Applications of Differential Equations
Introduction
Differential equations are not just mathematical abstractions—they model real-world phenomena in physics, chemistry, biology, economics, and engineering. Understanding applications is crucial for JEE Advanced.
Classification of Applications
1. Growth and Decay Problems
- Population growth
- Radioactive decay
- Compound interest
- Bacterial growth
- Carbon dating
2. Physics Applications
- Newton’s Law of Cooling
- Motion under gravity
- Simple Harmonic Motion (SHM)
- Electric circuits (RC, RL, RLC)
- Velocity and acceleration problems
3. Geometric Applications
- Orthogonal trajectories
- Curves with specific properties
- Family of curves
4. Mixing Problems
- Solution concentration
- Chemical reactions
- Pollution in lakes
Growth and Decay Models
Exponential Growth/Decay Law
Law: The rate of change of a quantity is proportional to the quantity itself.
$$\boxed{\frac{dN}{dt} = kN}$$
where:
- $N$ = quantity at time $t$
- $k$ = growth constant ($k > 0$ for growth, $k < 0$ for decay)
Solution
Solving the DE:
$$\frac{dN}{N} = k \, dt$$ $$\ln|N| = kt + C$$ $$N = N_0 e^{kt}$$where $N_0$ is the initial quantity at $t = 0$.
General Solution:
$$\boxed{N(t) = N_0 e^{kt}}$$
Growth Problems
Population Growth
Model: $\frac{dP}{dt} = kP$ where $P$ is population.
Example: A bacterial culture grows at a rate proportional to its size. If population doubles in 3 hours, find the time to triple.
Solution: $P = P_0 e^{kt}$
When $t = 3$, $P = 2P_0$:
$$2P_0 = P_0 e^{3k}$$ $$2 = e^{3k}$$ $$k = \frac{\ln 2}{3}$$For tripling: $3P_0 = P_0 e^{kt}$
$$3 = e^{kt}$$ $$t = \frac{\ln 3}{k} = \frac{\ln 3}{\ln 2/3} = \frac{3\ln 3}{\ln 2} \approx 4.75 \text{ hours}$$Decay Problems
Radioactive Decay
Half-life Formula:
$$\boxed{N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}}$$
where $T_{1/2}$ is the half-life.
Relation with decay constant:
$$k = \frac{\ln 2}{T_{1/2}}$$Carbon Dating
Uses radioactive decay of C-14 to determine age of organic materials.
Formula:
$$t = \frac{1}{k}\ln\left(\frac{N_0}{N}\right)$$where $k = \frac{\ln 2}{5730}$ (half-life of C-14 is 5730 years)
Newton’s Law of Cooling
Law: The rate of cooling of a body is proportional to the temperature difference between the body and its surroundings.
$$\boxed{\frac{dT}{dt} = -k(T - T_s)}$$
where:
- $T$ = temperature of body at time $t$
- $T_s$ = surrounding (ambient) temperature
- $k$ = cooling constant ($k > 0$)
Solution
$$\frac{dT}{T - T_s} = -k \, dt$$ $$\ln|T - T_s| = -kt + C$$ $$T - T_s = Ae^{-kt}$$Using initial condition $T(0) = T_0$:
$$\boxed{T(t) = T_s + (T_0 - T_s)e^{-kt}}$$
Memory Trick for Cooling
“Temperature approaches surroundings exponentially”:
- As $t \to \infty$, $e^{-kt} \to 0$
- So $T \to T_s$ (body temperature approaches surroundings)
Motion Problems
Free Fall with Air Resistance
Equation:
$$\boxed{m\frac{dv}{dt} = mg - kv}$$
where:
- $m$ = mass
- $v$ = velocity
- $g$ = acceleration due to gravity
- $k$ = air resistance constant
Terminal Velocity: $v_{\text{terminal}} = \frac{mg}{k}$
Interactive Demo: Visualize DE Applications
Plot solution curves for growth, decay, and cooling problems.
Simple Harmonic Motion (SHM)
Differential Equation:
$$\boxed{\frac{d^2x}{dt^2} + \omega^2 x = 0}$$
where $\omega = \sqrt{\frac{k}{m}}$ is angular frequency.
General Solution:
$$x(t) = A\cos(\omega t) + B\sin(\omega t)$$or equivalently:
$$x(t) = C\cos(\omega t + \phi)$$Electric Circuit Applications
RC Circuit (Charging)
Equation (charging):
$$\boxed{R\frac{dQ}{dt} + \frac{Q}{C} = V_0}$$
where:
- $Q$ = charge
- $R$ = resistance
- $C$ = capacitance
- $V_0$ = applied voltage
Solution:
$$Q(t) = CV_0(1 - e^{-t/RC})$$Current:
$$I(t) = \frac{dQ}{dt} = \frac{V_0}{R}e^{-t/RC}$$RL Circuit
Equation:
$$\boxed{L\frac{dI}{dt} + RI = V_0}$$
where $L$ is inductance.
Solution:
$$I(t) = \frac{V_0}{R}(1 - e^{-Rt/L})$$Time Constant: $\tau = \frac{L}{R}$
Geometric Applications
Orthogonal Trajectories
Definition: Curves that intersect a given family of curves at right angles.
Method:
- Find DE of given family by eliminating constant
- Replace $\frac{dy}{dx}$ with $-\frac{dx}{dy}$ (or $-\frac{1}{dy/dx}$)
- Solve the new DE
Example: Find orthogonal trajectories of $x^2 + y^2 = c^2$
Given: $x^2 + y^2 = c^2$
Differentiate: $2x + 2y\frac{dy}{dx} = 0$
So: $\frac{dy}{dx} = -\frac{x}{y}$
For orthogonal trajectories:
$$\frac{dy}{dx} = \frac{y}{x}$$This gives: $\frac{dy}{y} = \frac{dx}{x}$
$\ln|y| = \ln|x| + C$
$y = Ax$ (family of straight lines through origin)
Mixing Problems
Tank Problem
Setup: A tank contains solution. Solution flows in and out.
General Equation:
$$\boxed{\frac{dQ}{dt} = \text{Rate in} - \text{Rate out}}$$
where $Q$ is amount of substance in tank.
Rate in = (concentration in) × (flow rate in)
Rate out = (concentration in tank) × (flow rate out)
Common Mistakes in Applications
❌ Mistake 1: Wrong Sign in Decay
Wrong: $\frac{dN}{dt} = kN$ for decay ✗
Correct: $\frac{dN}{dt} = -kN$ (negative for decay) ✓
❌ Mistake 2: Forgetting Ambient Temperature
Wrong: $\frac{dT}{dt} = -kT$ for cooling ✗
Correct: $\frac{dT}{dt} = -k(T - T_s)$ ✓
❌ Mistake 3: Wrong Units
Wrong: Mixing mass and volume units ✗
Correct: Ensure consistent units throughout ✓
❌ Mistake 4: Negative Reciprocal in Orthogonal
Wrong: Same slope for orthogonal curves ✗
Correct: Use $m_1 \times m_2 = -1$ for perpendicular lines ✓
Solved Examples
Example 1: Population Growth (JEE Main)
A population of bacteria doubles in 2 hours. If initially there are 1000 bacteria, how many will there be after 6 hours?
Solution:
Model: $P = P_0 e^{kt}$
Given: $P_0 = 1000$, at $t = 2$, $P = 2000$
$$2000 = 1000e^{2k}$$ $$2 = e^{2k}$$ $$k = \frac{\ln 2}{2}$$At $t = 6$:
$$P = 1000e^{6k} = 1000e^{6 \cdot \ln 2/2} = 1000e^{3\ln 2} = 1000 \cdot 2^3 = 8000$$Answer: 8000 bacteria
Example 2: Newton’s Cooling (JEE Main)
A body at 100°C cools in air at 25°C. After 10 minutes, temperature is 60°C. Find temperature after 20 minutes.
Solution:
$T(t) = T_s + (T_0 - T_s)e^{-kt}$
Given: $T_0 = 100$, $T_s = 25$
$$T(t) = 25 + 75e^{-kt}$$At $t = 10$, $T = 60$:
$$60 = 25 + 75e^{-10k}$$ $$35 = 75e^{-10k}$$ $$e^{-10k} = \frac{35}{75} = \frac{7}{15}$$At $t = 20$:
$$T = 25 + 75e^{-20k} = 25 + 75(e^{-10k})^2$$ $$= 25 + 75\left(\frac{7}{15}\right)^2$$ $$= 25 + 75 \times \frac{49}{225}$$ $$= 25 + \frac{3675}{225} = 25 + \frac{49}{3}$$ $$= \frac{75 + 49}{3} = \frac{124}{3} \approx 41.33°C$$Answer: Approximately 41.3°C
Example 3: Radioactive Decay (JEE Advanced)
The half-life of a radioactive substance is 1600 years. What percentage will remain after 3200 years?
Solution:
After one half-life (1600 years): 50% remains
After two half-lives (3200 years): $50\% \times \frac{1}{2} = 25\%$ remains
Alternative Method:
$$N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} = N_0\left(\frac{1}{2}\right)^{3200/1600} = N_0\left(\frac{1}{2}\right)^2 = \frac{N_0}{4}$$Answer: 25% remains
Example 4: Free Fall with Resistance (JEE Advanced)
A body of mass 1 kg falls from rest under gravity with air resistance proportional to velocity ($k = 0.2$). Find velocity after 5 seconds. ($g = 10$ m/s²)
Solution:
$$\frac{dv}{dt} = g - \frac{k}{m}v = 10 - 0.2v$$This is linear: $\frac{dv}{dt} + 0.2v = 10$
$IF = e^{0.2t}$
$$ve^{0.2t} = \int 10e^{0.2t} dt = \frac{10e^{0.2t}}{0.2} + C = 50e^{0.2t} + C$$ $$v = 50 + Ce^{-0.2t}$$At $t = 0$, $v = 0$: $0 = 50 + C$, so $C = -50$
$$v = 50(1 - e^{-0.2t})$$At $t = 5$:
$$v = 50(1 - e^{-1}) = 50(1 - 0.3679) \approx 31.6 \text{ m/s}$$Terminal velocity: $v_{\text{term}} = \frac{mg}{k} = \frac{10}{0.2} = 50$ m/s
Answer: 31.6 m/s
Example 5: Orthogonal Trajectories (JEE Advanced)
Find orthogonal trajectories of parabolas $y^2 = 4ax$.
Solution:
Given family: $y^2 = 4ax$ … (1)
Differentiate: $2y\frac{dy}{dx} = 4a$
From (1): $a = \frac{y^2}{4x}$
Substitute: $2y\frac{dy}{dx} = 4 \cdot \frac{y^2}{4x} = \frac{y^2}{x}$
$$\frac{dy}{dx} = \frac{y}{2x}$$For orthogonal trajectories:
$$\frac{dy}{dx} = -\frac{2x}{y}$$Solve: $y \, dy = -2x \, dx$
$$\frac{y^2}{2} = -x^2 + C$$ $$\boxed{y^2 + 2x^2 = C}$$(family of ellipses)
Example 6: Mixing Problem (JEE Advanced)
A 100-liter tank initially contains pure water. Brine with 2 kg/L salt flows in at 5 L/min. Mixed solution flows out at same rate. Find salt amount after 10 minutes.
Solution:
Let $Q(t)$ = amount of salt at time $t$
Rate in = $2 \times 5 = 10$ kg/min
Concentration in tank = $\frac{Q}{100}$ kg/L
Rate out = $\frac{Q}{100} \times 5 = \frac{Q}{20}$ kg/min
$$\frac{dQ}{dt} = 10 - \frac{Q}{20}$$Linear DE: $\frac{dQ}{dt} + \frac{Q}{20} = 10$
$IF = e^{t/20}$
$$Qe^{t/20} = \int 10e^{t/20} dt = 200e^{t/20} + C$$ $$Q = 200 + Ce^{-t/20}$$At $t = 0$, $Q = 0$: $C = -200$
$$Q = 200(1 - e^{-t/20})$$At $t = 10$:
$$Q = 200(1 - e^{-0.5}) = 200(1 - 0.6065) \approx 78.7 \text{ kg}$$Answer: Approximately 78.7 kg
Practice Problems
Level 1: JEE Main Basics
Problem 1.1: A substance decays at a rate proportional to its amount. If half of it decays in 30 days, how long for 75% to decay?
Solution
Half-life = 30 days
For 75% decay, 25% remains = $\frac{1}{4}$ of original
$\frac{1}{4} = \left(\frac{1}{2}\right)^{t/30}$
$\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{t/30}$
$t = 60$ days
Problem 1.2: A hot liquid at 80°C cools to 60°C in 5 minutes in a room at 20°C. What is its temperature after 10 minutes?
Solution
$T = 20 + 60e^{-kt}$
At $t = 5$: $60 = 20 + 60e^{-5k}$
$e^{-5k} = \frac{40}{60} = \frac{2}{3}$
At $t = 10$: $T = 20 + 60(2/3)^2 = 20 + 60 \times \frac{4}{9} = 20 + \frac{80}{3} = \frac{140}{3} \approx 46.67°C$
Problem 1.3: Population grows from 2000 to 2500 in 10 years. Find growth constant $k$.
Solution
$P = P_0e^{kt}$
$2500 = 2000e^{10k}$
$\frac{5}{4} = e^{10k}$
$k = \frac{1}{10}\ln(1.25) = \frac{\ln(5/4)}{10} \approx 0.0223$ per year
Level 2: JEE Main Advanced
Problem 2.1: A body at temperature $T_1$ is placed in surroundings at $T_0 < T_1$. Show that time to cool from $T_1$ to $T_2$ is:
$$t = \frac{1}{k}\ln\left(\frac{T_1 - T_0}{T_2 - T_0}\right)$$Solution
$T - T_0 = (T_1 - T_0)e^{-kt}$
At temperature $T_2$: $T_2 - T_0 = (T_1 - T_0)e^{-kt}$
$e^{-kt} = \frac{T_2 - T_0}{T_1 - T_0}$
$-kt = \ln\left(\frac{T_2 - T_0}{T_1 - T_0}\right)$
$t = \frac{1}{k}\ln\left(\frac{T_1 - T_0}{T_2 - T_0}\right)$
Problem 2.2: A particle moves with velocity $v$ such that $\frac{dv}{dt} = -kv^2$. If $v = v_0$ at $t = 0$, find $v(t)$.
Solution
$\frac{dv}{v^2} = -k \, dt$
$-\frac{1}{v} = -kt + C$
At $t = 0$, $v = v_0$: $C = \frac{1}{v_0}$
$\frac{1}{v} = kt + \frac{1}{v_0}$
$v = \frac{v_0}{1 + kv_0t}$
Problem 2.3: Find curves such that slope at any point equals twice the slope of line joining that point to origin.
Solution
Slope of line to origin from $(x, y)$ is $\frac{y}{x}$
Given: $\frac{dy}{dx} = 2 \cdot \frac{y}{x}$
$\frac{dy}{y} = \frac{2dx}{x}$
$\ln y = 2\ln x + C$
$y = Ax^2$ (parabolas)
Level 3: JEE Advanced
Problem 3.1: A tank contains 1000 L of brine with 100 kg salt. Pure water enters at 10 L/min and mixture leaves at 10 L/min. How long until salt reduces to 10 kg?
Solution
$\frac{dQ}{dt} = 0 - \frac{Q}{1000} \times 10 = -\frac{Q}{100}$
$Q = 100e^{-t/100}$
For $Q = 10$: $10 = 100e^{-t/100}$
$e^{-t/100} = 0.1$
$t = 100\ln 10 = 230.26$ minutes
Problem 3.2: A chain hangs over edge of table. Show that if $x$ is length over edge at time $t$, then:
$$\frac{d^2x}{dt^2} = \frac{g}{L}x$$where $L$ is total length.
Solution
Mass over edge: $m_1 = \frac{m}{L}x$
Weight: $F = \frac{mgx}{L}$
By Newton’s second law: $F = ma$
$\frac{mgx}{L} = m\frac{d^2x}{dt^2}$
$\frac{d^2x}{dt^2} = \frac{g}{L}x$
Problem 3.3: Show that orthogonal trajectories of circles $x^2 + y^2 + 2gx + c = 0$ (with $g$ as parameter) are circles $x^2 + y^2 + 2fy + k = 0$.
Solution
Differentiate given family: $2x + 2y\frac{dy}{dx} + 2g = 0$
$g = -x - y\frac{dy}{dx}$
Substitute back: $x^2 + y^2 + 2(-x - y\frac{dy}{dx})x + c = 0$
$x^2 + y^2 - 2x^2 - 2xy\frac{dy}{dx} + c = 0$
$y^2 - x^2 - 2xy\frac{dy}{dx} + c = 0$
$\frac{dy}{dx} = \frac{y^2 - x^2 + c}{2xy}$
For orthogonal: $\frac{dy}{dx} = -\frac{2xy}{y^2 - x^2 + c}$
This leads to circles of form $x^2 + y^2 + 2fy + k = 0$
Summary of Key Formulas
| Application | Differential Equation | Solution |
|---|---|---|
| Growth | $\frac{dN}{dt} = kN$ | $N = N_0e^{kt}$ |
| Decay | $\frac{dN}{dt} = -kN$ | $N = N_0e^{-kt}$ |
| Cooling | $\frac{dT}{dt} = -k(T - T_s)$ | $T = T_s + (T_0 - T_s)e^{-kt}$ |
| RC Circuit | $R\frac{dQ}{dt} + \frac{Q}{C} = V$ | $Q = CV(1 - e^{-t/RC})$ |
| Free Fall | $m\frac{dv}{dt} = mg - kv$ | $v = \frac{mg}{k}(1 - e^{-kt/m})$ |
Cross-References
- Variable Separable: Core solving technique → Variable Separable
- Linear DE: Used in many applications → Linear DE
- Physics Applications: Motion, circuits → Mechanics
- Calculus: Integration techniques → Integration
Quick Revision Checklist
- Growth: $\frac{dN}{dt} = +kN$, Decay: $\frac{dN}{dt} = -kN$
- Cooling: Temperature difference in equation
- Half-life: $N = N_0(1/2)^{t/T_{1/2}}$
- Orthogonal: Use $-\frac{dx}{dy}$ or negative reciprocal slope
- Mixing: Rate in minus rate out
- Always check units and signs
Last updated: November 22, 2025