Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$.
• Integrating Factor (I.F.): For a linear first-order differential equation, the integrating factor is $e^{\int P(x) dx}$.
• Solution of Linear First-Order Differential Equation: The solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$, where $C$ is the constant of integration.

Step 1: Analyze the Given Differential Equation and Prepare for Substitution

The given differential equation is: $e^{\sin y}\cos y\frac{dy}{dx} + e^{\sin y}\cos x = \cos x$ We observe that the term $e^{\sin y}\cos y\frac{dy}{dx}$ resembles the derivative of $e^{\sin y}$ with respect to $x$. This suggests a substitution to simplify the equation.

Step 2: Introduce a Substitution

Let $t = e^{\sin y}$. Then, we find $\frac{dt}{dx}$ using the chain rule: $\frac{dt}{dx} = \frac{dt}{dy} \cdot \frac{dy}{dx} = e^{\sin y} \cos y \frac{dy}{dx}$ Substituting $t = e^{\sin y}$ and $\frac{dt}{dx} = e^{\sin y} \cos y \frac{dy}{dx}$ into the original equation, we get: $\frac{dt}{dx} + t\cos x = \cos x$

Step 3: Transform the Equation into a Linear First-Order Differential Equation

Rearranging the equation, we have: $\frac{dt}{dx} + (\cos x)t = \cos x$ This is a linear first-order differential equation in the form $\frac{dt}{dx} + P(x)t = Q(x)$, where $P(x) = \cos x$ and $Q(x) = \cos x$.

Step 4: Calculate the Integrating Factor

The integrating factor (I.F.) is given by: $\text{I.F.} = e^{\int P(x) dx} = e^{\int \cos x dx} = e^{\sin x}$

Step 5: Find the General Solution

The general solution is given by: $t \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$ Substituting the values, we have: $t \cdot e^{\sin x} = \int \cos x \cdot e^{\sin x} dx + C$ Let $u = \sin x$, so $du = \cos x dx$. Then, the integral becomes: $\int e^u du = e^u + C = e^{\sin x} + C$ Therefore, $t \cdot e^{\sin x} = e^{\sin x} + C$ $t = 1 + Ce^{-\sin x}$

Step 6: Substitute Back to Find y(x)

Substitute $t = e^{\sin y}$ back into the equation: $e^{\sin y} = 1 + Ce^{-\sin x}$

Step 7: Apply the Initial Condition y(0) = 0

Given $y(0) = 0$, we substitute $x = 0$ and $y = 0$ into the equation: $e^{\sin 0} = 1 + Ce^{-\sin 0}$ $e^0 = 1 + Ce^0$ $1 = 1 + C$ $C = 0$

Step 8: Obtain the Particular Solution

Since $C = 0$, the particular solution is: $e^{\sin y} = 1$ Taking the natural logarithm of both sides: $\sin y = \ln(1) = 0$ $\sin y = 0$ $y = n\pi, \quad n \in \mathbb{Z}$ Since $y(0) = 0$, we consider the solution $y = 0$ for all $x$.

Step 9: Evaluate the Expression

We need to find the value of: $1 + y\left( \frac{\pi}{6} \right) + \frac{\sqrt{3}}{2}y\left( \frac{\pi}{3} \right) + \frac{1}{\sqrt{2}}y\left( \frac{\pi}{4} \right)$ Since $y(x) = 0$ for all $x$: $1 + 0 + \frac{\sqrt{3}}{2}(0) + \frac{1}{\sqrt{2}}(0) = 1 + 0 + 0 + 0 = 1$

However, the correct answer is 0. There must be an error in our steps. Let's re-examine the solution from Step 5: $t e^{\sin x} = \int \cos x e^{\sin x} dx + C$ $t e^{\sin x} = e^{\sin x} + C$ $t = 1 + Ce^{-\sin x}$ $e^{\sin y} = 1 + Ce^{-\sin x}$ Using $y(0) = 0$, we have $e^{\sin 0} = 1 + Ce^{-\sin 0}$ $1 = 1 + C$ $C = 0$. So $e^{\sin y} = 1$, meaning $\sin y = 0$ and $y = 0$ for all $x$. The expression becomes: $1 + y(\pi/6) + \frac{\sqrt{3}}{2} y(\pi/3) + \frac{1}{\sqrt{2}} y(\pi/4) = 1 + 0 + 0 + 0 = 1$.

Let's go back to the original differential equation: ${e^{\sin y}}\cos y{{dy} \over {dx}} + {e^{\sin y}}\cos x = \cos x$ $\frac{dt}{dx} + t \cos x = \cos x$ $\frac{dt}{dx} = \cos x (1-t)$ $\frac{dt}{1-t} = \cos x dx$ $-\ln|1-t| = \sin x + C$ $\ln|1-t| = -\sin x - C$ $1-t = e^{-\sin x - C} = Ae^{-\sin x}$ for some constant A. $1-e^{\sin y} = Ae^{-\sin x}$ $e^{\sin y} = 1 - Ae^{-\sin x}$ When $x=0$, $y=0$. So $e^{\sin 0} = 1 - Ae^{-\sin 0}$. $1 = 1 - A$. Therefore, $A = 0$. This leads to the same solution $e^{\sin y} = 1$, so $y=0$.

There seems to be an error in the question itself or the given answer. Let's assume there's a typo and the initial condition is different. Suppose we want the final answer to be 0. Then: $1 + y(\pi/6) + \frac{\sqrt{3}}{2} y(\pi/3) + \frac{1}{\sqrt{2}} y(\pi/4) = 0$ $y(\pi/6) + \frac{\sqrt{3}}{2} y(\pi/3) + \frac{1}{\sqrt{2}} y(\pi/4) = -1$

Since the initial condition led to $y=0$, let's try separation of variables. $\frac{dt}{dx} = \cos x (1-t)$ $\int \frac{dt}{1-t} = \int \cos x dx$ $-\ln|1-t| = \sin x + C$ $\ln|1-t| = -\sin x - C$ $|1-t| = e^{-\sin x - C} = A e^{-\sin x}$ $1 - e^{\sin y} = A e^{-\sin x}$ $e^{\sin y} = 1 - A e^{-\sin x}$ $\sin y = \ln(1 - A e^{-\sin x})$ $y = \arcsin(\ln(1 - A e^{-\sin x}))$ $y(0) = 0$, so $0 = \arcsin(\ln(1 - A e^0))$ $0 = \arcsin(\ln(1 - A))$ $0 = \ln(1 - A)$ $1 = 1 - A$, so $A = 0$. This means we are back to $y=0$.

Let's try $y = \arcsin(0) = 0$.

Common Mistakes & Tips

• Be careful with the chain rule when differentiating composite functions.
• Remember to substitute back the original variable after solving the differential equation.
• Always check the initial conditions to find the particular solution.

Summary

We transformed the given differential equation into a linear first-order differential equation using a suitable substitution. After finding the integrating factor and solving for the general solution, we applied the initial condition to find the particular solution. However, the initial condition $y(0) = 0$ results in the trivial solution $y(x) = 0$, which yields a final answer of 1, not 0 as provided. There might be an error in the question or the expected answer.

The final answer is $\boxed{1}$.