Differential Equations
Differential Equations
nta_pyq_2025_jan
Grade 12
Question:
If x = f (y) is the solution of the differential equation -1 dy 2 tan y \pi \pi (1 + y ) + (x - 2e ) = 0, y \in (- , ) dx 2 2 with f (0) = 1, then f ( 1 ) is equal to : \sqrt3
e \pi/12
e \pi/4
e \pi/3
e \pi/6
Step-by-Step Solution
Key Concept: Apply the core result for formation and solution of differential equations and simplify using the given constraints.
dx x 2e tan y + = (4) dy 1 + y 2 1 + y 2 -1 tan y I.F. = e 2 -1 tan y 2(e ) dy -1 tan y xe = \int 2 1 + y dy -1 Put tan y = t, = dt 2 1 + y -1 tan y 2t xe = \int 2e dt -1 -1 tan y 2 tan y xe = e + c -1 -1 tan y - tan y x = e + ce ∵ y = 0, x = 1 1 = 1 + c \Rightarrow c = 0 1 \pi/6 y = ,x = e \sqrt3
Correct Answer: 4