Key Concepts and Formulas

• Exact Differential Form: Recognizing and utilizing exact differential forms like $d(xy) = x\,dy + y\,dx$ and $d(y/x) = \frac{x\,dy - y\,dx}{x^2}$ is crucial for solving these types of differential equations.
• Integration of $\frac{1}{u}$: $\int \frac{du}{u} = \ln|u| + C$, where $C$ is the constant of integration.
• Properties of Logarithms: $\ln(a) + \ln(b) = \ln(ab)$ and $e^{\ln(x)} = x$.

Step-by-Step Solution

Step 1: Transforming the Differential Equation

We are given the differential equation: $x \, dy - y \, dx + xy(x \, dy + y \, dx) = 0$ Our goal is to manipulate this equation into a form that can be easily integrated by recognizing exact differentials.

1. Recognizing Exact Differentials: Notice that $x \, dy + y \, dx = d(xy)$. We can substitute this directly into the equation: $x \, dy - y \, dx + xy \, d(xy) = 0$

2. Dividing by $y^2$: The term $x \, dy - y \, dx$ resembles the numerator of $d(x/y) = \frac{y \, dx - x \, dy}{y^2} = -\frac{x \, dy - y \, dx}{y^2}$. Dividing the entire equation by $y^2$ (assuming $y \neq 0$, which is valid due to the initial condition $y(1)=2$) gives: $\frac{x \, dy - y \, dx}{y^2} + \frac{xy \, d(xy)}{y^2} = 0$ $-\frac{y \, dx - x \, dy}{y^2} + \frac{x}{y} \, d(xy) = 0$ $-d\left(\frac{x}{y}\right) + \frac{x}{y} \, d(xy) = 0$ This step is crucial because it introduces the term $\frac{x}{y}$ in both parts of the equation.

Step 2: Separating Variables and Integrating

Now we have the simplified equation: $-d\left(\frac{x}{y}\right) + \frac{x}{y} \, d(xy) = 0$

1. Separating Variables: Rearrange the equation to separate the variables: $\frac{x}{y} \, d(xy) = d\left(\frac{x}{y}\right)$ Divide both sides by $\frac{x}{y}$ (assuming $x \neq 0$, which is valid due to the initial condition $x=1$): $d(xy) = \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}}$

2. Integrating Both Sides: Integrate both sides of the equation: $\int d(xy) = \int \frac{d\left(\frac{x}{y}\right)}{\frac{x}{y}}$ $xy = \ln\left|\frac{x}{y}\right| + C$ where $C$ is the constant of integration.

Step 3: Applying the Initial Condition

We are given the initial condition $y(1) = 2$. Substitute $x=1$ and $y=2$ into the general solution: $(1)(2) = \ln\left|\frac{1}{2}\right| + C$ $2 = \ln\left(\frac{1}{2}\right) + C$ $C = 2 - \ln\left(\frac{1}{2}\right) = 2 + \ln(2)$

Step 4: Substituting the Constant and Simplifying

Substitute the value of $C$ back into the general solution: $xy = \ln\left|\frac{x}{y}\right| + 2 + \ln(2)$ $xy = \ln\left|\frac{x}{y}\right| + \ln(e^2) + \ln(2)$ $xy = \ln\left(2e^2\left|\frac{x}{y}\right|\right)$

Step 5: Manipulating to Match the Given Form

Our goal is to get the equation in the form $\alpha|x| = |y|e^{xy - \beta}$.

1. Exponentiating Both Sides: Exponentiate both sides with base $e$: $e^{xy} = 2e^2\left|\frac{x}{y}\right|$ $e^{xy} = 2e^2 \frac{|x|}{|y|}$

2. Isolating $|y|$: Multiply both sides by $|y|$: $|y|e^{xy} = 2e^2|x|$

3. Rearranging to the Target Form: Divide both sides by $e^2$: $|y|e^{xy-2} = 2|x|$

Step 6: Identifying Parameters and Final Calculation

Comparing our derived solution $2|x| = |y|e^{xy-2}$ with the given form $\alpha|x| = |y|e^{xy-\beta}$, we can identify:

• $\alpha = 2$
• $\beta = 2$

Therefore, $\alpha + \beta = 2 + 2 = 4$.

Common Mistakes & Tips

• Incorrectly Applying Initial Conditions: Make sure to substitute the initial conditions correctly after integration to find the constant of integration.
• Sign Errors: Be careful with signs when manipulating the differential equation, especially when dealing with the quotient rule for differentials.
• Not Recognizing Exact Differentials: Practice recognizing common exact differentials to simplify the equation efficiently.

Summary

We solved the given differential equation by first recognizing exact differential forms and manipulating the equation to separate variables. After integrating and applying the initial condition, we obtained a solution that was then rearranged into the desired form $\alpha|x|=|y|e^{xy-\beta}$. By comparing the coefficients, we found $\alpha=2$ and $\beta=2$, resulting in $\alpha+\beta=4$.

The final answer is \boxed{4}.