1. Chapter at a Glance
Many laws of nature describe how things change rather than what they are. A population grows at a rate proportional to its size; a hot cup of tea cools at a rate proportional to how much hotter it is than the room; money in a bank grows with continuous interest. Each statement is an equation involving a derivative — a differential equation. Solving it gives the actual quantity as a function of time. This chapter covers the basic vocabulary (order, degree, solutions) and three methods for first-order equations: variables separable, homogeneous and linear.
- Differential equations; order and degree.
- General and particular solutions; verifying a solution.
- Variables separable method.
- Homogeneous differential equations (y = vx substitution).
- Linear first-order equations: integrating factor.
- Applications: growth, decay, interest.
2. Order and Degree
- Order: the order of the highest derivative present.
- Degree: the power of the highest-order derivative, when the equation is a polynomial in the derivatives. If it is not (e.g. sin(y′) appears), the degree is not defined.
| Equation | Order | Degree |
|---|---|---|
| dy/dx = cos x | 1 | 1 |
| (d²y/dx²)³ + (dy/dx)⁴ + y = 0 | 2 | 3 |
| y‴ + 2y″ + y′ = 0 | 3 | 1 |
| y″ + sin(y′) = 0 | 2 | not defined |
A general solution contains as many arbitrary constants as the order; a particular solution has specific values of the constants (usually from an initial condition).
Verify that y = a cos x + b sin x is a solution of y″ + y = 0.
Solution: y″ = −a cos x − b sin x = −y, so y″ + y = 0 ✓ (two constants, second order — a general solution).
3. Variables Separable
If dy/dx = g(x)h(y), rewrite as dy/h(y) = g(x) dx and integrate both sides: ∫dy/h(y) = ∫g(x)dx + C.
Solve dy/dx = (x + 1)/(2 − y), y ≠ 2.
Solution: (2 − y)dy = (x + 1)dx → 2y − y²/2 = x²/2 + x + C → x² + y² + 2x − 4y + C′ = 0.
A bacteria population grows at a rate proportional to its size. It doubles in 3 hours. In how many hours will it be 8 times the original? (Model: dP/dt = kP.)
Solution: dP/P = k dt → P = P₀e^(kt). Doubling: e^(3k) = 2. Eightfold: e^(kt) = 8 = 2³ = e^(9k) → t = 9 hours.
4. Homogeneous Equations
A function F(x, y) is homogeneous of degree n if F(λx, λy) = λⁿF(x, y). The equation dy/dx = F(x, y) is homogeneous if F is homogeneous of degree 0, i.e. F(x, y) = g(y/x).
Put y = vx, so dy/dx = v + x dv/dx. The equation becomes separable in v and x.
Solve (x − y) dy/dx = x + 2y.
Solution: dy/dx = (1 + 2v)/(1 − v). v + x v′ = (1 + 2v)/(1 − v) → x v′ = (1 + v + v²)/(1 − v). Separate: (1 − v)dv/(v² + v + 1) = dx/x. Integrating (write 1 − v = −½(2v + 1) + 3/2): −½ log(v² + v + 1) + √3 tan⁻¹((2v + 1)/√3) = log|x| + C. With v = y/x: ½ log(x² + xy + y²) = √3 tan⁻¹((2y + x)/(√3 x)) + C.
5. Linear Equations
dy/dx + P(x) y = Q(x). Integrating factor I.F. = e^(∫P dx). Then
Similarly dx/dy + P₁(y)x = Q₁(y) has I.F. = e^(∫P₁ dy).
Solve dy/dx − y = cos x.
Solution: P = −1, I.F. = e⁻ˣ. y e⁻ˣ = ∫e⁻ˣ cos x dx = e⁻ˣ(sin x − cos x)/2 + C → y = (sin x − cos x)/2 + Ceˣ.
Solve x dy/dx + 2y = x² (x ≠ 0).
Solution: dy/dx + (2/x)y = x. I.F. = e^(2 log x) = x². yx² = ∫x³ dx = x⁴/4 + C → y = x²/4 + C/x².
6. Solved Problems (Board Pattern)
Find the particular solution of dy/dx = −4xy², given y = 1 when x = 0.
Solution: dy/y² = −4x dx → −1/y = −2x² + C. x = 0, y = 1: C = −1. So 1/y = 2x² + 1 → y = 1/(2x² + 1).
Money in a bank grows continuously at 5% per year. In how many years does ₹1000 double? (log 2 ≈ 0.6931)
Solution: dP/dt = 0.05P → P = 1000e^(0.05t). 2 = e^(0.05t) → t = log 2/0.05 ≈ 13.86 years.
Find the curve through (0, −2) whose slope at any point (x, y) equals the sum of its coordinates, i.e. dy/dx = x + y.
Solution: dy/dx − y = x, I.F. e⁻ˣ: y e⁻ˣ = ∫x e⁻ˣ dx = −e⁻ˣ(x + 1) + C → y = −(x + 1) + Ceˣ. (0, −2): −2 = −1 + C → C = −1. y = −x − 1 − eˣ.
7. Master Formula Table
| Type | Form | Method |
|---|---|---|
| Separable | dy/dx = g(x)h(y) | ∫dy/h = ∫g dx |
| Homogeneous | dy/dx = g(y/x) | y = vx |
| Linear in y | y′ + Py = Q | I.F. = e^(∫P dx) |
| Linear in x | dx/dy + P₁x = Q₁ | I.F. = e^(∫P₁ dy) |
| Growth/decay | dP/dt = kP | P = P₀e^(kt) |
8. Master Summary Table
| Concept | Key point | Exam focus |
|---|---|---|
| Order, degree | Degree needs polynomial form | 1-mark questions |
| Solutions | Constants = order | Verify; find particular solution |
| Separable | Separate, integrate | Growth, decay, interest |
| Homogeneous | y = vx | 5-mark solving |
| Linear | Integrating factor | 5-mark solving |
9. Mind Map
- Order, degree
- General vs particular
- dy/h(y) = g dx
- P = P₀e^(kt)
- g(y/x)
- y = vx
- y′ + Py = Q
- I.F. e^(∫P)
10. Tips and Tricks
- First try to separate variables — it is the simplest method.
- Test homogeneity by replacing x, y with λx, λy.
- e^(log f(x)) = f(x) simplifies most integrating factors.
- Writing P with the wrong sign — put the equation in the form y′ + Py = Q first.
- Forgetting the constant of integration before applying the initial condition.
- Giving a degree when the equation is not polynomial in derivatives.
11. Important One-liners
- The order of a differential equation is a positive integer.
- The number of arbitrary constants in a general solution equals the order.
- Degree is undefined if derivatives appear inside sin, log, e, etc.
- For y′ + Py = Q, the integrating factor is e^(∫P dx).
- dP/dt = kP gives exponential growth (k > 0) or decay (k < 0).
12. Exam-pattern Questions with Detailed Answers
The order and degree of (y″)² + (y′)³ + sin y = 0 are:
- 2, 2
- 2, 3
- 3, 2
- 2, undefined
Show answer
Highest derivative y″ (order 2) with power 2; sin y is a function of y, not of a derivative, so degree is defined.
Answer: (A)
How many arbitrary constants are in the general solution of a third-order equation?
Show answer
Three.
Answer: 3
Find the integrating factor of dy/dx + y/x = x².
Show answer
e^(∫dx/x) = x.
Answer: x
Solve dy/dx = eˣ⁺ʸ.
Show answer
e⁻ʸ dy = eˣ dx → −e⁻ʸ = eˣ + C → eˣ + e⁻ʸ = C′.
Answer: eˣ + e⁻ʸ = C
Solve dy/dx + 2y = sin x.
Show answer
I.F. e²ˣ. y e²ˣ = ∫e²ˣ sin x dx = e²ˣ(2 sin x − cos x)/5 + C → y = (2 sin x − cos x)/5 + Ce⁻²ˣ.
Answer: y = (2 sin x − cos x)/5 + Ce⁻²ˣ
Show that dy/dx = (x + y)/x is homogeneous and solve it.
Show answer
(x + y)/x = 1 + y/x — homogeneous. v + xv′ = 1 + v → dv = dx/x → v = log|x| + C → y = x log|x| + Cx.
Answer: y = x log|x| + Cx
Solve (1 + x²) dy/dx = 1 + y², given y = 1 when x = 0.
Show answer
tan⁻¹y = tan⁻¹x + C; x = 0, y = 1 → C = π/4. tan⁻¹y − tan⁻¹x = π/4, i.e. (y − x)/(1 + xy) = 1 → y − x = 1 + xy.
Answer: y − x = 1 + xy
A radioactive substance decays at a rate proportional to the amount present. Half decays in 1600 years. Write the model and the formula for the amount.
Show answer
dA/dt = −kA → A = A₀e^(−kt), with e^(−1600k) = ½ → k = log 2/1600.
Answer: A = A₀e^(−(log 2/1600)t)
Solve x dy − y dx = √(x² + y²) dx (x > 0).
Show answer
dy/dx = (y + √(x² + y²))/x — homogeneous. y = vx: x v′ = √(1 + v²) → dv/√(1 + v²) = dx/x → log|v + √(1 + v²)| = log x + log C → y + √(x² + y²) = Cx².
Answer: y + √(x² + y²) = Cx²
Solve (1 + y²) dx = (tan⁻¹y − x) dy.
Show answer
dx/dy + x/(1 + y²) = tan⁻¹y/(1 + y²). I.F. = e^(tan⁻¹y). x e^(tan⁻¹y) = ∫ t eᵗ dt (t = tan⁻¹y) = (t − 1)eᵗ + C → x = tan⁻¹y − 1 + Ce^(−tan⁻¹y).
Answer: x = tan⁻¹y − 1 + Ce^(−tan⁻¹y)
Find the particular solution of dy/dx + y cot x = 4x cosec x, given y = 0 when x = π/2.
Show answer
I.F. = sin x. y sin x = ∫4x dx = 2x² + C. At x = π/2: 0 = π²/2 + C → C = −π²/2. y sin x = 2x² − π²/2.
Answer: y sin x = 2x² − π²/2
The population of a village increases continuously at a rate proportional to the number present. If it grew from 20,000 to 25,000 in 10 years (practice data), what will it be after another 10 years?
Show answer
P = P₀e^(kt); e^(10k) = 1.25. After 20 years: 20000 × 1.25² = 31,250.
Answer: 31,250
13. Quick Revision Summary
- Order = highest derivative; degree = its power (if polynomial in derivatives).
- General solution has as many constants as the order; particular uses conditions.
- Separable: separate and integrate; growth/decay P = P₀e^(kt).
- Homogeneous: y = vx.
- Linear: I.F. = e^(∫P dx), y·I.F. = ∫Q·I.F. dx + C.