to solve them. The method of separation of variables works whenever the DE can be written as $\dfrac{\text{d}y}{\text{d}x} = f(x)\,g(y)$ — that is, the right side is a product of a function of $x$ and a function of $y$.
If $\dfrac{\text{d}y}{\text{d}x} = f(x)\,g(y)$, then we separate by dividing by $g(y)$ and multiplying by $\text{d}x$:
$$\boxed{\;\frac{\text{d}y}{g(y)} = f(x)\,\text{d}x \quad\implies\quad \int \frac{1}{g(y)}\,\text{d}y = \int f(x)\,\text{d}x\;}$$The result is an equation relating $y$ and $x$ — the general solution, containing an arbitrary constant $C$.
Once separated, each side is integrated using the techniques from Week 6 (P3 §3.5). The type of integral depends on the form of $f(x)$ and $g(y)$:
$\bullet$ Power: $\int x^n\,\text{d}x = \dfrac{x^{n+1}}{n+1} + C$ ($n \neq -1$)
$\bullet$ Exponential: $\int e^{ax+b}\,\text{d}x = \dfrac{1}{a}\,e^{ax+b} + C$
$\bullet$ Trigonometric: $\int \sin(ax+b)\,\text{d}x = -\dfrac{1}{a}\cos(ax+b) + C$, $\int \cos(ax+b)\,\text{d}x = \dfrac{1}{a}\sin(ax+b) + C$
$\bullet$ Logarithmic: $\int \dfrac{1}{ax+b}\,\text{d}x = \dfrac{1}{a}\ln|ax+b| + C$
After integrating both sides, we obtain the general solution containing an arbitrary constant $C$ (or $A = e^C$ for exponential forms). Each value of $C$ gives a different member of a family of solution curves.
Example: solving $\dfrac{\text{d}y}{\text{d}x} = y$ gives $\ln|y| = x + C$, so $y = Ae^x$ where $A = e^C$. Different values of $A$ give different curves — all satisfying the same DE.
In groups of three you must approve each line before it is shown. At every stage, pick the line you believe comes next in the solution of $\dfrac{\text{d}y}{\text{d}x} = xy^{2}$. A correct pick reveals the line and unlocks the next stage; a wrong pick turns red and explains the trap. Only four wrong picks in the whole run earns you the top band.
Group rule: nobody clicks until all three members agree on the line. If you disagree, say why — that is where the learning is.
Two worked solutions, each with exactly one wrong line. Work on your own: click the line you believe is wrong, then choose the corrected line from the options. One wrong line per round — no guessing.
Solve the following differential equations by integration.
(i) $\dfrac{\text{d}y}{\text{d}x} = x^2$ (ii) $\dfrac{\text{d}y}{\text{d}x} = \cos x$ (iii) $\dfrac{\text{d}y}{\text{d}x} = e^x$ (iv) $\dfrac{\text{d}y}{\text{d}x} = \sqrt{x}$
👁 Reveal Solution
Integrate: $y = \int x^2\,\text{d}x = \dfrac{x^3}{3} + C$.
$\boxed{y = \dfrac{x^3}{3} + C}$
Integrate: $y = \int \cos x\,\text{d}x = \sin x + C$.
$\boxed{y = \sin x + C}$
Integrate: $y = \int e^x\,\text{d}x = e^x + C$.
$\boxed{y = e^x + C}$
Write $\sqrt{x} = x^{1/2}$. Integrate using the power rule: $y = \int x^{1/2}\,\text{d}x = \dfrac{x^{3/2}}{3/2} = \dfrac{2}{3}x^{3/2} + C$.
$\boxed{y = \dfrac{2}{3}x^{3/2} + C}$
Find the general solutions of the following differential equations by separating the variables.
(i) $\dfrac{\text{d}y}{\text{d}x} = xy^2$ (ii) $\dfrac{\text{d}y}{\text{d}x} = \dfrac{x^2}{y}$ (iii) $\dfrac{\text{d}y}{\text{d}x} = y$ (iv) $\dfrac{\text{d}y}{\text{d}x} = e^{x-y}$ (v) $\dfrac{\text{d}y}{\text{d}x} = \dfrac{y}{x}$ (vi) $\dfrac{\text{d}y}{\text{d}x} = x\sqrt{y}$ (vii) $\dfrac{\text{d}y}{\text{d}x} = y^2 \cos x$ (viii) $\dfrac{\text{d}y}{\text{d}x} = \dfrac{x(y^2 + 1)}{y(x^2 + 1)}$ (ix) $\dfrac{\text{d}y}{\text{d}x} = xe^y$ (x) $\dfrac{\text{d}y}{\text{d}x} = \dfrac{x \ln x}{y^2}$
👁 Reveal Solution
Separate: $\dfrac{\text{d}y}{y^2} = x\,\text{d}x$. Integrate: $\int y^{-2}\,\text{d}y = \int x\,\text{d}x \implies -\dfrac{1}{y} = \dfrac{x^2}{2} + C$.
Solve: $y = -\dfrac{1}{\frac{x^2}{2} + C}$ (absorbing the sign into $C$).
$\boxed{y = -\dfrac{1}{\dfrac{x^2}{2} + C}}$
Separate: $y\,\text{d}y = x^2\,\text{d}x$. Integrate: $\int y\,\text{d}y = \int x^2\,\text{d}x \implies \dfrac{y^2}{2} = \dfrac{x^3}{3} + C$.
$\boxed{y^2 = \dfrac{2x^3}{3} + C}$ (renaming the constant)
Separate: $\dfrac{\text{d}y}{y} = \text{d}x$. Integrate: $\ln|y| = x + C$.
Exponentiate: $y = e^{x+C} = e^C \cdot e^x = Ae^x$ where $A = e^C$.
$\boxed{y = Ae^x}$
Rewrite: $e^{x-y} = e^x \cdot e^{-y}$. Separate: $e^y\,\text{d}y = e^x\,\text{d}x$. Integrate: $\int e^y\,\text{d}y = \int e^x\,\text{d}x \implies e^y = e^x + C$.
Solve: $y = \ln(e^x + C)$.
$\boxed{y = \ln(e^x + C)}$
Separate: $\dfrac{\text{d}y}{y} = \dfrac{\text{d}x}{x}$. Integrate: $\ln|y| = \ln|x| + C$.
Exponentiate: $y = e^C \cdot |x| = Ax$ where $A = \pm e^C$.
$\boxed{y = Ax}$
Separate: $\dfrac{\text{d}y}{\sqrt{y}} = x\,\text{d}x$. Integrate: $\int y^{-1/2}\,\text{d}y = \int x\,\text{d}x \implies 2\sqrt{y} = \dfrac{x^2}{2} + C$.
So $\sqrt{y} = \dfrac{x^2}{4} + C$, giving $y = \left(\dfrac{x^2}{4} + C\right)^2$.
$\boxed{y = \left(\dfrac{x^2}{4} + C\right)^2}$
Separate: $\dfrac{\text{d}y}{y^2} = \cos x\,\text{d}x$. Integrate: $\int y^{-2}\,\text{d}y = \int \cos x\,\text{d}x \implies -\dfrac{1}{y} = \sin x + C$.
$\boxed{y = -\dfrac{1}{\sin x + C}}$
Split: $f(x) = \dfrac{x}{x^2+1}$, $g(y) = \dfrac{y^2+1}{y}$. Separate: $\dfrac{y}{y^2+1}\,\text{d}y = \dfrac{x}{x^2+1}\,\text{d}x$.
Integrate: $\dfrac{1}{2}\ln|y^2+1| = \dfrac{1}{2}\ln|x^2+1| + C$. Multiply by 2: $\ln(y^2+1) = \ln(x^2+1) + C$.
Exponentiate: $y^2+1 = A(x^2+1)$ where $A = e^C$.
$\boxed{y^2 + 1 = A(x^2 + 1)}$
Separate: $e^{-y}\,\text{d}y = x\,\text{d}x$. Integrate: $\int e^{-y}\,\text{d}y = \int x\,\text{d}x \implies -e^{-y} = \dfrac{x^2}{2} + C$.
So $e^{-y} = -\dfrac{x^2}{2} + C$, giving $y = -\ln\!\left(C - \dfrac{x^2}{2}\right)$.
$\boxed{y = -\ln\!\left(C - \dfrac{x^2}{2}\right)}$
Separate: $y^2\,\text{d}y = x\ln x\,\text{d}x$. Integrate: $\int y^2\,\text{d}y = \int x\ln x\,\text{d}x$.
Left: $\dfrac{y^3}{3}$. Right (by parts, $u = \ln x$, $\text{d}v = x\,\text{d}x$): $\dfrac{x^2}{2}\ln x - \int \dfrac{x}{2}\,\text{d}x = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4}$.
So $\dfrac{y^3}{3} = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C$, giving $y^3 = \dfrac{3x^2}{2}\ln x - \dfrac{3x^2}{4} + C$.
$\boxed{y^3 = \dfrac{3x^2}{2}\ln x - \dfrac{3x^2}{4} + C}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find the particular solution of each of the following differential equations.
(i) $\dfrac{\text{d}y}{\text{d}x} = x^2 - 1$, $y = 2$ when $x = 3$
(ii) $\dfrac{\text{d}y}{\text{d}x} = x^2 y$, $y = 1$ when $x = 0$
(iii) $\dfrac{\text{d}y}{\text{d}x} = x e^{-y}$, $y = 0$ when $x = 0$
(iv) $\dfrac{\text{d}y}{\text{d}x} = y^2$, $y = 1$ when $x = 1$
(v) $\dfrac{\text{d}y}{\text{d}x} = x(y + 1)$, $y = 0$ when $x = 1$
(vi) $\dfrac{\text{d}y}{\text{d}x} = y^2 \sin x$, $y = 1$ when $x = 0$
💡 Hint
For (i): no separation needed — just integrate $x^2 - 1$ and use $y=2$ at $x=3$ to find $C$.
For (ii)–(vi): separate variables, integrate, then substitute the initial condition to find $C$.
For (vi): $\int y^{-2}\,\text{d}y = -\dfrac{1}{y}$ and $\int \sin x\,\text{d}x = -\cos x$.
👁 Reveal Solution
Integrate: $y = \dfrac{x^3}{3} - x + C$. Substitute: $2 = \dfrac{27}{3} - 3 + C = 9 - 3 + C = 6 + C$, so $C = -4$.
$\boxed{y = \dfrac{x^3}{3} - x - 4}$
Separate: $\dfrac{\text{d}y}{y} = x^2\,\text{d}x$. Integrate: $\ln|y| = \dfrac{x^3}{3} + C$. Substitute: $\ln 1 = 0 + C \implies C = 0$.
$\boxed{y = e^{x^3/3}}$
Separate: $e^y\,\text{d}y = x\,\text{d}x$. Integrate: $e^y = \dfrac{x^2}{2} + C$. Substitute: $e^0 = 0 + C \implies C = 1$.
$\boxed{y = \ln\!\left(\dfrac{x^2}{2} + 1\right)}$
Separate: $\dfrac{\text{d}y}{y^2} = \text{d}x$. Integrate: $-\dfrac{1}{y} = x + C$. Substitute: $-\dfrac{1}{1} = 1 + C \implies C = -2$.
So $-\dfrac{1}{y} = x - 2$, giving $y = \dfrac{1}{2 - x}$.
$\boxed{y = \dfrac{1}{2 - x}}$
Separate: $\dfrac{\text{d}y}{y+1} = x\,\text{d}x$. Integrate: $\ln|y+1| = \dfrac{x^2}{2} + C$. Substitute: $\ln 1 = \dfrac{1}{2} + C \implies C = -\dfrac{1}{2}$.
So $\ln|y+1| = \dfrac{x^2 - 1}{2}$, giving $y + 1 = e^{(x^2-1)/2}$.
$\boxed{y = e^{(x^2-1)/2} - 1}$
Separate: $\dfrac{\text{d}y}{y^2} = \sin x\,\text{d}x$. Integrate: $-\dfrac{1}{y} = -\cos x + C$. Substitute: $-\dfrac{1}{1} = -\cos 0 + C \implies -1 = -1 + C \implies C = 0$.
So $-\dfrac{1}{y} = -\cos x$, giving $\dfrac{1}{y} = \cos x$, so $y = \dfrac{1}{\cos x} = \sec x$.
$\boxed{y = \sec x}$
A cold liquid at temperature $\theta^\circ\text{C}$, where $\theta < 20$, is standing in a warm room. The temperature of the liquid obeys the differential equation $\dfrac{\text{d}\theta}{\text{d}t} = 2(20 - \theta)$ where the time $t$ is measured in hours.
(i) Find the general solution of this differential equation.
(ii) Find the particular solution for which $\theta = 5$ when $t = 0$.
(iii) In this case, how long does the liquid take to reach a temperature of $18^\circ\text{C}$?
💡 Hint
For (i): separate as $\dfrac{\text{d}\theta}{20 - \theta} = 2\,\text{d}t$. The integral of $\dfrac{1}{20-\theta}$ involves $\ln|20-\theta|$ (don't forget the minus sign from the chain rule).
For (iii): set $\theta = 18$ in the particular solution and solve for $t$.
👁 Reveal Solution
Separate: $\dfrac{\text{d}\theta}{20 - \theta} = 2\,\text{d}t$. Integrate: $-\ln|20 - \theta| = 2t + C$.
So $\ln|20 - \theta| = -2t + C$ (absorbing the sign). Exponentiate: $20 - \theta = Ae^{-2t}$ where $A = e^C$.
$\boxed{\theta = 20 - Ae^{-2t}}$
Substitute: $5 = 20 - Ae^{0} = 20 - A$, so $A = 15$.
$\boxed{\theta = 20 - 15e^{-2t}}$
Set $\theta = 18$: $18 = 20 - 15e^{-2t}$, so $15e^{-2t} = 2$, giving $e^{-2t} = \dfrac{2}{15}$.
Take $\ln$: $-2t = \ln\!\left(\dfrac{2}{15}\right)$, so $t = -\dfrac{1}{2}\ln\!\left(\dfrac{2}{15}\right) = \dfrac{1}{2}\ln\!\left(\dfrac{15}{2}\right)$.
Numerically: $\dfrac{1}{2}\ln(7.5) \approx \dfrac{1}{2}(2.0149) \approx 1.01$ hours.
$\boxed{t = \dfrac{1}{2}\ln\!\left(\dfrac{15}{2}\right) \approx 1.01 \text{ hours}}$
A population of rabbits increases so that the number of rabbits $N$ (in hundreds), after $t$ years is modelled by the differential equation $\dfrac{\text{d}N}{\text{d}t} = N$.
(i) Find the general solution for $N$ in terms of $t$.
(ii) Find the particular solution for which $N = 10$ when $t = 0$.
(iii) What will happen to the number of rabbits when $t$ becomes very large? Why is this not a realistic model for anctual population of rabbits?
💡 Hint
For (i): separate $\dfrac{\text{d}N}{N} = \text{d}t$ and integrate.
For (ii): substitute $N = 10$ at $t = 0$ to find the constant.
For (iii): consider what happens to $e^t$ as $t \to \infty$. Think about real-world constraints on population growth.
👁 Reveal Solution
Separate: $\dfrac{\text{d}N}{N} = \text{d}t$. Integrate: $\ln|N| = t + C$.
Exponentiate: $N = Ae^t$ where $A = e^C$.
$\boxed{N = Ae^t}$
Substitute: $10 = Ae^{0} = A$, so $A = 10$.
$\boxed{N = 10e^t}$
As $t \to \infty$, $e^t \to \infty$, so $N \to \infty$. The model predicts unbounded exponential growth.
This is not realistic because real populations are limited by:
$\bullet$ Finite resources (food, water, shelter)
$\bullet$ Predators that increase as the population grows
$\bullet$ Disease that spreads more easily in dense populations
$\bullet$ Competition for territory and mates
A more realistic model would be the logistic equation $\dfrac{\text{d}N}{\text{d}t} = kN\!\left(1 - \dfrac{N}{K}\right)$, where $K$ is the carrying capacity.
An object is moving so that its velocity $v \left( = \dfrac{\text{d}s}{\text{d}t} \right)$ is inversely proportional to its displacement $s$ from a fixed point. If its velocity is $1 \, \text{m s}^{-1}$ when its displacement is $2 \, \text{m}$, find a differential equation to model the situation. Find the general solution of your differential equation.
💡 Hint 1
“Velocity is inversely proportional to displacement” means $v = \dfrac{k}{s}$ for some constant $k$.
Use the condition $v = 1$ when $s = 2$ to find $k$.
Then write $v = \dfrac{\text{d}s}{\text{d}t}$ and separate variables.
💡 Hint 2
From $v = 1$ at $s = 2$: $1 = \dfrac{k}{2}$, so $k = 2$. The DE is $\dfrac{\text{d}s}{\text{d}t} = \dfrac{2}{s}$.
Separate: $s\,\text{d}s = 2\,\text{d}t$. Integrate both sides.
👁 Reveal Solution
“Velocity is inversely proportional to displacement”: $v = \dfrac{k}{s}$ for constant $k$.
Use $v = 1$ when $s = 2$: $1 = \dfrac{k}{2} \implies k = 2$.
Since $v = \dfrac{\text{d}s}{\text{d}t}$, the differential equation is:
$$\frac{\text{d}s}{\text{d}t} = \frac{2}{s}$$
Separate: $s\,\text{d}s = 2\,\text{d}t$. Integrate: $\int s\,\text{d}s = \int 2\,\text{d}t \implies \dfrac{s^2}{2} = 2t + C$.
Multiply by 2: $s^2 = 4t + C$ (renaming the constant).
Since displacement is measured from a fixed point and $s = 2 > 0$ initially, take the positive root:
$\boxed{s = \sqrt{4t + C}}$
Verification: $\dfrac{\text{d}s}{\text{d}t} = \dfrac{4}{2\sqrt{4t+C}} = \dfrac{2}{\sqrt{4t+C}} = \dfrac{2}{s}$. ✓
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.