Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Convert the given differential equation to standard form.

The given differential equation is $(x - 1)\frac{dy}{dx} + 2xy = \frac{1}{x - 1}$. To get it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, we divide the entire equation by $(x - 1)$:

$\frac{dy}{dx} + \frac{2x}{x - 1}y = \frac{1}{(x - 1)^2}$

Now we can identify $P(x) = \frac{2x}{x - 1}$ and $Q(x) = \frac{1}{(x - 1)^2}$.

Step 2: Calculate the integrating factor (IF).

The integrating factor is given by $IF = e^{\int P(x) dx}$. We need to calculate $\int P(x) dx = \int \frac{2x}{x - 1} dx$. We can rewrite $\frac{2x}{x - 1}$ as $\frac{2(x - 1) + 2}{x - 1} = 2 + \frac{2}{x - 1}$.

Therefore, $\int \frac{2x}{x - 1} dx = \int (2 + \frac{2}{x - 1}) dx = 2x + 2\ln|x - 1| = 2x + \ln(x - 1)^2$ (since $x > 1$).

The integrating factor is then:

$IF = e^{2x + \ln(x - 1)^2} = e^{2x} \cdot e^{\ln(x - 1)^2} = (x - 1)^2 e^{2x}$

Step 3: Find the general solution.

The general solution is given by $y \cdot IF = \int (Q(x) \cdot IF) dx + C$. Substituting the values of $IF$ and $Q(x)$, we get:

$y(x - 1)^2 e^{2x} = \int \left(\frac{1}{(x - 1)^2} \cdot (x - 1)^2 e^{2x}\right) dx + C$

$y(x - 1)^2 e^{2x} = \int e^{2x} dx + C$

$y(x - 1)^2 e^{2x} = \frac{1}{2}e^{2x} + C$

$y = \frac{\frac{1}{2}e^{2x} + C}{(x - 1)^2 e^{2x}} = \frac{1}{2(x - 1)^2} + \frac{C}{(x - 1)^2 e^{2x}}$

Step 4: Use the initial condition to find the particular solution.

We are given that $y(2) = \frac{1 + e^4}{2e^4}$. Substituting $x = 2$ into the general solution, we get:

$\frac{1 + e^4}{2e^4} = \frac{1}{2(2 - 1)^2} + \frac{C}{(2 - 1)^2 e^{2(2)}}$

$\frac{1 + e^4}{2e^4} = \frac{1}{2} + \frac{C}{e^4}$

$\frac{1 + e^4}{2e^4} - \frac{1}{2} = \frac{C}{e^4}$

$\frac{1 + e^4 - e^4}{2e^4} = \frac{C}{e^4}$

$\frac{1}{2e^4} = \frac{C}{e^4}$

Therefore, $C = \frac{1}{2}$.

The particular solution is:

$y = \frac{1}{2(x - 1)^2} + \frac{1}{2(x - 1)^2 e^{2x}} = \frac{e^{2x} + 1}{2(x - 1)^2 e^{2x}}$

Step 5: Evaluate y(3).

We need to find $y(3)$. Substituting $x = 3$ into the particular solution, we get:

$y(3) = \frac{e^{2(3)} + 1}{2(3 - 1)^2 e^{2(3)}} = \frac{e^6 + 1}{2(2)^2 e^6} = \frac{e^6 + 1}{8e^6}$

Step 6: Compare with the given form and find alpha and beta.

We are given that $y(3) = \frac{e^\alpha + 1}{\beta e^\alpha}$. Comparing this with $y(3) = \frac{e^6 + 1}{8e^6}$, we can see that $\alpha = 6$ and $\beta = 8$.

Step 7: Calculate alpha + beta.

Finally, we calculate $\alpha + \beta = 6 + 8 = 14$.

Common Mistakes & Tips

• Careless Integration: Be careful when integrating $P(x)$ to find the integrating factor. A small mistake here can ruin the entire solution.
• Algebraic Errors: Pay close attention to algebraic manipulations, especially when simplifying the solution after finding the integrating factor.
• Don't Forget the Constant of Integration: Always remember to add the constant of integration ($C$) when performing indefinite integrals. Solve for C using the given initial condition.

Summary

We solved the given first-order linear differential equation by first converting it to standard form and then finding the integrating factor. Using the integrating factor, we obtained the general solution and then used the initial condition to find the particular solution. Finally, we evaluated the solution at $x=3$ and compared it with the given form to find $\alpha$ and $\beta$, and then calculated $\alpha + \beta$.

The final answer is \boxed{14}.