Question:

Step 1: Determine the initial conditions at $x=0$. We are given the equation $\sin x + \sin y = 1$. To find the value of $y$ when $x=0$, we substitute $x=0$ into the equation. $$ \sin(0) + \sin y = 1 $$ $$ 0 + \sin y = 1 $$ $$ \sin y = 1 $$ For $y$ in the first quadrant, this implies $y = \frac{\pi}{2}$. So, at $x=0$, we have $y=\frac{\pi}{2}$. Step 2: Differentiate the given equation implicitly with respect to $x$ to find $\frac{dy}{dx}$. Differentiating $\sin x + \sin y = 1$ with respect to $x$: $$ \frac{d}{dx}(\sin x) + \frac{d}{dx}(\sin y) = \frac{d}{dx}(1) $$ $$ \cos x + \cos y \cdot \frac{dy}{dx} = 0 $$ Now, we solve for $\frac{dy}{dx}$: $$ \cos y \cdot \frac{dy}{dx} = -\cos x $$ $$ \frac{dy}{dx} = -\frac{\cos x}{\cos y} $$ Let's evaluate $\frac{dy}{dx}$ at $x=0$ (where $y=\frac{\pi}{2}$): $$ \frac{dy}{dx} \Big|_{x=0} = -\frac{\cos(0)}{\cos(\frac{\pi}{2})} = -\frac{1}{0} $$ This indicates that $\frac{dy}{dx}$ is undefined at $x=0$, suggesting that we need to use series expansions to properly analyze the behavior near $x=0$. Step 3: Express $\sin y$ using the constraint and introduce a substitution for $y$ near $\frac{\pi}{2}$. From the given equation, $\sin y = 1 - \sin x$. We use the Maclaurin series expansion for $\sin x$: $$ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots = x - \frac{x^3}{6} + O(x^5) $$ Substituting this into the expression for $\sin y$: $$ \sin y = 1 - \left(x - \frac{x^3}{6} + O(x^5)\right) = 1 - x + \frac{x^3}{6} + O(x^5) $$ Since $y \to \frac{\pi}{2}$ as $x \to 0$, let $y = \frac{\pi}{2} + u$, where $u$ is a small quantity that approaches $0$ as $x \to 0$. Now, express $\sin y$ in terms of $u$: $$ \sin y = \sin\left(\frac{\pi}{2} + u\right) = \cos u $$ Using the Maclaurin series expansion for $\cos u$: $$ \cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6) $$ Step 4: Equate the two series expansions for $\sin y$ to find $u$ in terms of $x$. By equating the two expressions for $\sin y$ from Step 3: $$ 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6) = 1 - x + \frac{x^3}{6} + O(x^5) $$ Subtracting $1$ from both sides: $$ -\frac{u^2}{2} + \frac{u^4}{24} + O(u^6) = -x + \frac{x^3}{6} + O(x^5) $$ For small $x$, the dominant term on the right is $-x$. This implies that the dominant term for $-\frac{u^2}{2}$ must be $-x$. $$ -\frac{u^2}{2} \approx -x \implies u^2 \approx 2x $$ So, $u$ is of the order $\sqrt{x}$. We can approximate $u^4$ as $(2x)^2 = 4x^2$. $$ \frac{u^2}{2} = x - \frac{x^3}{6} + O(x^5) $$ $$ u^2 = 2x - \frac{x^3}{3} + O(x^5) $$ Now, take the square root to find $u$: $$ u = \sqrt{2x - \frac{x^3}{3} + O(x^5)} $$ Factor out $2x$: $$ u = \sqrt{2x \left(1 - \frac{x^2}{6} + O(x^4)\right)} $$ $$ u = \sqrt{2x} \left(1 - \frac{x^2}{6} + O(x^4)\right)^{1/2} $$ Using the generalized binomial theorem $(1+z)^\alpha \approx 1 + \alpha z$ for small $z$: $$ u = \sqrt{2x} \left(1 + \frac{1}{2}\left(-\frac{x^2}{6}\right) + O(x^4)\right) $$ $$ u = \sqrt{2x} \left(1 - \frac{x^2}{12} + O(x^4)\right) $$ $$ u = \sqrt{2x} - \frac{x^2\sqrt{2x}}{12} + O(x^{4}\sqrt{x}) $$ $$ u = \sqrt{2}x^{1/2} - \frac{\sqrt{2}}{12}x^{5/2} + O(x^{9/2}) $$ Step 5: Compute $\frac{dy}{dx}$ using the series expansion of $u$. Since $y = \frac{\pi}{2} + u$, we have $\frac{dy}{dx} = \frac{du}{dx}$. Differentiate the expression for $u$ found in Step 4: $$ \frac{dy}{dx} = \frac{d}{dx} \left(\sqrt{2}x^{1/2} - \frac{\sqrt{2}}{12}x^{5/2} + O(x^{9/2})\right) $$ $$ \frac{dy}{dx} = \sqrt{2} \cdot \frac{1}{2}x^{-1/2} - \frac{\sqrt{2}}{12} \cdot \frac{5}{2}x^{3/2} + O(x^{7/2}) $$ $$ \frac{dy}{dx} = \frac{\sqrt{2}}{2\sqrt{x}} - \frac{5\sqrt{2}}{24}x^{3/2} + O(x^{7/2}) $$ $$ \frac{dy}{dx} = \frac{1}{\sqrt{2x}} - \frac{5\sqrt{2}}{24}x^{3/2} + O(x^{7/2}) $$ Step 6: Compute $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$. Differentiate the expression for $\frac{dy}{dx}$ from Step 5: $$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{1}{\sqrt{2}}x^{-1/2} - \frac{5\sqrt{2}}{24}x^{3/2} + O(x^{7/2})\right) $$ $$ \frac{d^2y}{dx^2} = \frac{1}{\sqrt{2}} \cdot \left(-\frac{1}{2}\right)x^{-3/2} - \frac{5\sqrt{2}}{24} \cdot \frac{3}{2}x^{1/2} + O(x^{5/2}) $$ $$ \frac{d^2y}{dx^2} = -\frac{1}{2\sqrt{2}}x^{-3/2} - \frac{5\sqrt{2}}{16}x^{1/2} + O(x^{5/2}) $$ The dominant term as $x \to 0$ is the term with the most negative power of $x$: $$ \frac{d^2y}{dx^2} \approx -\frac{1}{2\sqrt{2}}x^{-3/2} $$ Step 7: Determine the constant $a$ such that $\lim_{x \to 0} x^a \frac{d^2y}{dx^2}$ is a non-zero finite value $L$. We have $\frac{d^2y}{dx^2} \approx -\frac{1}{2\sqrt{2}}x^{-3/2}$. We need to find $a$ such that the limit of $x^a \cdot \left(-\frac{1}{2\sqrt{2}}x^{-3/2}\right)$ as $x \to 0$ is a non-zero constant. $$ \lim_{x \to 0} x^a \left(-\frac{1}{2\sqrt{2}}x^{-3/2}\right) = \lim_{x \to 0} \left(-\frac{1}{2\sqrt{2}}x^{a - 3/2}\right) $$ For this limit to be finite and non-zero, the exponent of $x$ must be $0$: $$ a - \frac{3}{2} = 0 $$ $$ a = \frac{3}{2} $$ With this value of $a$, the limit $L = -\frac{1}{2\sqrt{2}}$, which is non-zero. Step 8: Calculate the value of $2a$. Given $a = \frac{3}{2}$, we calculate $2a$: $$ 2a = 2 \times \frac{3}{2} = 3 $$ The final answer is $\boxed{3}$.