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
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### Chapter Mix
Class 12 Maths: Differential Equations