Question

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

• A. $-\frac{5}{2}$
• B. $-\frac{5}{4}$
• C. $-3\sqrt{2} - 7$
• D. $-3\sqrt{3} - 7$

Hint:

Linear DEs involving trig terms often simplify nicely after substitution into the integral of the form $\int f(\sin x) \cos x dx$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a first-order linear differential equation. We convert it to standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and use an integrating factor.

Step 2: Detailed Explanation:
Divide by $\sec x$:
$\frac{dy}{dx} - 2\cos x \cdot y = (2 + 3\sin x)\cos x$.
Integrating factor $IF = e^{\int -2\cos x dx} = e^{-2\sin x}$.
Solution: $y \cdot e^{-2\sin x} = \int (2 + 3\sin x)\cos x \cdot e^{-2\sin x} dx$.
Let $u = -2\sin x \implies du = -2\cos x dx$.
\[ \int (2 - \frac{3u}{2}) e^u (-\frac{1}{2}) du = -\frac{1}{2} \int (2 - \frac{3u}{2}) e^u du \]
Integrating by parts:
\[ = -\frac{1}{2} [ (2 - \frac{3u}{2}) e^u - \int (-\frac{3}{2}) e^u du ] = -\frac{1}{2} e^u (2 - \frac{3u}{2} + \frac{3}{2}) = e^u (\frac{3u}{4} - \frac{7}{4}). \]
$y e^{-2\sin x} = e^{-2\sin x} (-\frac{3}{2}\sin x - \frac{7}{4}) + C \implies y = -\frac{3}{2}\sin x - \frac{7}{4} + C e^{2\sin x}$.
Given $y(0) = -7/4 \implies -7/4 = -7/4 + C \implies C = 0$.
So $y(x) = -\frac{3}{2}\sin x - \frac{7}{4}$.
For $x = \pi/6$: $y(\pi/6) = -\frac{3}{2}(1/2) - 7/4 = -3/4 - 7/4 = -10/4 = -5/2$.

Step 3: Final Answer:
$y(\pi/6) = -5/2$.