Key Concepts and Formulas

• Exact Differential: Recognizing and utilizing the differential of a product: $d(xy) = y\,dx + x\,dy$.
• Standard Integral: $\int \frac{du}{1-u^2} = \frac{1}{2} \log \left|\frac{1+u}{1-u}\right| + C$.
• Logarithm Properties: $a \log(x) = \log(x^a)$ and $\log(a) + \log(b) = \log(ab)$

Step-by-Step Solution

Step 1: Transforming the Differential Equation

We are given the differential equation: $(1-x^2y^2)dx = ydx + xdy$

Our goal is to rewrite the equation in a form that can be easily integrated. We recognize that $ydx + xdy = d(xy)$. Substituting this into the equation, we get: $(1-(xy)^2)dx = d(xy)$ Now, we separate variables: $dx = \frac{d(xy)}{1-(xy)^2}$

Step 2: Integrating Both Sides

Integrating both sides of the equation: $\int dx = \int \frac{d(xy)}{1-(xy)^2}$ Let $u = xy$, so $du = d(xy)$. The integral becomes: $\int dx = \int \frac{du}{1-u^2}$ Integrating both sides: $x = \frac{1}{2} \log \left|\frac{1+u}{1-u}\right| + C$ Substituting back $u = xy$: $x = \frac{1}{2} \log \left|\frac{1+xy}{1-xy}\right| + C$

Step 3: Applying the Initial Condition

We are given that when $x=1$, $y=2$. We use this information to find the constant of integration, $C$. Substituting $x=1$ and $y=2$ into the general solution: $1 = \frac{1}{2} \log \left|\frac{1+(1)(2)}{1-(1)(2)}\right| + C$ $1 = \frac{1}{2} \log \left|\frac{3}{-1}\right| + C$ $1 = \frac{1}{2} \log(3) + C$ Solving for $C$: $C = 1 - \frac{1}{2} \log(3)$

Step 4: Finding the Particular Solution and Solving for xy

Substitute the value of $C$ back into the general solution: $x = \frac{1}{2} \log \left|\frac{1+xy}{1-xy}\right| + 1 - \frac{1}{2} \log(3)$ Rearrange the equation: $x - 1 + \frac{1}{2} \log(3) = \frac{1}{2} \log \left|\frac{1+xy}{1-xy}\right|$ Multiply by 2: $2(x-1) + \log(3) = \log \left|\frac{1+xy}{1-xy}\right|$ Using logarithm properties: $\log(e^{2(x-1)}) + \log(3) = \log \left|\frac{1+xy}{1-xy}\right|$ $\log(3e^{2(x-1)}) = \log \left|\frac{1+xy}{1-xy}\right|$ Exponentiating both sides: $3e^{2(x-1)} = \left|\frac{1+xy}{1-xy}\right|$

When $x=1$ and $y=2$, $\frac{1+xy}{1-xy} = \frac{1+2}{1-2} = \frac{3}{-1} = -3 < 0$. Since the expression is negative at the initial condition, we have: $\frac{1+xy}{1-xy} = -3e^{2(x-1)}$ $1+xy = -3e^{2(x-1)}(1-xy)$ $1+xy = -3e^{2(x-1)} + 3e^{2(x-1)}xy$ $xy - 3e^{2(x-1)}xy = -3e^{2(x-1)} - 1$ $xy(1 - 3e^{2(x-1)}) = -3e^{2(x-1)} - 1$ $xy = \frac{-3e^{2(x-1)} - 1}{1 - 3e^{2(x-1)}} = \frac{3e^{2(x-1)} + 1}{3e^{2(x-1)} - 1}$

Step 5: Finding the Value of α

We are given that when $x=2$, $y=\alpha$. Substitute $x=2$ and $y=\alpha$ into the equation for $xy$: $2\alpha = \frac{3e^{2(2-1)} + 1}{3e^{2(2-1)} - 1} = \frac{3e^2 + 1}{3e^2 - 1}$ $\alpha = \frac{3e^2 + 1}{2(3e^2 - 1)} = \frac{1+3e^2}{2(3e^2-1)}$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when manipulating the differential equation and when dealing with the absolute value in the logarithm.
• Constant of Integration: Don't forget to include the constant of integration, $C$, after integrating. It is crucial for finding the particular solution.
• Logarithm Properties: Remember the properties of logarithms, as they are essential for simplifying the solution.

Summary

We solved the given differential equation by recognizing the exact differential $d(xy)$, separating variables, integrating, using the initial condition to find the constant of integration, and finally solving for $y$ in terms of $x$. We then substituted $x=2$ to find the value of $\alpha$.

The final answer is \boxed{\frac{1+3 e^{2}}{2\left(3 e^{2}-1\right)}}, which corresponds to option (A).