Key Concepts and Formulas

• Exact Differential Equations: A differential equation of the form $M(x,y) \, dx + N(x,y) \, dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The solution is given by $f(x,y) = C$, where $df = M \, dx + N \, dy$.
• Total Differential: The total differential of a function $f(x,y)$ is given by $df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy$.
• Integration: $\int d(f(x,y)) = f(x,y) + C$.

Step-by-Step Solution

Step 1: Rearrange the Differential Equation

We are given the differential equation: $(e^y+1) \cos x \, dx + e^y \sin x \, dy = 0$ We expand the first term and rearrange to group similar terms. This helps in recognizing a possible total differential. $e^y \cos x \, dx + \cos x \, dx + e^y \sin x \, dy = 0$ Grouping the terms involving $e^y$: $(e^y \cos x \, dx + e^y \sin x \, dy) + \cos x \, dx = 0$

Why this step? Expanding and regrouping allows us to isolate expressions that might resemble known total differentials. The terms $e^y \cos x \, dx$ and $e^y \sin x \, dy$ suggest a connection to the product rule involving $e^y$ and $\sin x$.

Step 2: Identify the Total Differential

Consider the total differential of $e^y \sin x$: $d(e^y \sin x) = \frac{\partial}{\partial x}(e^y \sin x) \, dx + \frac{\partial}{\partial y}(e^y \sin x) \, dy$ $d(e^y \sin x) = e^y \cos x \, dx + e^y \sin x \, dy$ This matches the grouped terms in the previous step. Substituting this back into the equation: $d(e^y \sin x) + \cos x \, dx = 0$

Why this step? Recognizing that part of the equation is the total differential of $e^y \sin x$ simplifies the integration process considerably.

Step 3: Integrate to Find the General Solution

Integrate both sides of the equation: $\int d(e^y \sin x) + \int \cos x \, dx = \int 0 \, dx$ $e^y \sin x + \sin x = C$ where $C$ is the constant of integration.

Why this step? Integration is the inverse of differentiation. Integrating the differential forms gives us the general solution, which represents the relationship between $y$ and $x$.

Step 4: Apply Initial Conditions

The solution passes through the point $(\frac{\pi}{2}, 0)$. Substitute $x = \frac{\pi}{2}$ and $y = 0$ into the general solution: $e^0 \sin\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = C$ Since $e^0 = 1$ and $\sin\left(\frac{\pi}{2}\right) = 1$: $(1)(1) + 1 = C$ $C = 2$

Why this step? The initial condition allows us to find a particular solution by determining the value of the constant of integration, $C$.

Step 5: Formulate the Specific Solution

Substitute $C = 2$ back into the general solution: $e^y \sin x + \sin x = 2$ Factoring out $\sin x$: $(e^y + 1) \sin x = 2$

Why this step? This specific solution represents the curve $y(x)$ that passes through $(\pi/2, 0)$ and satisfies the differential equation.

Step 6: Evaluate the Required Expression

We need to find $e^{y(\frac{\pi}{6})}$. Substitute $x = \frac{\pi}{6}$ into the particular solution: $(e^{y(\frac{\pi}{6})} + 1) \sin\left(\frac{\pi}{6}\right) = 2$ Since $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$: $(e^{y(\frac{\pi}{6})} + 1) \cdot \frac{1}{2} = 2$ Multiply both sides by 2: $e^{y(\frac{\pi}{6})} + 1 = 4$ Subtract 1 from both sides: $e^{y(\frac{\pi}{6})} = 3$

Why this step? This step provides the final answer by using the derived particular solution to compute the value of $e^y$ at $x = \frac{\pi}{6}$.

Common Mistakes & Tips:

• Always look for total differentials to simplify the solution.
• Don't forget the constant of integration when integrating.
• Be careful when substituting initial conditions to find the constant of integration.

Summary:

By recognizing the exact differential $d(e^y \sin x)$, we simplified the given differential equation and found the general solution. Applying the initial condition allowed us to determine the particular solution, and finally, we evaluated $e^{y(\pi/6)}$ to obtain the answer.

The final answer is \boxed{3}.