Q.The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
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Differential Equation Modeling: From Intuition to Precision
Imagine you're watching a cup of hot coffee cool down on your desk. It starts hot, then gradually loses heat to the room. But here's the key question: how fast does it cool at any given moment? The answer isn't a fixed number — it depends on how hot the coffee is right now. The hotter it is, the faster it cools. This is the core idea behind differential equation modeling: the rate of change of a quantity depends on the quantity itself.
The Intuition First
Suppose the room is at 20∘C and your coffee starts at 90∘C. Newton's Law of Cooling says the rate at which the coffee cools is proportional to the temperature difference between the coffee and the room. So when the coffee is 90∘C, the difference is 70∘C and it cools fast; when it's 40∘C, the difference is only 20∘C and it cools slowly. The rate changes as the temperature changes.
If we let T(t) be the temperature at time t, then "rate of change of temperature" is dtdT, and the statement above becomes:
dtdT=−k(T−20)
The minus sign is because the temperature is decreasing; the constant k depends on the cup, the liquid, etc. This is a differential equation — an equation that involves a function and its derivative.
The Precise Statement
A differential equation is any equation that contains an unknown function and one or more of its derivatives. The goal is to find the function itself (here, T(t)) that satisfies the equation.
dxdy=f(x,y)
This is the general form of a first-order ordinary differential equation. The unknown is y(x), and the equation tells you how y changes at every point based on x and y itself.
Modeling means taking a real-world situation and translating it into a differential equation:
- Identify the quantity you want to study (population, temperature, drug concentration, voltage).
- State the rate law in words: "The population grows at a rate proportional to its size."
- Translate into math: dtdP=kP.
- Add initial conditions: P(0)=P0 (the starting value).
Why This Matters
Without differential equations, you'd have to guess the whole future behavior of a system. With them, you get a precise rule that governs every instant of change. The solution to dtdP=kP is P(t)=P0ekt — exponential growth. The solution to the cooling equation is T(t)=20+70e−kt — exponential decay toward room temperature.
Most real-world models are not this simple. But every model starts the same way: observe how something changes, express that change in terms of the thing itself, and write it as a differential equation.
A Simple Example to Try
A bacteria culture doubles every hour. If you start with 100 bacteria:
Step 1: Quantity = population P(t). …
Concept: Related Rates — we relate the rate of change of volume to the rate of change of radius using the chain rule.
Step 1: Volume of a sphere: V=34πr3. Given dtdV=k (constant).
Step 2: Differentiate: dtdV=4πr2dtdr=k. So r2dtdr=4πk, a constant.
Step 3: Integrate: ∫r2dr=4πk∫dt gives 3r3=4πkt+C. …
The radius after t seconds is r(t)=(63t+27)1/3 units.
The volume of a sphere is V=34πr3, and it changes at a constant rate, so dtdV=k for some constant k.
Integrate. Since dtd(34πr3)=k,
34πr3=kt+C.
Use the two data points.
- At t=0,r=3: 34π(27)=C⇒C=36π.
- At t=3,r=6: 34π(216)=3k+36π⇒288π=3k+36π⇒k=84π. …
Method: "Rate is constant" modelling with the chain rule
Use this when a quantity (like a sphere's volume) changes at a constant rate and you must find another quantity (its radius) over time.
Steps
Step 1: Express the changing quantity and set its rate constant
Write the volume V=34πr3 and state dtdV=k (constant), so V=kt+V0.
Step 2: Use given data to find the constants
Substitute the initial value and one later value (here r=3 at t=0, r=6 at t=3) to solve for k and V0. …
Common Mistakes
Mistake 1: Assuming the radius changes at a constant rate
Why it's wrong: it is the volume that changes at a constant rate, not the radius; dtdr is not constant. Correct approach: set dtdV=k with V=34πr3.
Mistake 2: Using an exponential model …
- KCET 2025Set A-11 markMCQQ.If ‘a’ and ‘b’ are the order and degree respectively of the differentiable equation. (dx2d2y)2+(dxdy)3+x4=0, then a−b= (A) 1 (B) 2 (C) −1 (D) 0
›Reveal solutionSolution
Order = highest derivative present (=2); degree = power of that highest derivative once the equation is polynomial in derivatives (=2); so a−b=0.
Step 1 — Recall the two definitions (they are often confused).
- Order = the order of the highest-order derivative occurring in the equation.
- Degree = the exponent of that highest-order derivative, after the equation has been made free of radicals and fractional powers of derivatives — i.e. once it is a polynomial in the derivatives.
Step 2 — Inspect the equation.
(dx2d2y)2+(dxdy)3+x4=0
There are no radicals, no fractional powers and no derivatives inside transcendental functions — it is already a polynomial in dxdy and dx2d2y. So both definitions apply directly.
Step 3 — Order.
The derivatives present are dxdy (order 1) and dx2d2y (order 2). The highest is the second derivative, so
a=2. …
- KCET 2022Set C-41 markMCQQ.The sum of the degree and order of the differential equation (1+y12)2/3=y2 is (A) 6 (B) 5 (C) 7 (D) 4
›Reveal solutionSolution
The key idea is to rewrite the equation so the highest derivative appears without a fractional exponent, then read off the order and degree. The sum is 5.
The problem gives a differential equation in a form where the highest derivative y2 (the second derivative) is already isolated on one side. But the left side has a fractional exponent, which makes the degree ambiguous. The degree of a differential equation is defined only when the equation is a polynomial in the derivatives — meaning every derivative must appear with a whole-number exponent. So our first job is to clear that fractional power.
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Identify the order. The highest derivative present is y2=dx2d2y. That makes the order 2. This is immediate and doesn't change no matter how we rewrite the equation.
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Remove the fractional exponent to find the degree. The equation is
(1+y12)2/3=y2
where y1=dxdy. To make the left side a polynomial in y1 and y2, raise both sides to the power 3:
[(1+y12)2/3]3=y23
which simplifies to
(1+y12)2=y23.
Now expand if you like — but you don't need to. The equation is already a polynomial in the derivatives: y2 appears with exponent 3, and y1 appears with exponent 2 (inside the square). The highest exponent on the highest derivative y2 is 3. That is the degree. …
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- COMEDK 2022Set 20221 markMCQQ.The solution of the differential equation dx2d2y=0 represents (A) all circles in a plane (B) all straight lines in a plane (C) all parabolas in a plane (D) all ellipses in a plane
›Reveal solutionSolution
This two-parameter family y = ax + b is precisely the family of all straight lines in the plane (non-vertical ones; a circle/parabola/ellipse would need a non-zero second derivative somewhere).
Concept: Integrate twice.
d²y/dx² = 0 → dy/dx = a (constant) → y = ax + b, with a, b arbitrary constants. …
- KCET 2021Set A-11 markMCQQ.Solution of Differential Equation xdy−ydx=0 represents (A) A rectangular Hyperbola (B) Parabola whose vertex is at origin (C) Straight line passing through origin (D) A circle whose centre is origin
›Reveal solutionSolution
xdy−ydx=0 separates to ydy=xdx, whose solution y=cx is the family of lines through the origin.
Step 1 — Separate the variables.
xdy−ydx=0⟹xdy=ydx⟹ydy=xdx(x=0, y=0).
The equation is variable-separable (and equivalently homogeneous of degree 0, since dxdy=xy depends only on y/x).
Step 2 — Integrate both sides.
∫ydy=∫xdx⟹log∣y∣=log∣x∣+log∣c∣.
Step 3 — Exponentiate.
∣y∣=∣c∣∣x∣⟹y=cx
Step 4 — Identify the curve. y=cx is a one-parameter family of straight lines of slope c, every one of which passes through (0,0). …
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