1. Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x)g(y)$ can be separated into $h(y)dy = k(x)dx$, where $h(y) = \frac{1}{g(y)}$ and $k(x) = f(x)$.
• Integration: The process of finding the antiderivative of a function.
• Initial Conditions: Values of the dependent variable ($y$) at specific values of the independent variable ($x$), used to determine the constant of integration.
• Exponential Properties: $e^{a+b} = e^a \cdot e^b$, $e^{\log_e x} = x$, $a \log_b c = \log_b c^a$, $e^{-\log_e x} = \frac{1}{x}$.

2. Step-by-Step Solution

Step 1: Separate the Variables

We are given the differential equation: $dy = e^{\alpha x + y} dx$ Our goal is to rewrite this equation in the form $h(y)dy = k(x)dx$.

First, we rewrite the equation as: $\frac{dy}{dx} = e^{\alpha x + y}$ Using the property of exponents $e^{a+b} = e^a \cdot e^b$, we get: $\frac{dy}{dx} = e^{\alpha x} \cdot e^y$ Now, we separate the variables by dividing both sides by $e^y$ and multiplying both sides by $dx$: $\frac{dy}{e^y} = e^{\alpha x} dx$ Using the property $\frac{1}{e^y} = e^{-y}$, we obtain the separated form: $e^{-y} dy = e^{\alpha x} dx$ Explanation: We manipulate the differential equation to isolate terms involving $y$ and $dy$ on one side and terms involving $x$ and $dx$ on the other side. This allows us to integrate each side independently.

Step 2: Integrate Both Sides

Now that the variables are separated, we integrate both sides of the equation: $\int e^{-y} dy = \int e^{\alpha x} dx$ Integrating each side, we get:

• Left side: $\int e^{-y} dy = -e^{-y}$

• Right side: Since $\alpha \in N$, $\alpha$ is a natural number, and therefore a constant. $\int e^{\alpha x} dx = \frac{e^{\alpha x}}{\alpha} + C'$

Combining the results, we get: $-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C'$ We can rewrite this as: $-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C \quad \text{.....(1)}$ where $C$ is the constant of integration. Explanation: Integration is the reverse of differentiation. We add the constant of integration $C$ because the derivative of a constant is zero, so any constant could be added to the antiderivative.

Step 3: Apply the Initial Conditions

We are given two initial conditions:

1. $y(\log_e 2) = \log_e 2$
2. $y(0) = \log_e \left( \frac{1}{2} \right)$

We will use these conditions to find the values of $C$ and $\alpha$.

Using the first initial condition: $x = \log_e 2$ and $y = \log_e 2$. Substitute these values into equation (1): $-e^{-\log_e 2} = \frac{e^{\alpha \log_e 2}}{\alpha} + C$ Using the properties of logarithms and exponentials, we simplify:

• $e^{-\log_e 2} = e^{\log_e (2^{-1})} = \frac{1}{2}$
• $e^{\alpha \log_e 2} = e^{\log_e (2^\alpha)} = 2^\alpha$ Substituting these simplified terms back into the equation: $-\frac{1}{2} = \frac{2^\alpha}{\alpha} + C \quad \text{.....(2)}$

Using the second initial condition: $x = 0$ and $y = \log_e \left( \frac{1}{2} \right) = -\log_e 2$. Substitute these values into equation (1): $-e^{-(-\log_e 2)} = \frac{e^{\alpha \cdot 0}}{\alpha} + C$ Simplifying:

• $e^{-(-\log_e 2)} = e^{\log_e 2} = 2$
• $e^{\alpha \cdot 0} = e^0 = 1$ Substituting these simplified terms back into the equation: $-2 = \frac{1}{\alpha} + C \quad \text{.....(3)}$ Explanation: Each initial condition provides a specific point $(x, y)$ that the solution must satisfy. Substituting these points into the general solution gives us equations with $C$ and $\alpha$.

Step 4: Solve for $\alpha$

We have the following system of equations:

1. $-\frac{1}{2} = \frac{2^\alpha}{\alpha} + C$
2. $-2 = \frac{1}{\alpha} + C$

Subtract equation (3) from equation (2) to eliminate $C$: $\left(-\frac{1}{2}\right) - (-2) = \left(\frac{2^\alpha}{\alpha} + C\right) - \left(\frac{1}{\alpha} + C\right)$ $\frac{3}{2} = \frac{2^\alpha}{\alpha} - \frac{1}{\alpha}$ $\frac{3}{2} = \frac{2^\alpha - 1}{\alpha}$ Multiply both sides by $2\alpha$: $3\alpha = 2(2^\alpha - 1)$ $3\alpha = 2^{\alpha+1} - 2$ Now we test integer values for $\alpha$ since $\alpha \in N$:

• If $\alpha = 1$: $3(1) = 3$ and $2^{1+1} - 2 = 2^2 - 2 = 4 - 2 = 2$. Since $3 \neq 2$, $\alpha \neq 1$.

• If $\alpha = 2$: $3(2) = 6$ and $2^{2+1} - 2 = 2^3 - 2 = 8 - 2 = 6$. Since $6 = 6$, $\alpha = 2$.

• If $\alpha = 3$: $3(3) = 9$ and $2^{3+1} - 2 = 2^4 - 2 = 16 - 2 = 14$. Since $9 \neq 14$, $\alpha \neq 3$.

Since the exponential term grows faster than the linear term, $\alpha = 2$ is the only solution. Explanation: We solve for $\alpha$ by substituting possible natural number values until we find a solution that satisfies the equation.

3. Common Mistakes & Tips

• Forgetting the constant of integration: Always include the constant of integration, $C$, when performing indefinite integrals.
• Sign errors: Be careful with negative signs, especially when integrating $e^{-y}$.
• Algebraic manipulation: Double-check your algebraic manipulations when separating variables and solving for constants.

4. Summary

We solved the given differential equation by first separating the variables, then integrating both sides. We then used the given initial conditions to find the value of the constant of integration and the parameter $\alpha$. By substituting natural number values for $\alpha$ into the resulting equation, we found that $\alpha = 2$.

5. Final Answer

The final answer is $\boxed{2}$.