Differential Equations
Differential Equations
nta_pyq_2025_jan
Grade 12
Question:
If for the solution curve y = f (x) of the differential equation dy , 2+sec x + (tan x)y = dx 2 (1+2 sec x) -\pi \sqrt3 x \in ( 2 , \pi 2 ),f ( \pi 3 ) = 10 , then f ( \pi 4 ) is equal to : \sqrt3+1
10(4+\sqrt3) 5-\sqrt3
2\sqrt2 9\sqrt3+3
10(4+\sqrt3) 4-\sqrt2
14
Step-by-Step Solution
Key Concept: Apply the core result for formation and solution of differential equations and simplify using the given constraints.
If e \int tan xdx = e ln(sec x) = sec x (4) \therefore y ⋅ sec x = \int { 2 + sec x } sec xdx 2 (1 + 2 sec x) 2 2 cos x + 1 1 - t = \int dx Let cos x = 2 2 (cos x + 2) 1 + t 2 1-t 2( ) + 1 2 1+t = \int 2dt 2 2 1-t ( + 2) 2 1+t 2 2 2 - 2t + 1 + t = \int \times 2dt 2 2 2 (1 - t + 2 + 2t ) 2 3 - t = 2\int dt 2 2 (t + 3) Let t + 3 t = u 3 (1 - ) dt = du 2 t du = -2 \int 2 u 2 y ⋅ (sec x) = + c u 2 y ⋅ sec x = + c. . . (I ) 3 t + t \pi x 1 At x = , t = tan = 3 2 \sqrt3 \sqrt3 2 2 ⋅ = + c 10 1 +3\sqrt3 \sqrt3 \sqrt3 2\sqrt 3 2. = + c \Rightarrow C = 0 10 10 \pi x At x = , t = tan = \sqrt2 - 1 4 2 2 \therefore y ⋅ \sqrt2 = 3 \sqrt2 - 1 + \sqrt2-1 2(\sqrt2 - 1) y ⋅ \sqrt2 = 6 - 2\sqrt 2 \sqrt2(\sqrt2 - 1) 1 2\sqrt 2 - 1 y = = \times 2(3 - \sqrt2) \sqrt2 7 4 - \sqrt2 = 14
Correct Answer: 4