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

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Solution

Q.
Let y = y(x) be the solution of the differential equation $\frac{d y}{d x} + 3 (\tan^{2} x) y + 3 y = \sec^{2} x, y (0) = \frac{1}{3} + e^{3}$ . Then $y (\frac{π}{4})$ is equal to [2025]

1 $\frac{4}{3} + e^{3}$

2 $\frac{4}{3}$

3 $\frac{2}{3}$

4 $\frac{2}{3} + e^{3}$

Ans.
(2)

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

$\Rightarrow \frac{d y}{d x} + 3 (\sec^{2} x) y = \sec^{2} x$

$I . F . = e^{3 \int \sec^{2} x d x} = e^{3 \tan x}$

$∴$ Solution is given by $e^{3 \tan x} y = \int e^{3 \tan x} \sec^{2} x d x$

$\Rightarrow e^{3 \tan x} y = \frac{e^{3 \tan x}}{3} + C$ ... (i)

$∵ y (0) = \frac{1}{3} + e^{3} \Rightarrow C = e^{3}$

Using equation (i), we get

$e^{3 \tan x} y = \frac{e^{3 \tan x}}{3} + e^{3}$

$∴ y (\frac{π}{4}) = \frac{\frac{e^{3}}{3} + e^{3}}{e^{3}} = \frac{4}{3}$

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