Key Concepts and Formulas

• Trigonometric Identity: $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$
• Integration: $\int \sec^2(x) \, dx = \tan(x) + C$ and $\int \sin(x) \, dx = -\cos(x) + C$
• Implicit Differentiation and the Chain Rule

Step-by-Step Solution

Step 1: Simplify the Differential Equation

We are given the differential equation: $\sec y \frac{dy}{dx} - \sin(x + y) - \sin(x - y) = 0$ We simplify using the trigonometric identity $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$, where $A = x$ and $B = y$. This gives us: $\sin(x+y) + \sin(x-y) = 2\sin x \cos y$ Substituting this back into the differential equation, we get: $\sec y \frac{dy}{dx} - 2\sin x \cos y = 0$ Now, we isolate the derivative term: $\sec y \frac{dy}{dx} = 2\sin x \cos y$ Why: This step uses a standard trigonometric identity to simplify the equation into a more manageable form, preparing it for separation of variables.

Step 2: Separate the Variables

We separate the variables by dividing both sides by $\cos y$ and multiplying by $dx$: $\frac{\sec y}{\cos y} dy = 2\sin x dx$ Since $\sec y = \frac{1}{\cos y}$, we have $\frac{\sec y}{\cos y} = \frac{1}{\cos^2 y} = \sec^2 y$. Thus, the equation becomes: $\sec^2 y \, dy = 2\sin x \, dx$ Why: Separating variables allows us to integrate each side independently, which is a standard technique for solving differential equations of this form.

Step 3: Integrate Both Sides

We integrate both sides of the equation: $\int \sec^2 y \, dy = \int 2\sin x \, dx$ The integral of $\sec^2 y$ is $\tan y$, and the integral of $2\sin x$ is $-2\cos x$. Therefore, we have: $\tan y = -2\cos x + C$ where $C$ is the constant of integration. Why: Integration is the inverse operation of differentiation. We integrate to find the relationship between $y$ and $x$. The constant of integration accounts for all possible solutions.

Step 4: Apply the Initial Condition

We are given the initial condition $y(0) = 0$. Substituting $x = 0$ and $y = 0$ into the general solution, we get: $\tan(0) = -2\cos(0) + C$ Since $\tan(0) = 0$ and $\cos(0) = 1$, we have: $0 = -2(1) + C$ Solving for $C$, we find $C = 2$. Thus, the particular solution is: $\tan y = -2\cos x + 2$ Why: The initial condition allows us to find a unique solution (particular solution) that satisfies both the differential equation and the given point.

Step 5: Implicit Differentiation and Evaluation at $x = \frac{\pi}{2}$

We differentiate the particular solution implicitly with respect to $x$: $\frac{d}{dx}(\tan y) = \frac{d}{dx}(-2\cos x + 2)$ $\sec^2 y \frac{dy}{dx} = 2\sin x$ Now we need to find $y'(\frac{\pi}{2})$. We first find $y$ when $x = \frac{\pi}{2}$: $\tan y = -2\cos\left(\frac{\pi}{2}\right) + 2$ Since $\cos(\frac{\pi}{2}) = 0$, we have $\tan y = 2$. Then, we have $\sec^2 y = 1 + \tan^2 y = 1 + 2^2 = 5$. Substituting $x = \frac{\pi}{2}$ and $\sec^2 y = 5$ into the differentiated equation, we get: $5 \frac{dy}{dx} = 2\sin\left(\frac{\pi}{2}\right)$ Since $\sin(\frac{\pi}{2}) = 1$, we have: $5 \frac{dy}{dx} = 2$ Thus, $5y'\left(\frac{\pi}{2}\right) = 2$. Why: Implicit differentiation is used because we have $y$ defined implicitly as a function of $x$. We evaluate at $x=\pi/2$ to get the value of the expression we want.

Step 6: Final Answer

The problem asks for the value of $5y'\left(\frac{\pi}{2}\right)$, which we found to be 2.

Common Mistakes & Tips

• Be careful with the signs when integrating trigonometric functions.
• Remember to use the chain rule when differentiating implicitly.
• Don't forget to use the initial condition to find the particular solution.

Summary

We solved the differential equation by separating variables, integrating both sides, and using the initial condition to find the particular solution. We then used implicit differentiation to find the derivative and evaluated it at the given point. The value of $5y'\left(\frac{\pi}{2}\right)$ is 2.

Final Answer The final answer is \boxed{2}.