Key Concepts and Formulas

• Product Rule for Differentiation: $\frac{d}{dx}[u(x)v(x)] = u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}$
• Integration as the Inverse of Differentiation: $\int \frac{d}{dx} f(x) \, dx = f(x) + C$
• Using Initial Conditions: Substitute given values of $x$ and $y$ into the general solution to find the constant of integration, $C$.

Step-by-Step Solution

Step 1: Recognize the Product Rule in Reverse We are given the differential equation: $(2 + \sin x)\frac{dy}{dx} + (y + 1)\cos x = 0$ Our goal is to identify if the left-hand side can be expressed as the derivative of a product. This allows us to directly integrate and bypass more complex methods. Let's consider $u(x) = (2 + \sin x)$ and $v(x) = (y + 1)$. Then, $\frac{du}{dx} = \cos x$ and $\frac{dv}{dx} = \frac{dy}{dx}$. Substituting these into the product rule, we get: $u(x)\frac{dv}{dx} + v(x)\frac{du}{dx} = (2 + \sin x)\frac{dy}{dx} + (y + 1)\cos x$ This matches the given differential equation perfectly.

Step 2: Apply the Reverse Product Rule Since we have identified $u(x)$ and $v(x)$ such that the given equation is in the form $u(x)\frac{dv}{dx} + v(x)\frac{du}{dx} = 0$, we can rewrite it as the derivative of their product: $\frac{d}{dx}[(2 + \sin x)(y + 1)] = 0$ Why this step? This transformation simplifies the differential equation, indicating that the expression $(2 + \sin x)(y + 1)$ is a constant with respect to $x$.

Step 3: Integrate both sides of the equation To eliminate the derivative, we integrate both sides with respect to $x$: $\int \frac{d}{dx}[(2 + \sin x)(y + 1)] \, dx = \int 0 \, dx$ This yields: $(2 + \sin x)(y + 1) = C$ where $C$ is the constant of integration. Why this step? Integration is the inverse of differentiation. Applying it allows us to find the general solution of the differential equation.

Step 4: Use the initial condition to find the constant of integration We are given the initial condition $y(0) = 1$. This means when $x = 0$, the value of $y$ is $1$. Substitute these values into our general solution to find $C$. Substitute $x = 0$ and $y = 1$ into the equation $(2 + \sin x)(y + 1) = C$: $(2 + \sin 0)(1 + 1) = C$ Since $\sin 0 = 0$: $(2 + 0)(2) = C$ $4 = C$ Why this step? The initial condition pins down a specific solution from the family of solutions represented by the general solution.

Step 5: Find the particular solution Now that we have the value of $C$, we can write the particular solution: $(2 + \sin x)(y + 1) = 4$ This equation describes the specific relationship between $y$ and $x$ for the given problem.

Step 6: Evaluate $y\left(\frac{\pi}{2}\right)$ We need to find the value of $y$ when $x = \frac{\pi}{2}$. Substitute $x = \frac{\pi}{2}$ into the particular solution: $(2 + \sin \frac{\pi}{2})(y + 1) = 4$ Since $\sin \frac{\pi}{2} = 1$: $(2 + 1)(y + 1) = 4$ $3(y + 1) = 4$ Now, solve for $y$: $y + 1 = \frac{4}{3}$ $y = \frac{4}{3} - 1$ $y = \frac{4}{3} - \frac{3}{3}$ $y = \frac{1}{3}$

Step 7: Check if there is any typo in the question and solution Since the correct answer is given as -2/3, let's recheck if the question has any sign error. The given equation is: $(2 + \sin x)\frac{dy}{dx} + (y + 1)\cos x = 0$ The initial condition is y(0) = 1. We found the particular solution to be $(2 + \sin x)(y + 1) = 4$ Substituting $x = \frac{\pi}{2}$ $(2 + 1)(y + 1) = 4$ $3(y + 1) = 4$ $y + 1 = \frac{4}{3}$ $y = \frac{4}{3} - 1 = \frac{1}{3}$ There seems to be an error in the given correct answer. Let's assume that the equation is $(2 + \sin x)\frac{dy}{dx} - (y + 1)\cos x = 0$ Then $\frac{d}{dx}[(2 + \sin x)(y + 1)] = 2\cos x(y+1) + (2 + \sin x)\frac{dy}{dx}$ So the equation becomes $(2 + \sin x)\frac{dy}{dx} = (y + 1)\cos x$ $\frac{dy}{y+1} = \frac{\cos x}{2 + \sin x}dx$ $\int \frac{dy}{y+1} = \int \frac{\cos x}{2 + \sin x}dx$ $\ln(y+1) = \ln(2 + \sin x) + C$ $y + 1 = e^C(2 + \sin x)$ $y + 1 = k(2 + \sin x)$ Using y(0) = 1, $1 + 1 = k(2 + 0)$ $2 = 2k \implies k = 1$ $y + 1 = 2 + \sin x$ $y = 1 + \sin x$ $y(\frac{\pi}{2}) = 1 + 1 = 2$ Let's consider another possibility, the equation is: $(2 + \sin x)\frac{dy}{dx} + (y - 1)\cos x = 0$ The initial condition is y(0) = 1. Then $\frac{d}{dx}[(2 + \sin x)(y - 1)] = (2 + \sin x)\frac{dy}{dx} + (y - 1)\cos x = 0$ $(2 + \sin x)(y - 1) = C$ Using y(0) = 1 $(2 + 0)(1 - 1) = C \implies C = 0$ $(2 + \sin x)(y - 1) = 0$ Since $2 + \sin x \ne 0$, $y - 1 = 0 \implies y = 1$ Then $y(\frac{\pi}{2}) = 1$

Let's assume the initial condition is y(0) = -1. Then C = 0, y = -1.

Let us assume that the original equation is correct and the correct answer is -1/3.

$(2 + \sin x)(y + 1) = 4$ $3(y + 1) = 4$ $y + 1 = 4/3$ $y = 1/3$ Then we want y = -1/3.

$(2 + \sin x)(y + a) = C$

Since we derived the correct solution to the original question, and the correct answer is 1/3.

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always include $+C$ when integrating an indefinite integral.
• Incorrectly Applying Initial Conditions: Ensure you substitute the correct values for $x$ and $y$ at the given point.
• Double check the question statement.

Summary

The given differential equation was solved by recognizing the reverse product rule. The particular solution was found to be $(2 + \sin x)(y + 1) = 4$. Substituting $x = \frac{\pi}{2}$ into this solution, we found that $y\left(\frac{\pi}{2}\right) = \frac{1}{3}$. There appears to be an error in the provided correct answer.

The final answer is $\boxed{1/3}$, which corresponds to option (D).