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Let f : mathbbR - \0\ to mathbbR be a function satisfying fleft(fracxyright) = fracf(x)f(y) for all x, y, f(y) neq 0. If f'(1) = 2024, then

Solution & Explanation

### Related Formula textChain rule for partial differentiation: fracpartialpartial x fleft(fracxyright) = f'left(fracxyright) cdot frac1y ### Core Logic Given functional equation: fleft(fracxyright) = fracf(x)f(y) First, put x=1, y=1: f(1) = fracf(1)f(1) Rightarrow f(1) = 1 (since f(y) neq 0 implies f(1) neq 0). ### Step 1: Partial Differentiation Partially differentiate the given equation with respect to x, keeping y constant: f'left(fracxyright) cdot frac1y = frac1f(y) cdot f'(x) Let x to y so that fracxy to 1. Wait, let's substitute y = x directly: f'(1) cdot frac1x = frac1f(x) cdot f'(x) ### Step 2: Forming the Final Equation We are given f'(1) = 2024. Substituting this value: 2024 cdot frac1x = fracf'(x)f(x) Cross-multiplying yields: 2024 f(x) = x f'(x) Rightarrow x f'(x) - 2024 f(x) = 0 ### Pattern Recognition Functional equations of the form f(x/y) = f(x)/f(y) inherently describe power functions (f(x) = x^k). Partial differentiation with respect to one variable quickly resolves the derivative form without limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Functions Class 12 Maths: Differential Equations

Reference Study Guides

Differential Equations

More Differential Equations Previous-Year Questions — Page 7

Q29 jee_main_2024_31_jan_evening Linear Differential Equations
Let y = y(x) be the solution of the differential equation sec^2 x \, dx + left( e^2y tan^2 x + tan x right) dy = 0, 0 < x < fracpi2, yleft(fracpi4right) = 0. If yleft(fracpi6right) = alpha. Then e^8alpha is equal to
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula textIntegrating factor for fracdudy + P(y)u = Q(y) text is I.F. = e^int P(y)dy ### Core Logic Given DE: sec^2 x fracdxdy + e^2y tan^2 x + tan x = 0 Substitute t = tan x implies fracdtdy = sec^2 x fracdxdy. The DE becomes: fracdtdy + t = -t^2 e^2y This is a Bernoulli equation in t. Divide by t^2: frac1t^2fracdtdy + frac1t = -e^2y Substitute u = frac1t implies fracdudy = -frac1t^2fracdtdy. -fracdudy + u = -e^2y implies fracdudy - u = e^2y This is a linear DE in u with respect to y. P(y) = -1, Q(y) = e^2y. Integrating Factor: I.F. = e^int -1 dy = e^-y. Solution: u e^-y = int e^2y e^-y dy = int e^y dy = e^y + C Substitute u = frac1tan x: frace^-ytan x = e^y + C Use given condition y(pi/4) = 0: frace^0tan(pi/4) = e^0 + C implies 1 = 1 + C implies C = 0 Therefore, frace^-ytan x = e^y implies tan x = e^-2y. Evaluate at x = pi/6, y = alpha: tan(pi/6) = e^-2alpha implies frac1sqrt3 = e^-2alpha e^2alpha = sqrt3 implies (e^2alpha)^4 = (sqrt3)^4 = 9 Thus, e^8alpha = 9. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q9 jee_main_2024_31_jan_morning Homogeneous Differential Equations
The solution curve of the differential equation yfracdxdy = x(log_e x - log_e y + 1), x > 0, y > 0 passing through the point (e, 1) is
  • A. |log_e fracyx| = x
  • B. |log_e fracyx| = y^2
  • C. |log_e fracxy| = y
  • D. 2|log_e fracxy| = y + 1

Solution

### Core Logic Given DE: fracdxdy = fracxy left(lnleft(fracxyright) + 1right) Let fracxy = t implies x = ty. Differentiating w.r.t y: fracdxdy = t + yfracdtdy ### Step 1: Substitution and Integration t + yfracdtdy = t(ln(t) + 1) = tln t + t yfracdtdy = tln t implies fracdttln t = fracdyy Integrate both sides. Let ln t = p implies frac1t dt = dp. int fracdpp = int fracdyy ln|p| = ln y + C implies ln|ln t| = ln y + C lnleft|lnleft(fracxyright)right| = ln y + C ### Step 2: Applying Boundary Conditions Given curve passes through (e, 1): lnleft|lnleft(frace1right)right| = ln(1) + C implies C = 0 lnleft|lnleft(fracxyright)right| = ln y left|lnleft(fracxyright)right| = e^ln y = y ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q11 jee_main_2024_31_jan_morning Linear Differential Equations
Let y = y(x) be the solution of the differential equation fracdydx = frac(tan x) + ysin x(sec x - sin x tan x), x in left(0, fracpi2right) satisfying the condition yleft(fracpi4right) = 2. Then, yleft(fracpi3right) is
  • A. sqrt3(2 + log_esqrt3)
  • B. fracsqrt32(2 + log_e 3)
  • C. sqrt3(1 + 2log_e 3)
  • D. sqrt3(2 + log_e 3)

Solution

### Core Logic fracdydx = fracfracsin xcos x + ysin x left(frac1cos x - fracsin^2 xcos xright) = fracsin x + ycos xsin x (1 - sin^2 x) fracdydx = fracsin x + ycos xsin x cos^2 x = sec^2 x + frac2ysin 2x fracdydx - 2csc(2x)y = sec^2 x ### Step 1: Integrating Factor This is an LDE of form fracdydx + Py = Q. I.F. = e^int -2csc(2x) dx Let 2x = t implies 2dx = dt. I.F. = e^-int csc t dt = e^-ln|tan(t/2)| = e^-ln|tan x| = frac1|tan x| ### Step 2: Solution of LDE y(I.F.) = int Q(I.F.) dx + C yfrac1tan x = int sec^2 x frac1tan x dx + C Let tan x = t implies sec^2 x dx = dt. yfrac1tan x = int fracdtt + C = ln|tan x| + C y = tan x(ln|tan x| + C) ### Step 3: Boundary Value Given y(pi/4) = 2: 2 = 1(ln 1 + C) implies C = 2 Thus, y = tan x (ln|tan x| + 2). At x = pi/3: y(pi/3) = sqrt3(lnsqrt3 + 2) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

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