MHT-CET Maths · Differential Equations

Growth, Decay, and Continuous Models

When a quantity changes at a rate proportional to itself, it grows or decays exponentially. Set up dP/dt = kP, solve to P = P0 e^{kt}, fix k from two data points, and answer — the recurring MHT-CET application of differential equations.

All Differential Equations notes MHT-CET Maths strategy

Why this matters

This is the single densest applied subtopic in the chapter: 33 PYQs sit here (10 HARD, 17 MODERATE, 6 EASY), and MHT-CET repeats the same handful of stories — bacteria/population growth, radioactive/half-life decay, continuous bank compounding, moisture loss, and the special square-root and surface-area rate models — almost verbatim across years. Master one clean template (write the rate law, separate, integrate, fix the constant, fix k from a second data point) and you can answer every one. The traps are all in the setup: k is negative for decay, 'doubles' means P/P0 = 2 (not +2), and a percentage rate must become a decimal.

Concept 1 of 7

The Modelling Step — Rate Proportional to Quantity

Intuition

Almost every word problem in this subtopic hides the same sentence: 'the rate of change is proportional to the amount present.' Translate that one phrase into an equation — rate means dP/dt, proportional to the amount means kP — and the differential equation writes itself. This first step is where the marks are won or lost.

Definition

The phrase 'rate of change of P is proportional to P' translates directly to

\dfrac{dP}{dt} = kP.

$k > 0$ gives growth (population, bacteria, invested principal).
$k < 0$ gives decay (radioactivity, moisture loss, cooling) — write it as $\dfrac{dP}{dt} = -kP$ with $k>0$ to keep signs honest.

This is a separable, first-order, first-degree equation. Separate the variables and integrate: $\displaystyle\int\dfrac{dP}{P} = \int k\,dt$ , giving $\log P = kt + c$ .

Rate proportional to quantity

\dfrac{dP}{dt} = kP \quad\Longrightarrow\quad \log P = kt + c

Pthe changing quantity (mass, population, amount)
kproportionality constant — positive for growth, negative for decay
ttime

Worked example

Translate into a differential equation: 'a colony grows at a rate proportional to its size P', and separate the variables ready to integrate.

'Rate' is $\dfrac{dP}{dt}$ ; 'proportional to its size' is $kP$ . So $\dfrac{dP}{dt} = kP$ .
Separate: $\dfrac{dP}{P} = k\,dt$ .
Integrate both sides: $\displaystyle\int\dfrac{dP}{P} = \int k\,dt \Rightarrow \log P = kt + c$ .

Answer:

\dfrac{dP}{dt} = kP

, which integrates to

\log P = kt + c

Practice this conceptself-check · 4 quick reps

Try it yourself

A substance loses mass at a rate proportional to the mass present. Write the differential equation and integrate it.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Model: 'population grows proportional to itself'.
2.
Model: 'radioactive mass decays proportional to mass'.
3.
Separate $\dfrac{dP}{dt}=kP$ .
4.
Integrate $\dfrac{dP}{P}=k\,dt$ .

Decay carries a negative sign

'Rate of reduction / decay / loss' means

\dfrac{dP}{dt} = -kP

, not

+kP

. Dropping the minus makes the quantity grow — the answer then heads the wrong way. Keep

k>0

and put the sign in front explicitly.

'Proportional to' is not 'equal to'

'Rate proportional to P' introduces a constant

k

; it is

\dfrac{dP}{dt}=kP

, not

\dfrac{dP}{dt}=P

. You must determine

k

later from a data point — never assume

k=1

Drill 1 more on the modelling step — rate proportional to quantity

Concept 2 of 7

The Exponential Solution P = P0 e^{kt} and Finding k

Intuition

Once you have dP/dt = kP, the answer is always the same shape: P starts at P0 and multiplies by e^{kt}. To pin the curve down you need TWO facts — the starting value gives P0, and a second (time, value) pair gives k. Everything after that is substitution.

Definition

Solving $\dfrac{dP}{dt} = kP$ with $P(0)=P_0$ gives the master formula:

P = P_0\,e^{kt}.

Recipe:

The initial value fixes $P_0$ .
A second data point $(t_1, P_1)$ fixes $k$ : $\dfrac{P_1}{P_0} = e^{kt_1} \Rightarrow k = \dfrac{1}{t_1}\log\dfrac{P_1}{P_0}$ .
Often you never need $k$ alone — dividing two instances of $P_0 e^{kt}$ cancels $P_0$ , and the ratio form $\dfrac{P_2}{P_1}=e^{k(t_2-t_1)}$ does all the work.

Exponential growth/decay solution

P = P_0\,e^{kt}, \qquad k = \dfrac{1}{t_1}\log\!\dfrac{P_1}{P_0}

P_0value at $t=0$
krate constant, found from a second data point

Worked example

A quantity follows

\dfrac{dP}{dt}=kP

. It is 500 at

t=0

and 1500 at

t=4

. Find

P

t=8

General solution: $P = 500\,e^{kt}$ .
At $t=4$ : $1500 = 500\,e^{4k} \Rightarrow e^{4k} = 3$ .
At $t=8$ : $P = 500\,e^{8k} = 500\,(e^{4k})^2 = 500\cdot 3^2 = 4500$ .

Answer:

P(8) = 4500

Practice this conceptself-check · 4 quick reps

Try it yourself

The population of a town increases proportionally to itself, from 40,000 to 80,000 in 40 years. Find the population after another 40 years.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$P = P_0 e^{kt}$ , $P_0=10$ , $e^{5k}=2$ . Find $P$ at $t=15$ .
2.
$e^{2k}=\tfrac32$ . Value of $e^{4k}$ ?
3.
$P_0=4$ lakh grows to 6 lakh in 20 yr. $e^{20k}=$ ?
4.
Solve $\dfrac{dy}{dx}=y$ , $y(0)=3$ . Find $y(\log 2)$ .

From the bank · past-year question

Example 2Differential EquationsMODERATE

The population of a town increases at a rate proportional to the population at that time. If the population increases from forty thousand to eighty thousand in 20 years, then the population in another 40 years will be

[Q123 · 22 April Shift I · 2025]

Cancel $P_0$ by dividing — don't solve for k first

For 'grows from A to B in time

T

, find value after another

T

', you never need

k

P_0

numerically.

e^{kT} = B/A

, and after

2T

the value is

A(B/A)^2

. Chasing

k = \tfrac1T\log(B/A)

and re-exponentiating wastes time and invites arithmetic slips.

The extra time is measured from the start

'In ANOTHER 40 years' when the first stage was already 40 years means the total elapsed time is

t=80

, so

P=P_0 e^{80k}

. Reading it as

t=40

again halves the exponent and drops a doubling.

Drill 4 more on the exponential solution p = p0 e^{kt} and finding k

Concept 3 of 7

Population and Bacteria — Doubling Time and Percentage Growth

Intuition

Bacteria and population problems are pure exponential growth with a story. If the quantity DOUBLES every fixed period, you can skip calculus and just count doublings; if it grows by a PERCENTAGE, convert the percent to a multiplying factor (1 + rate) first. Both reduce to P = P0 e^{kt}.

Definition

For growth $P = P_0 e^{kt}$ :

**Doubling in period $T$ :** $e^{kT}=2$ . After $n$ such periods $(t=nT)$ , $P = P_0\cdot 2^{n}$ . No logs needed when $t$ is a whole multiple of $T$ .
Percentage increase: 'increases by $p\%$ in time $T$ ' means $\dfrac{P(T)}{P_0} = 1 + \dfrac{p}{100}$ . A 20% rise is a factor $\tfrac65$ ; a 10% rise is $\tfrac{11}{10}$ . Then $e^{kT} = 1+\tfrac{p}{100}$ fixes $k$ .
**Finding the start $P_0$ :** given two later readings, divide to get $k$ , then back-substitute one reading to recover $P_0$ .

Doubling growth

e^{kT} = 2 \quad\Longrightarrow\quad P = P_0\,2^{\,t/T}

Tdoubling time
t/Tnumber of doubling periods elapsed

Worked example

Bacteria grow proportionally to their number and double every 8 hours. If the original number is

N

, how many are present after 24 hours?

Doubling time $T=8$ h. In 24 hours there are $24/8 = 3$ doubling periods.
Each period multiplies by 2: $N \cdot 2^3 = 8N$ .

Answer:

8N

Practice this conceptself-check · 4 quick reps

Try it yourself

A culture starts at 1000 bacteria and rises 20% in 2 hours. When does it reach 2000?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Doubles every 5 h. Factor after 20 h?
2.
Convert 'rises 25%' to a growth factor.
3.
$N=N_0 e^{kt}$ : $10^4$ at $t=3$ , $4\cdot10^4$ at $t=5$ . Find $k$ .
4.
Same data: find $N_0$ .

From the bank · past-year question

Example 3Differential EquationsEASY

The bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 8 hours, then the number of bacteria in 24 hours will be

[Q113 · 11th May Shift 1 · 2023]

'Doubles' means the ratio is 2, not '+2'

'The population doubles' sets

\dfrac{P}{P_0}=2

(so

e^{kT}=2

) — it does not mean

P = P_0 + 2

. Similarly 'triples' is a factor 3. Always translate the word into a multiplying ratio.

Turn a percentage into a factor before touching k

A '20% increase' is a factor of

\tfrac65

(i.e.

1.2

), NOT

0.20

. Write

e^{kT} = 1 + \tfrac{p}{100}

; using

0.20

instead of

1.20

corrupts

k

and every later value.

Drill 4 more on population and bacteria — doubling time and percentage growth

Concept 4 of 7

Radioactive Decay and Half-Life

Intuition

Radioactivity is exponential decay: mass falls at a rate proportional to itself. The half-life h is the time to lose half the mass, so it plays exactly the role a doubling time plays for growth — after n half-lives, only (1/2)^n of the mass survives. The rate constant is k = log 2 / h.

Definition

Decay model $\dfrac{dm}{dt} = -km$ with solution $m = m_0 e^{-kt}$ .

Half-life link: at $t=h$ , $m = \tfrac{m_0}{2}$ , so $e^{-kh}=\tfrac12 \Rightarrow k = \dfrac{\log 2}{h}$ .
**After $n$ half-lives** $(t = nh)$ : $m = m_0\left(\tfrac12\right)^{n}$ . Just count half-lives when $t$ is a whole multiple of $h$ .
Initial decay rate: $\left.\dfrac{dm}{dt}\right|_{t=0} = -km_0 = -\dfrac{m_0\log 2}{h}$ — negative because mass is falling.

Half-life rate constant

k = \dfrac{\log 2}{h}, \qquad m = m_0\left(\tfrac12\right)^{t/h}

hhalf-life — time to lose half the mass
m_0initial mass at $t=0$

Worked example

A radioactive sample has half-life 5 years and initial mass 64 g. How much is left after 15 years?

Number of half-lives: $15/5 = 3$ .
Each half-life halves the mass: $64 \to 32 \to 16 \to 8$ g.
Equivalently $m = 64\left(\tfrac12\right)^3 = 8$ g.

Answer:

8

Practice this conceptself-check · 4 quick reps

Try it yourself

A substance of half-life

h

days and initial mass

m_0

has what initial decay rate?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Half-life 1600 yr, start 60 g. Amount after 3200 yr?
2.
Half-life 15 min. Fraction left after 30 min?
3.
Rate constant for half-life $h$ ?
4.
27 g decays to 8 g in 3 h. $m$ after 1 more hour?

From the bank · past-year question

Example 4Differential EquationsMODERATE

A radioactive substance, with initial mass

m_0

, has a half-life of

h

days. Then its initial decay rate is given by

[Q140 · 13th May Shift 2 · 2024]

The initial decay rate is negative

Because mass decreases,

\dfrac{dm}{dt}=-km_0 = -\dfrac{m_0\log 2}{h}

carries a MINUS sign. The positive-looking distractor

\dfrac{m_0}{h}\log 2

is the classic trap — decay rates are negative.

Count half-lives only when time is a whole multiple

The shortcut

m = m_0(\tfrac12)^{t/h}

is easiest when

t/h

is an integer (3 half-lives, etc.). If

t

is not a clean multiple (e.g. 27 g → 8 g in 3 h, then 1 more hour), fall back to

m = m_0 e^{-kt}

with

e^{-k}

found from the given step.

Drill 6 more on radioactive decay and half-life

Concept 5 of 7

Continuous Compounding of Money

Intuition

When a principal grows 'continuously' at rate r, it obeys the same growth law as a population — dA/dt = rA — so A = P e^{rt}. The only new wrinkle is that the rate is quoted as a percentage per year and must be written as a decimal (10% ⇒ r = 0.1).

Definition

Continuous growth of a principal:

\dfrac{dA}{dt} = rA \quad\Longrightarrow\quad A = P\,e^{rt},

where

P

is the amount invested at

t=0

and

r

is the annual rate written as a decimal.

'Doubles in $T$ years' gives $e^{rT}=2$ , so $rT = \log 2$ — used to find either $r$ or a doubling-based amount.
'Rate $x\%$ , doubles in $T$ ': $\dfrac{x}{100}\cdot T = \log 2 \Rightarrow x = \dfrac{100\log 2}{T}$ .

Continuous compounding

A = P\,e^{rt}, \qquad e^{rT} = 2 \text{ if the principal doubles in } T

Pprincipal invested at $t=0$
rannual rate as a decimal ( $x\% \to x/100$ )

Worked example

₹2000 is invested at 10% per year compounded continuously. What is the amount after 5 years? (

e^{0.5}=1.648

)

$A = P e^{rt}$ with $P=2000$ , $r = 0.10$ , $t=5$ .
$rt = 0.10\times 5 = 0.5$ , so $A = 2000\,e^{0.5} = 2000\times 1.648$ .
$A = 3296$ .

Answer:₹3296

Practice this conceptself-check · 4 quick reps

Try it yourself

The principal doubles in 10 years under continuous compounding. Find the rate

x\%

. (

\log 2 = 0.6931

)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
$P=200$ , doubles to 400 in 6 yr. $A$ at 33 yr?
2.
Write 8% per year as $r$ .
3.
Doubles in 20 yr. Rate $x\%$ ? ( $\log 2=0.693$ )
4.
$A = P e^{rt}$ : $e^{6r}=2$ . $A/P$ at $t=12$ ?

From the bank · past-year question

Example 5Differential EquationsMODERATE

The money invested in a company is compounded continuously. If Rs. 200 invested today becomes Rs. 400 in 6 years, then at the end of 33 years it will become Rs.

[Q104 · 14th May Shift 1 · 2024]

Convert the % rate to a decimal

10% per year is

r = 0.10

, not

r = 10

. Using 10 blows the exponent up by a factor of 100. Write

r = x/100

every time.

Continuous compounding uses $e^{rt}$ , not $(1+r)^t$

The word 'continuously' means

A = Pe^{rt}

. Reaching for the annual-compound formula

A = P(1+r)^t

or simple interest is the intended distractor — it gives a different (wrong) number.

Drill 4 more on continuous compounding of money

Concept 6 of 7

Moisture Loss and General First-Order Rate Models

Intuition

Moisture-loss problems are ordinary exponential decay dressed as a story: 'loses half in the first hour' fixes the rate exactly like a half-life. A slightly harder cousin adds a constant term — dP/dt = kP − c — which is still separable but needs a substitution before you integrate.

Definition

Pure proportional loss (moisture, cooling of the simplest kind): $\dfrac{dP}{dt} = -kP \Rightarrow \log\dfrac{P}{N} = -kt$ . 'Loses half in the first hour' gives $k = \log 2$ ; then solve for the time to lose any fraction. Mixed model with a constant $\dfrac{dP}{dt} = kP - c$ : rewrite as $\dfrac{dP}{kP - c} = dt$ and integrate to $\dfrac1k\log|kP - c| = t + C$ . Fix $C$ from $P(0)$ , then substitute the target $P$ . (For $\dfrac{dP}{dt}=0.5P-450$ , $kP-c = 0.5(P-900)$ , so $2\log|P-900| = t + C$ .)

Fraction-lost time (pure decay)

\log\dfrac{P}{N} = -kt, \qquad k = \log 2 \text{ if half is lost in unit time}

Ninitial content at $t=0$
P/Nfraction remaining

Worked example

A wet sheet loses half its moisture in the first hour, at a rate proportional to the moisture present. In how many hours is 99% lost?

$\dfrac{dP}{dt} = -kP \Rightarrow \log\dfrac{P}{N} = -kt$ . Half lost in 1 hour: $\log\tfrac12 = -k \Rightarrow k = \log 2$ .
99% lost means $P = 0.01N$ , i.e. $\dfrac{P}{N} = \tfrac{1}{100}$ .
$\log\tfrac{1}{100} = -kt \Rightarrow t = \dfrac{\log 100}{k} = \dfrac{2\log 10}{\log 2}$ .

Answer:

t = \dfrac{2\log 10}{\log 2}

hours.

Practice this conceptself-check · 4 quick reps

Try it yourself

Solve

\dfrac{dP}{dt} = 0.5P - 450

with

P(0) = 850

, and find the time when

P = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Loses half in 1 h. Time to lose 90%?
2.
$k$ if half is lost per hour.
3.
Factor $0.5P - 450$ .
4.
Integrate $\dfrac{dP}{P-900}=0.5\,dt$ .

From the bank · past-year question

Example 6Differential EquationsMODERATE

A wet substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the open air loses half its moisture during the first hour, then the time

t

, in which 99% of the moisture will be lost, is

[Q146 · 4th May Shift 1 · 2023]

'99% lost' means the fraction LEFT is 0.01

Losing 99% leaves 1%: use

P/N = 0.01

, giving

\log 100

on top. Plugging in 0.99 (the fraction lost) instead of 0.01 (the fraction remaining) inverts the ratio and the answer.

The constant term needs factoring before you separate

\dfrac{dP}{dt}=0.5P-450

is NOT

\dfrac{dP}{P}=0.5\,dt

. Factor to

0.5(P-900)

first, so the variable that integrates cleanly is

P-900

, not

P

Drill 2 more on moisture loss and general first-order rate models

Concept 7 of 7

Special-Rate Models — Square-Root and Surface-Area Decay

Intuition

Two recurring non-standard rate laws break the 'proportional to itself' mould. Assets/tanks that change at a rate proportional to the SQUARE ROOT of the amount give dx/dt = −k√x, which integrates to 2√x = −kt + c (a straight line in √x). A raindrop evaporating proportionally to its SURFACE AREA collapses, after using V and S, to the simple linear dr/dt = −k.

Definition

Square-root rate (assets shrinking, tank draining): $\dfrac{dx}{dt} = -k\sqrt{x}$ . Separate $\dfrac{dx}{\sqrt x} = -k\,dt$ and integrate:

2\sqrt{x} = -kt + c.

Fix

c

from

x(0)

, fix

k

from a second reading, then set

x=0

for 'empties / bankrupt'. Surface-area rate (spherical raindrop evaporating):

\dfrac{dV}{dt} = -kS

. With

V = \tfrac43\pi r^3,\ S = 4\pi r^2

, the

4\pi r^2

cancels and it reduces to

\dfrac{dr}{dt} = -k

, so the radius falls linearly:

r = -kt + c

Square-root and surface-area models

\dfrac{dx}{dt} = -k\sqrt{x} \Rightarrow 2\sqrt{x} = -kt + c; \qquad \dfrac{dV}{dt} = -kS \Rightarrow \dfrac{dr}{dt} = -k

2\sqrt{x} = -kt+cthe integrated square-root law — linear in $\sqrt{x}$
dr/dt = -ksurface-area evaporation ⇒ radius shrinks at a constant rate

Worked example

A person's assets fall at a rate proportional to

\sqrt{\text{assets}}

. They drop from ₹25 lakh to ₹6.25 lakh in 2 years. When is he bankrupt?

$\dfrac{dx}{dt} = -k\sqrt x \Rightarrow 2\sqrt x = -kt + c$ .
At $t=0,\ x=25$ : $2\cdot 5 = c \Rightarrow c = 10$ .
At $t=2,\ x=6.25$ : $2\cdot 2.5 = -2k + 10 \Rightarrow 5 = -2k+10 \Rightarrow k = \tfrac52$ .
Bankrupt when $x=0$ : $0 = -\tfrac52 T + 10 \Rightarrow T = 4$ years.

Answer:

T = 4

years

Practice this conceptself-check · 4 quick reps

Try it yourself

Water drains from a tank at a rate proportional to

\sqrt{\text{depth}}

. Depth is 16 m at

t=0

and 4 m at

t=2

h. Find the depth at

t=3.5

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Integrate $\dfrac{dx}{dt}=-k\sqrt x$ .
2.
Assets 10 lakh → 10000 in 3 yr ( $\sqrt{}$ rate). Bankrupt when?
3.
Raindrop, $dV/dt \propto S$ . What law does $r$ obey?
4.
Raindrop $r=3$ at $t=0$ , $r=2$ at $t=1$ . $r(t)$ ?

From the bank · past-year question

Example 7Differential EquationsHARD

The assets of a person are reduced in his business such that the rate of reduction is proportional to the square root of the existing assets. If the assets were initially ₹10,00,000 and due to loss they reduce to ₹10,000 after 3 years, then the number of years required for the person to go bankrupt will be

[Q144 · 16th May Shift 1 · 2023]

$\int \dfrac{dx}{\sqrt x} = 2\sqrt x$ , not $\log\sqrt x$

The square-root rate separates to

x^{-1/2}\,dx

, whose integral is

2\sqrt x

— a power-rule integral, NOT a logarithm. Reflexively writing

\log

(as for

dx/dt = -kx

) is the number-one error in these problems.

Surface-area evaporation makes the RADIUS linear

For a raindrop, 'proportional to surface area' plus

V=\tfrac43\pi r^3

forces

\dfrac{dr}{dt}=-k

, so

r = c - kt

is linear in

t

. Trying to make the volume or radius exponential misses the cancellation of

4\pi r^2

Drill 6 more on special-rate models — square-root and surface-area decay

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (7)

The Modelling Step — Rate Proportional to Quantity
Rate proportional to quantity
$\dfrac{dP}{dt} = kP \quad\Longrightarrow\quad \log P = kt + c$
The Exponential Solution P = P0 e^{kt} and Finding k
Exponential growth/decay solution
$P = P_0\,e^{kt}, \qquad k = \dfrac{1}{t_1}\log\!\dfrac{P_1}{P_0}$
Population and Bacteria — Doubling Time and Percentage Growth
Doubling growth
$e^{kT} = 2 \quad\Longrightarrow\quad P = P_0\,2^{\,t/T}$
Radioactive Decay and Half-Life
Half-life rate constant
$k = \dfrac{\log 2}{h}, \qquad m = m_0\left(\tfrac12\right)^{t/h}$
Continuous Compounding of Money
Continuous compounding
$A = P\,e^{rt}, \qquad e^{rT} = 2 \text{ if the principal doubles in } T$
Moisture Loss and General First-Order Rate Models
Fraction-lost time (pure decay)
$\log\dfrac{P}{N} = -kt, \qquad k = \log 2 \text{ if half is lost in unit time}$
Special-Rate Models — Square-Root and Surface-Area Decay
Square-root and surface-area models
$\dfrac{dx}{dt} = -k\sqrt{x} \Rightarrow 2\sqrt{x} = -kt + c; \qquad \dfrac{dV}{dt} = -kS \Rightarrow \dfrac{dr}{dt} = -k$

Watch out for (14)

Decay carries a negative sign
→ The Modelling Step — Rate Proportional to Quantity
'Proportional to' is not 'equal to'
→ The Modelling Step — Rate Proportional to Quantity
Cancel $P_0$ by dividing — don't solve for k first
→ The Exponential Solution P = P0 e^{kt} and Finding k
The extra time is measured from the start
→ The Exponential Solution P = P0 e^{kt} and Finding k
'Doubles' means the ratio is 2, not '+2'
→ Population and Bacteria — Doubling Time and Percentage Growth
Turn a percentage into a factor before touching k
→ Population and Bacteria — Doubling Time and Percentage Growth
The initial decay rate is negative
→ Radioactive Decay and Half-Life
Count half-lives only when time is a whole multiple
→ Radioactive Decay and Half-Life
Convert the % rate to a decimal
→ Continuous Compounding of Money
Continuous compounding uses $e^{rt}$ , not $(1+r)^t$
→ Continuous Compounding of Money
'99% lost' means the fraction LEFT is 0.01
→ Moisture Loss and General First-Order Rate Models
The constant term needs factoring before you separate
→ Moisture Loss and General First-Order Rate Models
$\int \dfrac{dx}{\sqrt x} = 2\sqrt x$ , not $\log\sqrt x$
→ Special-Rate Models — Square-Root and Surface-Area Decay
Surface-area evaporation makes the RADIUS linear
→ Special-Rate Models — Square-Root and Surface-Area Decay

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Differential EquationsMODERATE

The rate of increase of the population of a city is proportional to the population present at that instant. In the period of 40 years the population increased from 30,000 to 40,000. At any time

t

the population is

(a)(b)^{\frac{t}{40}}

. Then the values of a and

b

are respectively

[Q111 · 25 April Shift II · 2025]

Example 2Differential EquationsMODERATE

The rate at which the population of a city increases varies as the population. In a period of 20 years, the population increased from 4 lakhs to 6 lakhs. In another 20 years the population will be

[Q104 · 25 April Shift I · 2025]

Example 3Differential EquationsHARD

In a certain culture of bacteria, the rate of increase is proportional to the number present. If there are

10^4

at the end of 3 hours and

4\cdot10^4

at the end of 5 hours, then there were _____ in the beginning.

[Q145 · 9th May Shift 2 · 2023]

Example 4Differential EquationsEASY

An ice ball melts at the rate which is proportional to the amount of ice at that instant. Half of the quantity of ice melts in 15 minutes.

x_0

is the initial quantity of ice. If after 30 minutes the amount of ice left is

kx_0

, then the value of k is

[Q104 · 11th May Shift 2 · 2023]

Example 5Differential EquationsMODERATE

The money invested in a company is compounded continuously. ₹ 400 invested today becomes ₹800 in 6 years, then at the end of 33 years, it will become.

(\sqrt{2}= 1.4142)

[Q131 · 20 April Shift I · 2025]

Drill every past-year question on this subtopic

33 questions from the bank — paginated, with cart and Word-export support.

Drill the 33 questions

Drill every past-year question on this subtopic

Related notes