📘 Calculus · Chapter MN15🎯 NDA Level : High Priority
A differential equation (DE) relates a function to its derivatives. Unlike algebraic equations whose solutions are numbers, a DE’s solution is a function. For NDA, this chapter covers three standard solution methods — variable separable, homogeneous, and linear (integrating factor) — plus important physical applications like exponential growth/decay and Newton’s Law of Cooling.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Finding order and degree of a given differential equation; (2) Forming a DE by eliminating arbitrary constants from a family of curves; (3) Solving DEs by variable separation: f(y) dy = g(x) dx, then integrate both sides; (4) Identifying homogeneous DEs (dy/dx = F(y/x)) and solving by substitution y = vx; (5) Solving linear DEs dy/dx + P(x)y = Q(x) using the integrating factor e∫P dx; (6) Growth and decay: dN/dt = kN, solution N = N₀ekt; (7) Newton’s Law of Cooling: dT/dt = −k(T − Tenv).
Topics at a Glance
① Order & Degree
Identifying, formation from family of curves
② Variable Separable
f(y) dy = g(x) dx → integrate both sides
③ Homogeneous DE
F(y/x); substitute y = vx
④ Linear DE & IF
dy/dx + Py = Q; IF = e∫P dx
⑤ Growth & Decay
dN/dt = kN; N = N₀ekt
⑥ Newton’s Law of Cooling
dT/dt = −k(T − Ts)
1. Order, Degree & Formation of DEs
1.1
Definitions: Order, Degree & General Solution
Order = highest derivative; Degree = its power after clearing radicals
⚡ Order, Degree & Key Definitions
ORDER of a DE:
The ORDER is the highest order derivative present in the equation.
d³y/dx³ + ... → order 3. dy/dx + ... → order 1.
DEGREE of a DE:
The DEGREE is the power of the highest-order derivative,
AFTER the equation is made free from radicals and fractions
in the derivatives (i.e., expressed as a polynomial in derivatives).
Example 1: (d²y/dx²)² + (dy/dx)³ + y = 0 → Order 2, Degree 2
Example 2: dy/dx = √(1 + y²) → (dy/dx)² = 1+y² → Order 1, Degree 2
Example 3: e^(dy/dx) = x → Degree is NOT defined
(exponential of derivative cannot be expressed as polynomial)
GENERAL SOLUTION:
Contains arbitrary constants (one constant per order).
e.g., y = Aeˣ + Be⁻ˣ is the general solution of d²y/dx² − y = 0.
PARTICULAR SOLUTION:
Obtained by assigning specific values to arbitrary constants
using initial/boundary conditions.
When the degree is asked, always rationalise first. The degree of e^(dy/dx) or sin(dy/dx) type DEs is undefined (transcendental). This is a direct NDA MCQ trap.
Differential Equation
Order
Degree
Note
dy/dx + 2y = x
1
1
Standard linear, 1st order
(d²y/dx²)² + (dy/dx)³ = 0
2
2
Power on highest derivative
dy/dx = √(1 + (dy/dx)²)
1
2
Square both sides first
d³y/dx³ + sin(d²y/dx²) = 0
3
Not defined
Transcendental function of derivative
y² + (dy/dx)³ = 1
1
3
Power 3 on dy/dx
1.2
Formation of a Differential Equation
Eliminate the arbitrary constants by differentiating as many times as there are constants
To form a DE from a family of curves with n arbitrary constants, differentiate n times and eliminate the constants.
1
Count the number of arbitrary constants (n).
2
Differentiate the equation n times to get n+1 equations total.
3
Eliminate all n arbitrary constants algebraically from these equations.
4
The resulting equation (involving x, y, and dy/dx etc.) is the required DE. Its order = n.
Worked Example — Formation
Form the DE for the family y = Aex + Be−x.
Two constants (A, B) → differentiate twice.
dy/dx = Aex − Be−x. d²y/dx² = Aex + Be−x = y.
Eliminate A and B: d²y/dx² − y = 0. (Order 2, Degree 1.)
Worked Example — Formation (One Constant)
Form the DE for y = cx + c² (c is arbitrary).
Differentiate: dy/dx = c → c = dy/dx.
Substitute back: y = (dy/dx)x + (dy/dx)².
DE: y = x(dy/dx) + (dy/dx)². (Order 1, Degree 2 — Clairaut’s form.)
📝 TOPIC-WISE PYQ
Order & Degree — NDA-Pattern Questions
Q1. The order and degree of (d²y/dx²)² + 3(dy/dx) + 2y = 0 are respectively:
(a) 2, 1 (b) 2, 2 (c) 1, 2 (d) 1, 1
Answer: (b) 2, 2
Highest derivative: d²y/dx² → Order = 2.
Power of d²y/dx² in the equation = 2 → Degree = 2.
Q2. The degree of the DE [1 + (dy/dx)²]3/2 = d²y/dx² is:
(a) 2 (b) 3 (c) 4 (d) Not defined
Answer: (a) 2
Square both sides: [1+(dy/dx)²]³ = (d²y/dx²)².
Highest derivative d²y/dx² has power 2 → Degree = 2.
Q3. The DE of all circles centred at the origin is:
(a) x + y(dy/dx) = 0 (b) x − y(dy/dx) = 0 (c) y + x(dy/dx) = 0 (d) x = y²
2. Methods of Solving First-Order, First-Degree DEs
2.1
Three Standard Types at a Glance
Identify the type first — then apply the correct method
Type 1
Variable Separable
Form: dy/dx = f(x)⋅g(y) or equivalently: h(y) dy = g(x) dx Method: Separate, then integrate both sides.
Type 2
Homogeneous
Form: dy/dx = F(y/x) Both f(x,y) and g(x,y) same degree in x,y. Method: Let y = vx, then separate v and x.
Type 3
Linear
Form: dy/dx + P(x)y = Q(x) Coefficients of y and dy/dx depend only on x. Method: Multiply by IF = e∫P dx.
2.2
Variable Separable Method
Separate all y-terms to one side, all x-terms to the other, then integrate
⚡ Variable Separable — Method & Form
A DE is variable separable if it can be written as:
g(y) dy = f(x) dx OR dy/dx = f(x) / h(y)
Steps:
1. Rearrange so all y and dy are on one side, x and dx on the other.
2. Integrate both sides independently:
∫ g(y) dy = ∫ f(x) dx + C
3. The result is the general solution.
Recognition:
dy/dx = (function of x only) / (function of y only)
Example: dy/dx = x²/y → y dy = x² dx → separable
Counter: dy/dx = x + y → NOT separable (x and y mixed on one side)
The constant C appears after integrating. Always add +C on one side only (since it’s arbitrary). For NDA, the most common form is dy/dx = ex−y or dy/dx = xy type.
Worked Example — Variable Separable
Solve: dy/dx = ex−y.
Rewrite: dy/dx = ex ⋅ e−y.
Separate: ey dy = ex dx.
Integrate: ey = ex + C.
General solution: ey − ex = C.
Worked Example — With Initial Condition
Solve dy/dx = x/y, given y(0) = 2.
Separate: y dy = x dx. Integrate: y²/2 = x²/2 + C → y² − x² = k (where k = 2C).
Apply y(0) = 2: 4 − 0 = k → k = 4.
Particular solution: y² − x² = 4.
📝 TOPIC-WISE PYQ
Variable Separable — NDA-Pattern Questions
Q4. The solution of dy/dx = y sin x is:
(a) y = Cecos x (b) y = Ce−cos x (c) y = Cesin x (d) ln y = cos x + C
Answer: (b) y = Ce−cos x
Separate: dy/y = sin x dx. Integrate: ln|y| = −cos x + C₁.
y = eC₁ e−cos x = Ce−cos x.
A DE dy/dx = f(x,y)/g(x,y) is homogeneous if f and g are homogeneous functions of the same degree — meaning f(tx, ty) = tnf(x,y). This always leads to dy/dx being a function of y/x only.
⚡ Homogeneous DE — Method
Identify: dy/dx = F(y/x) [depends only on the ratio y/x]
Substitution:
Let y = vx, so dy/dx = v + x(dv/dx)
After substitution:
v + x(dv/dx) = F(v)
x(dv/dx) = F(v) − v [v and x are now separated!]
dv / [F(v) − v] = dx / x
Integrate both sides, then back-substitute v = y/x.
Check if homogeneous:
Replace x→tx, y→ty: if f(tx,ty) = tⁿ f(x,y) for all t, it’s homogeneous degree n.
Or simply: try to write dy/dx as a function of (y/x) alone.
After substituting y=vx, the equation in v and x is always variable separable. The final answer must replace v with y/x. For NDA, the substitution step is usually the part tested.
Worked Example — Homogeneous DE
Solve: x dy/dx = y + x, with y = 0 when x = 1.
dy/dx = y/x + 1 = F(y/x). ✓ Homogeneous.
Let y = vx: v + x dv/dx = v + 1 → x dv/dx = 1.
Separate: dv = dx/x. Integrate: v = ln|x| + C.
Substitute v = y/x: y/x = ln|x| + C → y = x(ln|x| + C).
Apply y(1) = 0: 0 = 1(0 + C) → C = 0. Solution: y = x ln x.
📝 TOPIC-WISE PYQ
Homogeneous DE — NDA-Pattern Questions
Q6. Which of the following is a homogeneous DE?
(a) dy/dx = x + y (b) dy/dx = (x²+y²)/xy (c) dy/dx = 2x + 1 (d) dy/dx = y/x + x
Answer: (b) dy/dx = (x²+y²)/xy
Divide numerator and denominator by x²: dy/dx = (1 + (y/x)²) / (y/x).
This is entirely a function of y/x → homogeneous.
Q7. For the homogeneous DE (x² + xy) dy = (x² + y²) dx, the substitution used is:
(a) y = vx² (b) x = vy (c) y = vx (d) v = x + y
Answer: (c) y = vx
All homogeneous DEs use y = vx (or x = vy if it’s more convenient). The substitution converts the equation into a separable one.
Rearrange: dy/(1+y²) = −dx/(1+x²).
Integrate: tan⁻¹(y) = −tan⁻¹(x) + C.
tan⁻¹(x) + tan⁻¹(y) = C.
Using addition formula: if C = tan⁻¹(k), then (x+y)/(1−xy) = k → x + y = k(1 − xy).
Or equivalently: tan⁻¹(x) + tan⁻¹(y) = C. Trap: Students forget the negative sign when moving dx to the right, leading to a wrong sign in the answer.
🤯 T2. Solve: (x + y) dy/dx = 1.
This looks non-separable in y, but switch perspective: treat x as function of y.
dx/dy = x + y. Rearrange: dx/dy − x = y. This is a linear DE in x!
IF = e∫(−1)dy = e−y.
Multiply: d/dy(xe−y) = ye−y.
Integrate: xe−y = ∫ye−ydy = −ye−y − e−y + C.
x = −y − 1 + Cey → x + y + 1 = Cey. Key insight: When x and y cannot be separated and dy/dx method fails, try dx/dy — treat x as the dependent variable.
2.4
Linear Differential Equations — Integrating Factor Method
The most important method for NDA — memorise the IF formula and solution structure
A linear DE of first order has the form dy/dx + P(x)y = Q(x), where P and Q are functions of x only. The method uses a clever multiplying factor called the Integrating Factor (IF).
⚡ Linear DE — Complete Solution Method
Standard form: dy/dx + P(x) y = Q(x)
Step 1: Identify P(x) and Q(x) from the standard form.
(If not in standard form, divide through by the coefficient of dy/dx.)
Step 2: Compute the Integrating Factor (IF):
IF = e∫P(x) dx (integrate P, no +C needed here)
Step 3: Multiply both sides by IF:
d/dx [y ⋅ IF] = Q(x) ⋅ IF (left side collapses to a single derivative!)
Step 4: Integrate both sides:
y ⋅ IF = ∫ Q(x) ⋅ IF dx + C
Step 5: Solve for y (divide by IF if needed).
Summary formula:
y ⋅ e∫P dx = ∫ Q ⋅ e∫P dx dx + C
Also valid for x as dependent variable:
dx/dy + P(y) x = Q(y) → IF = e∫P(y)dy
x ⋅ IF = ∫ Q(y) ⋅ IF dy + C
The magic of the IF is that it turns the left side into an exact derivative: d(y⋅IF)/dx = Q⋅IF. This is the whole trick. The most common IFs in NDA are: e∫(1/x)dx = x; e∫2x dx = ex²; e∫k dx = ekx.
Worked Example — Linear DE with IF
Solve: dy/dx + y/x = x².
P(x) = 1/x, Q(x) = x².
IF = e∫(1/x)dx = eln x = x.
Multiply: d/dx(xy) = x ⋅ x² = x³.
Integrate: xy = x⁴/4 + C.
Solution: y = x³/4 + C/x.
Worked Example — Linear DE (Bernoulli type)
Solve: dy/dx − 3y = e2x.
P = −3, Q = e2x.
IF = e∫(−3)dx = e−3x.
Multiply: d/dx(y e−3x) = e2x ⋅ e−3x = e−x.
Integrate: y e−3x = −e−x + C.
Solution: y = −e2x + Ce3x.
📝 TOPIC-WISE PYQ
Linear DE — NDA-Pattern Questions
Q8. The integrating factor of dy/dx + 2y = e3x is:
(a) ex (b) e2x (c) e3x (d) e−2x
Answer: (b) e2x
P(x) = 2. IF = e∫2 dx = e2x.
Q9. Solve: dy/dx + y cot x = 2 cos x.
(a) y sin x = sin² x + C (b) y = sin x + C (c) y = 2 cos x + C (d) y sin x = cos x + C
Answer: (a) y sin x = sin² x + C
P = cot x, IF = e∫cot x dx = eln sin x = sin x.
d/dx(y sin x) = 2 cos x sin x = sin 2x.
y sin x = ∫ sin 2x dx = −cos 2x/2 + C = (1−2cos²x)/2+C′.
More directly: y sin x = ∫2 cos x ⋅ sin x dx = ∫sin 2x dx = sin² x + C.
Q10. The IF of x(dy/dx) − y = x² is:
(a) x (b) 1/x (c) ln x (d) ex
Answer: (b) 1/x
Divide by x: dy/dx − y/x = x. P = −1/x.
IF = e∫(−1/x)dx = e−ln x = 1/x.
🔥 TRICKY QUESTIONS
Linear DE — IF Traps & Reduction
🤯 T3. Solve the Bernoulli equation: dy/dx + y = y².
This is Bernoulli’s equation (dy/dx + P(x)y = Q(x)yⁿ, n=2). Divide by y²:
y−2 dy/dx + y−1 = 1.
Let v = y−1 = 1/y. Then dv/dx = −y−2 dy/dx.
So: −dv/dx + v = 1 → dv/dx − v = −1. (Linear in v!)
IF = e∫(−1)dx = e−x.
d/dx(ve−x) = −e−x. Integrate: ve−x = e−x + C.
v = 1 + Cex → 1/y = 1 + Cex → y = 1/(1 + Cex). Bernoulli: always divide by yⁿ, substitute v = y1−n, reduce to linear.
🤯 T4. If ∫0x f(t) dt = x + ∫x1 t f(t) dt, find f(x).
Differentiate both sides using the Fundamental Theorem:
f(x) = 1 + [−x f(x)] (applying Leibniz rule: d/dx ∫x1 tf(t) dt = −x f(x)).
f(x) = 1 − x f(x) → f(x)(1+x) = 1 → f(x) = 1/(1+x). This is an integral equation that reduces to an algebraic equation after differentiation using FTC.
3. Applications — Growth, Decay & Newton’s Law of Cooling
3.1
Exponential Growth/Decay & Newton’s Law of Cooling
Two direct application models — both solved by variable separation
🌿
Exponential Growth & Decay
DE: dN/dt = kN (k > 0: growth; k < 0: decay)
Separate: dN/N = k dt. Integrate: ln N = kt + C. Solution: N = N₀ekt where N₀ = N at t=0 (initial value). Half-life T1/2: ekT = 1/2 → T = −ln 2/k.
❄
Newton’s Law of Cooling
DE: dT/dt = −k(T − Ts) (k > 0)
Ts = surrounding/ambient temperature (constant). Separate: dT/(T−Ts) = −k dt. Integrate: ln|T−Ts| = −kt + C. Solution: T − Ts = Ae−kt A = T₀ − Ts = initial excess temperature.
Fig 1: Exponential growth (green, k > 0) and decay (red, k < 0). Both start at N₀. Half-life T½ is where the decay curve crosses N₀/2.
⚡ Growth/Decay & Cooling — Key Results
EXPONENTIAL GROWTH/DECAY:
DE: dN/dt = kN
Solution: N(t) = N₀ ekt (N₀ = initial amount at t=0)
Doubling time (growth): td = ln 2 / k
Half-life (decay): T½ = ln 2 / |k|
NEWTON’S LAW OF COOLING:
DE: dT/dt = −k(T − Ts)
Solution: T(t) = Ts + (T₀ − Ts) e−kt
As t → ∞: T(t) → Ts (body reaches surrounding temperature)
At t=0: T = T₀ (initial temperature)
RADIOACTIVE DECAY (specific form):
DE: dA/dt = −λA (λ = decay constant)
Solution: A = A₀ e−λt
Half-life: T½ = ln 2 / λ = 0.693 / λ
In all these problems: (1) Write the DE from the physical law. (2) Solve by variable separation. (3) Apply initial conditions to find A or N₀. (4) Apply the target condition to find the time or amount asked.
Worked Example — Radioactive Decay
A radioactive substance decays so that the rate of decay is proportional to the amount present. If initially there is 100g and after 2 years 90g remains, find the amount after 4 years.
Q12. A population grows at rate proportional to itself. If it was 1 lakh in 2010 and 2 lakh in 2020, what will it be in 2030?
(a) 3 lakh (b) 4 lakh (c) 2.5 lakh (d) 5 lakh
Answer: (b) 4 lakh
N = N₀ekt. At t=10: 2 = e10k → e10k = 2.
At t=20: N = e20k = (e10k)² = 4 lakh. Population doubles every 10 years.
🔥 TRICKY QUESTIONS
Applications — Model Setup & Rate Traps
🤯 T5. The half-life of a radioactive substance is 10 years. What fraction remains after 25 years?
T½ = 10 years → k = ln2/10.
A(25) = A₀e−k⋅25 = A₀e−25 ln2/10 = A₀ ⋅ 2−2.5 = A₀ / 22.5 = A₀ / (4√2).
Fraction remaining = 1/(4√2) = √2/8 ≈ 0.177 (about 17.7%). Method: Use A/A₀ = (1/2)t/T½ = (1/2)25/10 = (1/2)2.5. This avoids finding k explicitly.
🤯 T6. Water leaks from a tank so that the rate of flow is proportional to the square root of the volume remaining. If the tank has 100 litres and 75 litres remain after 5 minutes, when is it empty?
dV/dt = −k√V. Separate: dV/√V = −k dt → 2√V = −kt + C.
At t=0: 2√100 = C → C = 20.
At t=5: 2√75 = −5k + 20 → 2(5√3) = −5k + 20 → 10√3 = 20 − 5k → 5k = 20−10√3 → k = (20−10√3)/5 = 4−2√3.
Empty when V=0: 0 = −kt + 20 → t = 20/k = 20/(4−2√3) = 10/(2−√3).
Rationalise: 10(2+√3)/((4−3)) = 10(2+√3) ≈ 10(2+1.73) = 37.3 minutes. This “square root leakage” is a Torricelli’s theorem problem. The key is correct separation and applying two conditions sequentially.
📝 Master Formula Sheet — MN15 Differential Equations
All critical formulae for rapid pre-exam revision.
📜 Order & Degree
Order = highest derivative present
Degree = power of highest derivative (after rationalising)
This material is for personal NDA exam preparation only.
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