For each the differential equations given, find the general solution :
$\frac{d y}{d x} + 2 y sin x$

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Solution

$\frac{d y}{d x} + 2 y sin x = 0$

$\Rightarrow \int \frac{d y}{y} = \int - 2 sin x d x$

$\Rightarrow ln y = 2 cos x + c$

$\Rightarrow y = e^{2 cos x + c}$

$\Rightarrow y = k e^{2 cos x}$

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\frac{d y}{d x} + \frac{y}{x} = x^{2}

x l o g x \frac{d y}{d x} + y = \frac{2}{x} l o g x

\frac{d y}{d x} + (sec x) y = tan x (0 \leq x \leq \frac{π}{2})

(1 + x^{2}) d y + 2 x y d x = cot x d x (x \neq 0)

For each of the given differential equation find the general solution.
$\frac{d y}{d x} + y s e c x = t a n x (0 \leq x \leq \frac{π}{2}) .$

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