Applications — Growth/decay, orthogonal trajectories
Differential Equations
12
JEE Qs
8%
Hard
75
min
Master the precise steps for both growth/decay (setting up DE, initial conditions) and orthogonal trajectories (forming DE of given family, replacement, solving new DE) to avoid common pitfalls.
🧮 Key Formulas
✅ Key Points for JEE
- 1For growth/decay problems, correctly set up the differential equation dN/dt = kN and determine the sign of 'k' based on whether it's growth (k>0) or decay (k<0). Pay close attention to initial conditions to find the constant of integration.
- 2The first step in finding orthogonal trajectories is to form the differential equation of the *given* family of curves by differentiating and eliminating the arbitrary constant(s).
- 3To obtain the differential equation of the orthogonal family, replace dy/dx with -dx/dy (or -1/(dy/dx)) in the DE of the given family. Then, solve this new differential equation.
- 4For certain families of curves (e.g., circles centered at origin, spirals), converting to polar coordinates (x=r cosθ, y=r sinθ) can significantly simplify the process of finding the orthogonal trajectories. Remember to convert the final solution back to Cartesian if required.
- 5The final step involves solving the new differential equation, which often reduces to a variable separable or linear differential equation. Accuracy in integration is crucial.
⚠️ Common Mistakes
- ✕Incorrectly setting up the differential equation for growth/decay, especially the sign of 'k' or misinterpreting initial conditions.
- ✕Algebraic errors while eliminating the arbitrary constant(s) to form the differential equation of the given family of curves.
- ✕Incorrectly replacing dy/dx with -dx/dy (e.g., using -dy/dx instead of its reciprocal negative) for orthogonal trajectories.
- ✕Errors in solving the resulting differential equation for the orthogonal family, particularly during integration or applying limits/initial conditions.
📝 Practice Questions
See allQ12.Let f : R →R be a twice differentiable function such that f(x + y) = f(x)f(y) for all x, y ∈R. If f ′(0) = 4a and f satisfies f ′′(x) −3af ′(x) −f(x) = 0, a > 0, then the area of the region R = {(x, y) ∣0 ≤y ≤f(ax), 0 ≤x ≤2} is: (1) e2 −1 (2) e2 + 1 (3) e4 + 1 (4) e4 −1
Q24.Let y = f(x) be the solution of the differential equation dydx + x2−1xy = √1−x2x6+4x f(0) = 0. If 6 ∫1/2−1/2 f(x)dx = 2π −α then α2 is equal to _______ .
Q2. Let x = x(y) be the solution of the differential equation y2 dx + (x −1y )dy (1) 1 2 + e (2) 3 + e (3) 3 −e (4) 32 + e
Q9. Let f(x) be a real differentiable function such that f(0) = 1 and f(x + y) = f(x)f ′(y) + f ′(x)f(y) for all x, y ∈R. Then ∑100n=1 loge f(n) is equal to : (1) 2525 (2) 5220 (3) 2384 (4) 2406
Q16.If x = f(y) is the solution of the differential equation (1 + y2) + (x −2etan−1 y) dydx is equal to : f(0) = 1, then f ( √31 ) (1) eπ/12 (2) eπ/4 (3) eπ/3 (4) eπ/6
Q6. Let a curve y = f(x) pass through the points (0, 5) and (loge 2, k). If the curve satisfies the differential equation 2(3 + y)e2xdx −(7 + e2x)dy = 0, then k is equal to (1) 4 (2) 32 (3) 8 (4) 16
NCERT Chapters
- Class 12 Mathematics Ch 9: Differential Equations