Key Concepts and Formulas

• Separation of Variables: A method to solve first-order differential equations by isolating variables: $f(y) dy = g(x) dx$.
• Integration: Reversing the process of differentiation to find a function from its derivative.
• Logarithm Properties: $\int \frac{1}{u} du = \ln|u| + C$ and $e^{\ln(x)} = x$ and $\ln(a) + \ln(b) = \ln(ab)$.

Step-by-Step Solution

Step 1: Separate the Variables

We are given the differential equation: ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$ Our goal is to isolate terms with $y$ and $dy$ on one side, and terms with $x$ and $dx$ on the other.

First, subtract $e^x$ from both sides: ${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x}$ Now, multiply both sides by $(2+y)$ and $dx$, and divide both sides by $(5+e^x)$: $dy = \frac{-e^x}{{{5 + {e^x}}}} (2+y) dx$ $\frac{dy}{2+y} = \frac{-e^x}{5+e^x} dx$ Now the variables are separated.

Step 2: Integrate Both Sides

Integrate both sides of the equation with respect to their respective variables: $\int \frac{1}{2+y} dy = \int \frac{-e^x}{5+e^x} dx$ Let's evaluate each integral:

• Left Hand Side (LHS) Integral: $\int \frac{1}{2+y} dy = \ln|2+y| + C_1$ Explanation: This is a standard integral. The integral of $1/(ax+b)$ is $(1/a)\ln|ax+b|$.

• Right Hand Side (RHS) Integral: $\int \frac{-e^x}{5+e^x} dx$ Use the substitution method: Let $u = 5+e^x$, then $du = e^x dx$. $\int \frac{-1}{u} du = -\ln|u| + C_2 = -\ln|5+e^x| + C_2$ Since $5+e^x > 0$ for all real $x$, we can drop the absolute value: $-\ln(5+e^x) + C_2$ Explanation: The substitution method simplifies the integral.

Combining the results, we have: $\ln|2+y| = -\ln(5+e^x) + C$ where $C = C_2 - C_1$.

Step 3: Simplify the General Solution

Simplify the expression using logarithm properties. $\ln|2+y| + \ln(5+e^x) = C$ Using the logarithm property $\ln A + \ln B = \ln(AB)$: $\ln(|2+y|(5+e^x)) = C$ Exponentiate both sides: $e^{\ln(|2+y|(5+e^x))} = e^C$ $|2+y|(5+e^x) = e^C$ Let $K = e^C$, where $K$ is a positive constant. $|2+y|(5+e^x) = K$ Since the initial condition is $y(0)=1$, we have $2+y=3>0$. We will assume $2+y$ remains positive. Therefore, we can remove the absolute value: $(2+y)(5+e^x) = K$

Step 4: Apply the Initial Condition

We are given the initial condition $y(0) = 1$. Substitute $x=0$ and $y=1$ into the equation: $(2+1)(5+e^0) = K$ $(3)(5+1) = K$ $(3)(6) = K$ $K = 18$

Step 5: Write the Particular Solution

Substitute $K=18$ back into the equation: $(2+y)(5+e^x) = 18$ Solve for $y$: $2+y = \frac{18}{5+e^x}$ $y = \frac{18}{5+e^x} - 2$

Step 6: Evaluate y at the Desired Point

Find the value of $y(\log_e 13)$. Substitute $x = \log_e 13$ into the particular solution: $y(\log_e 13) = \frac{18}{5+e^{\log_e 13}} - 2$ Using the property $e^{\log_e A} = A$: $y(\log_e 13) = \frac{18}{5+13} - 2$ $y(\log_e 13) = \frac{18}{18} - 2$ $y(\log_e 13) = 1 - 2$ $y(\log_e 13) = -1$

Common Mistakes & Tips

• Always check the sign when separating variables.
• Remember the constant of integration when performing indefinite integrals.
• Use logarithm properties to simplify the general solution before applying the initial condition.

Summary

We solved the given differential equation using separation of variables. We integrated both sides, applied the initial condition to find the particular solution, and then evaluated the solution at $x = \log_e 13$. The final value of $y(\log_e 13)$ is -1.

Final Answer

The final answer is \boxed{-1}, which corresponds to option (A).