Solving first-order differential equations by separation of variables, modelling with DEs.
Answer at least 3 of 3 correctly to complete this section.
A differential equation (DE) is an equation involving a derivative. You have already seen simple ones like dxdy=4x−1, which you solved by direct integration. But what about dxdy=2y or dxdy=xy? Here the rate of change depends on y itself — direct integration does not work.
Differential equations are everywhere in the real world. Population growth, radioactive decay, Newton’s law of cooling, the spread of diseases, mixing problems — all are modelled by DEs. At A-Level, you need to solve first-order separable differential equations and apply them in modelling contexts.
The good news: the method is always the same. Separate the variables, integrate both sides, find the constant. Master these steps and you can tackle any separable DE the exam throws at you.
A separable differential equation can be written in the form:
dxdy=f(x)⋅g(y)
The right-hand side is a product of a function of x only and a function of y only. To solve it:
g(y)1dy=f(x)dx
∫g(y)1dy=∫f(x)dx
Solve dxdy=3x2y.
Separate: Divide both sides by y and multiply by dx:
y1dy=3x2dx
Integrate:
∫y1dy=∫3x2dx
ln∣y∣=x3+c
Rearrange: Taking exponentials of both sides:
∣y∣=ex3+c=ec⋅ex3
Since ec is a positive constant, write A=±ec to absorb the sign:
y=Aex3
where A is an arbitrary constant (A=0).
Why just one constant? If you write +c1 on the left and +c2 on the right, they combine: ln∣y∣+c1=x3+c2 gives ln∣y∣=x3+(c2−c1). The two constants merge into one, so you only need one c from the start.
Solve dxdy=yx.
Separate: Multiply both sides by y:
ydy=xdx
Integrate:
∫ydy=∫xdx
2y2=2x2+c
y2=x2+k(where k=2c)
This is the general solution. It describes a family of hyperbolas (or circles, depending on the sign of k).
Answer at least 3 of 3 correctly to complete this section.
The general solution contains an arbitrary constant. An initial condition (or boundary condition) pins down the value of that constant, giving a particular solution.
Solve dxdy=4xy, given that y=3 when x=0.
Separate and integrate:
y1dy=4xdx
ln∣y∣=2x2+c
Use the initial condition y=3 when x=0:
ln3=2(0)+c⟹c=ln3
So ln∣y∣=2x2+ln3, which gives:
ln∣y∣−ln3=2x2⟹ln3y=2x2
3y=e2x2⟹y=3e2x2
Tip: It is often easier to find c before rearranging. Once you have ln∣y∣=2x2+ln3, substituting the numbers is straightforward. If you rearrange to y=Ae2x2 first, you can also substitute: 3=Ae0=A, giving A=3.
The rate of change of a quantity N is proportional to N itself. Given that N=50 when t=0, and N=150 when t=2, find N in terms of t.
The DE is dtdN=kN for some constant k>0.
Separate and integrate:
N1dN=kdt⟹lnN=kt+c
(Taking N>0 since it is a quantity.)
Use N=50 when t=0: ln50=c.
So lnN=kt+ln50, giving N=50ekt.
Use N=150 when t=2:
150=50e2k⟹e2k=3⟹2k=ln3⟹k=2ln3
Therefore:
N=50etln3/2=50⋅3t/2
Solve dxdy=xy+1, given that y=0 when x=1.
Separate:
y+11dy=x1dx
Integrate:
ln∣y+1∣=ln∣x∣+c
Use y=0, x=1: ln1=ln1+c, so c=0.
Therefore ln∣y+1∣=ln∣x∣, giving y+1=x (taking the positive case since y=0 when x=1), so y=x−1.
Always check: Substitute back into the original DE. dxdy=1 and xy+1=x(x−1)+1=1 ✓. Also check the condition: when x=1, y=0 ✓.
Answer at least 3 of 3 correctly to complete this section.
The simplest model: “the rate of change is proportional to the current value.”
dtdP=kP
The solution is always P=P0ekt, where P0 is the initial value.
A population of bacteria doubles every 3 hours. Initially there are 200 bacteria. Find the population after 10 hours.
Model: dtdP=kP, with P(0)=200.
General solution: P=200ekt.
Using “doubles every 3 hours”: P(3)=400.
400=200e3k⟹e3k=2⟹k=3ln2
Therefore: P=200etln2/3=200⋅2t/3.
At t=10: P=200⋅210/3=200⋅23.33…≈2016 bacteria.
“The rate of cooling is proportional to the temperature difference between the object and its surroundings.”
dtdθ=−k(θ−θ0)
where θ is the object’s temperature, θ0 is the surrounding temperature, and k>0.
A cup of tea at 90°C is placed in a room at 20°C. After 5 minutes, the tea is at 70°C. Find the temperature after 15 minutes.
dtdθ=−k(θ−20)
Separate: θ−201dθ=−kdt
Integrate: ln∣θ−20∣=−kt+c
Initial condition (t=0, θ=90): ln70=c
So ln(θ−20)=−kt+ln70, giving θ−20=70e−kt, i.e.:
θ=20+70e−kt
Use θ=70 when t=5:
70=20+70e−5k⟹50=70e−5k⟹e−5k=75
−5k=ln75⟹k=−51ln75=51ln57
At t=15:
θ=20+70e−15k=20+70(e−5k)3=20+70(75)3
=20+70⋅343125=20+3438750=20+25.5…≈45.5°C
Exam tip: Notice the trick of writing e−15k=(e−5k)3. This avoids having to calculate k as a decimal. Examiners reward exact answers.
In an exam, you may be asked to comment on the limitations of a model:
Answer at least 3 of 3 correctly to complete this section.
Now that you have practised the core separation technique and seen modelling applications, we can tackle DEs where the integration step requires partial fractions. This combines two techniques you already know.
Solve dxdy=y(1−y), given that 0<y<1.
Separate:
y(1−y)1dy=dx
To integrate the left side, use partial fractions:
y(1−y)1=y1+1−y1
(You can verify: y1+1−y1=y(1−y)1−y+y=y(1−y)1 ✓)
Integrate:
∫(y1+1−y1)dy=∫1dx
lny−ln(1−y)=x+c
ln(1−yy)=x+c
Why this appears here, not earlier: The separation step is the same as always. But integrating y(1−y)1 requires partial fractions — an algebraic technique you need to have practised separately before combining it with DEs.
Answer at least 1 of 1 correctly to complete this section.
The rate of increase of a plant’s height h cm is modelled by dtdh=h6, where t is time in weeks. When t=0, h=4.
(a) Solve the differential equation to find h in terms of t.
(b) Find the height of the plant after 6 weeks.
(c) State one limitation of this model.
(a) Separate: hdh=6dt.
Integrate: 2h2=6t+c.
Use h=4, t=0: 216=c, so c=8.
2h2=6t+8⟹h2=12t+16
Since h>0: h=12t+16.
(b) At t=6: h=72+16=88=222≈9.38 cm.
(c) The model predicts the plant grows without limit as t→∞ (since h=12t+16→∞), which is unrealistic. In reality, the plant reaches a maximum height.
Water containing 5 grams per litre of salt flows into a tank at 2 litres per minute. The well-mixed solution flows out at the same rate. The tank holds 100 litres and initially contains pure water. Let S be the mass of salt in the tank after t minutes.
Before jumping to the DE, work through each part of the model:
Step 1 — What is the rate in?
Salt concentration in = 5 g/L. Flow rate in = 2 L/min. So:
Rate in=5×2=10 g/min
Step 2 — What is the rate out?
The tank contains S grams in 100 litres, so the concentration is 100S g/L. Flow rate out = 2 L/min. So:
Rate out=100S×2=50S g/min
Step 3 — Write the DE.
dtdS=rate in−rate out=10−50S=50500−S
Step 4 — Solve.
Separate:
500−S1dS=501dt
Integrate:
−ln∣500−S∣=50t+c
Use S=0, t=0: −ln500=c.
−ln(500−S)=50t−ln500
ln500−S500=50t
500−S500=et/50
500−S=500e−t/50⟹S=500(1−e−t/50)
As t→∞, S→500 g, which makes sense: the tank approaches the same concentration as the inflow (5 g/L × 100 L = 500 g).
Answer at least 3 of 4 correctly to complete this section.
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