Differential Equations
Practice JEE Main Differential Equations PYQs online. Linear differential equations (LDE), integrating factor (IF), homogeneous equations & step-by-step solutions.
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Differential Equations:
- Order and Degree of Differential Equations
- Variable Separable Method
- Homogeneous Differential Equations (y = v·x substitution)
- Linear Differential Equations: dy/dx + P(x)·y = Q(x)
- Integrating Factor IF = e^(∫ P dx) and General Solution y · IF = ∫ Q · IF dx
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Differential Equations questions in under 45 seconds:
LDE Solution: y · e^(∫ P dx) = ∫ Q · e^(∫ P dx) dx + C.
Bernoulli’s Equation: dy/dx + P·y = Q·yⁿ reduces to LDE with substitution z = y^(1-n).
Exact differential shortcut: d(xy) = x dy + y dx, d(y/x) = (x dy - y dx)/x².
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 1 + x²
- Step 1: Put the differential equation in standard first-order linear form: dy/dx + P(x) y = Q(x).
- Step 2: Divide by (1 + x²): dy/dx + [2x / (1 + x²)] y = 4x² / (1 + x²).
- Step 3: Identify P(x) = 2x / (1 + x²).
- Step 4: Integrating Factor IF = e^(∫ P dx) = e^(∫ 2x/(1 + x²) dx) = e^(ln(1 + x²)) = 1 + x².
⚡ Speed Tip: Standard IF: ∫ 2x/(1+x²) dx = ln(1+x²), so e^(ln(1+x²)) = 1 + x² directly.
💡 View Step-by-Step Solution & Answer
Correct Answer: x + y = v
- Step 1: Notice that the linear expression (x + y) appears in both numerator and denominator.
- Step 2: Let v = x + y. Differentiating with respect to x gives dv/dx = 1 + dy/dx ⇒ dy/dx = dv/dx - 1.
- Step 3: Substitute into the differential equation: dv/dx - 1 = (v + 1)/(v - 1).
- Step 4: dv/dx = (v + 1)/(v - 1) + 1 = 2v / (v - 1), which is easily solved by variable separation.
⚡ Speed Tip: Linear sum substitution: Whenever a1x + b1y + c1 shares coefficients with denominator, substitute v = ax + by.
💡 View Step-by-Step Solution & Answer
Correct Answer: π²/16 + π√2/4
- Step 1: Standard linear equation with P(x) = tan x, Q(x) = 2x + x² tan x.
- Step 2: IF = e^(∫ tan x dx) = e^(ln|sec x|) = sec x.
- Step 3: General solution: y · sec x = ∫ (2x sec x + x² sec x tan x) dx = ∫ d/dx(x² sec x) dx = x² sec x + C.
- Step 4: Apply y(0) = 1: 1 · sec(0) = 0 + C ⇒ C = 1. Thus y = x² + cos x. At x = π/4: y(π/4) = (π/4)² + cos(π/4) = π²/16 + 1/√2.
⚡ Speed Tip: Exact Derivative Integration: 2x sec x + x² sec x tan x is the exact derivative of x² sec x.
❓ Frequently Asked Questions (FAQs)
What type of differential equation is most common in JEE Main?
First-order Linear Differential Equation (dy/dx + Py = Q) with integrating factor IF = e^(∫ P dx) appears in over 70% of shifts.