Let y = y(x) be the solution of the differential equation such that . Then is equal to __________. [2025]

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Solution

Q.
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 __________. [2025]

Ans.
(21)

We have, $\frac{d y}{d x} + (2 \sec^{2} x) y = 2 \sec^{2} x + 3 \tan x \sec^{2} x$

$I . F . = e^{\int 2 \sec^{2} x d x} = e^{2 \tan x}$

$∴$ Solution of differential equation is given by

$y \cdot e^{2 \tan x} = \int 2 \sec^{2} x \cdot e^{2 \tan x} d x + \int 3 \tan x \sec^{2} x \cdot e^{2 \tan x} d x$

Put $\tan x = t \Rightarrow \sec^{2} x d x = d t$

$∴ y e^{2 \tan x} = \int 2 e^{2 t} d t + 3 \int t e^{2 t} d t$

$= \frac{2 e^{2 t}}{2} + 3 [\frac{t e^{2 t}}{2} - \int \frac{e^{2 t}}{2} d t]$

$= e^{2 t} + \frac{3}{2} t e^{2 t} - \frac{3}{4} e^{2 t} + c$

$\Rightarrow y e^{2 \tan x} = e^{2 \tan x} + \frac{3}{2} \tan x e^{2 \tan x} - \frac{3}{4} e^{2 \tan x} + c$

$\Rightarrow y = \frac{1}{4} + \frac{3}{2} \tan x + c e^{- 2 \tan x}$

Now, $y (0) = \frac{5}{4} \Rightarrow \frac{5}{4} = \frac{1}{4} + c \Rightarrow c = 1$

$∴ y = \frac{1}{4} + \frac{3}{2} \tan x + e^{- 2 \tan x}$

$\Rightarrow y (\frac{π}{4}) = \frac{1}{4} + \frac{3}{2} + e^{- 2} = \frac{7}{4} + e^{- 2}$

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

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