Differential Equations

🟢 Lite — Quick Review (1h–1d)

Core Types & Formulas

Type	Equation	Solution Form
Variable separable	dy/dx = f(x)g(y)	∫dy/g(y) = ∫f(x)dx
Linear (1st order)	dy/dx + Py = Q	IF = e^(∫Pdx), solution: y·IF = ∫Q·IF dx
Bernoulli	dy/dx + Py = Qyⁿ	Substitute z = y^(1-n)
Clairaut’s	y = px + f(p)	General: y = cx + f(c), Singular via dp/dx = 0
Homogeneous	dy/dx = f(y/x)	Substitute v = y/x

Integrating Factor (IF): μ(x) = e^(∫P dx) for linear DE dy/dx + P(x)y = Q(x)

Orthogonal Trajectories: For family F(x,y,c)=0, replace dy/dx → −dx/dy, then solve.

⚡ GATE Quick-Check: Know when to use separation of variables vs linear — it’s the most common mistake.

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🟡 Standard — Regular Study (2d–2mo)

First-Order Differential Equations

Variable Separable

Rewrite as f(y)dy = g(x)dx, integrate both sides.

Example: dy/dx = xy → (1/y)dy = x dx → ln|y| = x²/2 + C

Linear Differential Equations

Standard form: dy/dx + P(x)y = Q(x)

Integrating Factor: μ(x) = exp(∫P dx)

Solution: $$y \cdot \mu(x) = \int Q(x) \cdot \mu(x) dx + C$$

Example (GATE 2020): dy/dx + y/x = x² → IF = e^(∫1/x dx) = x → d/dx(yx) = x³ → yx = x⁴/4 + C → y = x³/4 + C/x

Bernoulli Equation

Form: dy/dx + P(x)y = Q(x)yⁿ, where n ≠ 0, 1

Reduction: Let z = y^(1-n), then dz/dx + (1-n)P(x)z = (1-n)Q(x)

Example: dy/dx + y/x = y² → here n=2 → z = y^(−1) → dz/dx − z/x = −1 → solve with IF = 1/x → …

Clairaut’s Equation

Form: y = px + f(p), where p = dy/dx

General solution: Replace p with arbitrary constant c: y = cx + f(c) — a family of straight lines

Singular solution: Eliminate p from (i) y = px + f(p) and (ii) x + f’(p) = 0 → gives envelope curve

Orthogonal Trajectories (OT)

For family F(x,y,c)=0:

1. Differentiate to get dy/dx
2. Replace dy/dx with −dx/dy (perpendicular slope)
3. Solve the resulting DE

Example: Family y = cx² → dy/dx = 2cx → OT: replace with −dx/dy = 1/(2cx) → leads to solving via separation…

⚡ Common Trap: Don’t forget constant of integration C — GATE often tests whether you track it correctly.

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🔴 Extended — Deep Study

Integrating Factor — Detailed Derivation

For dy/dx + Py = Q:

• Multiply both sides by IF = e^(∫Pdx): $$e^{\int P dx}\frac{dy}{dx} + P e^{\int P dx}y = Q e^{\int P dx}$$
• Left side = d/dx[y·e^(∫Pdx)], so: $$\frac{d}{dx}[y \cdot e^{\int P dx}] = Q \cdot e^{\int P dx}$$
• Integrate: y·e^(∫Pdx) = ∫Q·e^(∫Pdx) dx + C

GATE Pattern: Questions often give P(x) as a rational function — practice integrating rational functions quickly.

Exact Equations

Test: M(x,y)dx + N(x,y)dy = 0 is exact if ∂M/∂y = ∂N/∂x

Solution: Find ψ(x,y) such that ∂ψ/∂x = M and ∂ψ/∂y = N → ψ(x,y) = C

IF for non-exact: If (∂M/∂y − ∂N/∂x)/N depends only on x → IF = exp(∫(∂M/∂y − ∂N/∂x)/N dx)

Bernoulli — Complete Treatment

Problem: dy/dx + (1/x)y = y²/x³

Solution:

• Here n=2, so z = y^(−1)
• dz/dx − (1/x)z = −1/x³
• IF = exp(∫−1/x dx) = exp(−ln|x|) = 1/x
• d/dx(z/x) = −1/x⁴ → z/x = ∫−1/x⁴ dx = 1/(3x³) + C
• z = x²/3 + Cx → 1/y = x²/3 + Cx

Orthogonal Trajectories — Worked Example

Family: x² + y² = c (circles centred at origin)

1. Differentiate: 2x + 2yy’ = 0 → y’ = −x/y
2. OT condition: y’_(OT) = +y/x (perpendicular slope = reciprocal positive)
3. dy/dx = y/x → separate: dy/y = dx/x → ln|y| = ln|x| + C → y = kx

Result: OT of circles x² + y² = c is the family of straight lines y = kx through origin. ✓ Makes sense: radial lines are perpendicular to circles.

GATE Previous-Year Highlights

Year	Problem	Key Concept
2019	Solve dy/dx + 2y = e^(−2x)	Linear with IF = e^(2x)
2020	Bernoulli: dy/dx + y/x = y²	Substitution z = 1/y
2021	Orthogonal trajectories of y = ce^x	OT: dy/dx = −e^x/y → separable
2022	Clairaut’s: y = px + p²	General: y = cx + c², Singular: from x + 2p = 0
2023	Exact DE: (e^x sin y + 2x) dx + (e^x cos y + 2y) dy = 0	Check ∂M/∂y = ∂N/∂x

⚡ GATE Warning: Bernoulli’s equation is frequently tested with n=2 (y²). If you see y² or 1/y in the equation, immediately try Bernoulli’s substitution.

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