JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x

such that

y (0) = \frac{5}{4}

. Then

12 (y (\frac{π}{4}) - e^{- 2})

is equal to ______.

Answer: 21

Solution

The equation is first-order linear with integrating factor

e^{\int 2 s e c^{2} x d x} = e^{2 t a n x}

. With

u = tan x

, the right side integrates as

\int (2 + 3 u) e^{2 u} d u = e^{2 u} (\frac{1}{4} + \frac{3 u}{2})

, so

y = \frac{1}{4} + \frac{3}{2} tan x + C e^{- 2 t a n x}

. Applying

y (0) = \frac{5}{4}

gives

C = 1

. At

x = \frac{π}{4}

y = \frac{1}{4} + \frac{3}{2} + e^{- 2} = \frac{7}{4} + e^{- 2}

, so

12 (y (\frac{π}{4}) - e^{- 2}) = 12 \cdot \frac{7}{4} = 21

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