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(NONE) The given differential equation is $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$.
Dividing by $\sec x$, we get $\frac{dy}{dx} - 2y \cos x = 2 \co

Vedclass · Accepted Answer

$2 - \frac{9e}{4}$

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(NONE) The given differential equation is $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$.
Dividing by $\sec x$, we get $\frac{dy}{dx} - 2y \cos x = 2 \cos x + 3 \sin x \cos x$.
This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = -2 \cos x$ and $Q = 2 \cos x + 3 \sin x \cos x$.
The integrating factor $I.F. = e^{\int P dx} = e^{\int -2 \cos x dx} = e^{-2 \sin x}$.
The general solution is $y \cdot I.F. = \int Q \cdot I.F. dx + C$.
$y e^{-2 \sin x} = \int (2 \cos x + 3 \sin x \cos x) e^{-2 \sin x} dx + C$.
Let $u = -2 \sin x$, then $du = -2 \cos x dx$, so $\cos x dx = -\frac{du}{2}$.
The integral becomes $\int (-1 - \frac{3}{2} u) e^u dx = \int (-1 - \frac{3}{2} u) e^u (-\frac{du}{2}) = \int (\frac{1}{2} + \frac{3}{4} u) e^u du = \frac{1}{2} e^u + \frac{3}{4} (u e^u - e^u) + C = \frac{3}{4} u e^u - \frac{1}{4} e^u + C$.
Substituting back $u = -2 \sin x$: $y e^{-2 \sin x} = \frac{3}{4} (-2 \sin x) e^{-2 \sin x} - \frac{1}{4} e^{-2 \sin x} + C = -\frac{3}{2} \sin x e^{-2 \sin x} - \frac{1}{4} e^{-2 \sin x} + C$.
$y = -\frac{3}{2} \sin x - \frac{1}{4} + C e^{2 \sin x}$.
Given $y(0) = -\frac{7}{4}$, we have $-\frac{7}{4} = 0 - \frac{1}{4} + C \Rightarrow C = -\frac{6}{4} = -\frac{3}{2}$.
So, $y(x) = -\frac{3}{2} \sin x - \frac{1}{4} - \frac{3}{2} e^{2 \sin x}$.
At $x = \frac{\pi}{6}$, $\sin x = \frac{1}{2}$, so $y(\frac{\pi}{6}) = -\frac{3}{2}(\frac{1}{2}) - \frac{1}{4} - \frac{3}{2} e^{2(1/2)} = -\frac{3}{4} - \frac{1}{4} - \frac{3e}{2} = -1 - \frac{3e}{2}$.

Let $y=y(x)$ be the solution of the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3 \sin x$, where $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$ and $y(0) = -\frac{7}{4}$. Then $y(\frac{\pi}{6})$ is equal to:

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