Key Concepts and Formulas

• First-Order Linear Differential Equation (LDE): A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$.
• Integrating Factor (I.F.): For a linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$, the integrating factor is given by I.F. $= e^{\int P(x) dx}$. The solution is then given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$.
• Substitution: Transforming a differential equation into a simpler form by replacing a function of the dependent variable with a new variable. The chain rule is key here: $\frac{d}{dx}f(y) = f'(y)\frac{dy}{dx}$

Step-by-Step Solution

Step 1: Rearrange the Given Differential Equation

We begin by rearranging the given differential equation to a more convenient form. The goal is to isolate $\frac{dy}{dx}$ and look for opportunities for substitution.

• The given equation is: $2 x^{2} y \mathrm{~d} y-\left(1-x y^{2}\right) \mathrm{d} x=0$
• Move the $dx$ term to the right-hand side: $2 x^{2} y \mathrm{~d} y = \left(1-x y^{2}\right) \mathrm{d} x$
• Divide both sides by $dx$: $2 x^{2} y \frac{\mathrm{d} y}{\mathrm{~d} x} = 1-x y^{2}$
• Divide both sides by $x^2$ (since $x > 0$): $2 y \frac{\mathrm{d} y}{\mathrm{~d} x} = \frac{1}{x^2} - \frac{y^{2}}{x}$
• Rearrange to group the terms containing $y$: $2 y \frac{\mathrm{d} y}{\mathrm{~d} x} + \frac{y^{2}}{x} = \frac{1}{x^2}$ Why this step? This arrangement highlights the term $2y \frac{dy}{dx}$, which is the derivative of $y^2$. This strongly suggests a substitution $t = y^2$.

Step 2: Apply the Substitution

We introduce a substitution to transform the equation into a linear differential equation.

• Let $t = y^2$. Then, differentiating with respect to $x$ gives: $\frac{\mathrm{d} t}{\mathrm{~d} x} = 2y \frac{\mathrm{d} y}{\mathrm{~d} x}$ Why this substitution? This substitution simplifies the equation by replacing $2y\frac{dy}{dx}$ with $\frac{dt}{dx}$, making it linear.

Step 3: Transform the Equation

Substitute $t = y^2$ and $\frac{dt}{dx} = 2y \frac{dy}{dx}$ into the rearranged equation from Step 1:

• Substituting into $2 y \frac{\mathrm{d} y}{\mathrm{~d} x} + \frac{y^{2}}{x} = \frac{1}{x^2}$ yields: $\frac{\mathrm{d} t}{\mathrm{~d} x} + \frac{t}{x} = \frac{1}{x^2}$ This is a first-order linear differential equation in the form $\frac{dt}{dx} + P(x)t = Q(x)$, where $P(x) = \frac{1}{x}$ and $Q(x) = \frac{1}{x^2}$.

Step 4: Calculate the Integrating Factor

We calculate the integrating factor for the linear differential equation.

• The integrating factor is given by: $\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx}$
• Since $x > 0$, $\int \frac{1}{x} \, dx = \ln x$. Therefore, $\text{I.F.} = e^{\ln x} = x$ Why this step? The integrating factor transforms the left-hand side of the equation into an exact derivative.

Step 5: Find the General Solution

We find the general solution of the linear differential equation.

• The general solution is given by: $t \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C$
• Substituting $t$, $\text{I.F.} = x$, and $Q(x) = \frac{1}{x^2}$: $t x = \int \frac{1}{x^2} \cdot x \, dx + C$ $t x = \int \frac{1}{x} \, dx + C$
• Since $x > 0$, $\int \frac{1}{x} \, dx = \ln x$. Thus, $t x = \ln x + C$
• Substitute back $t = y^2$: $y^2 x = \ln x + C$

Step 6: Apply the Initial Condition

We use the initial condition $y(2) = \sqrt{\ln 2}$ to find the particular solution.

• Substitute $x = 2$ and $y = \sqrt{\ln 2}$ into the general solution: $(\sqrt{\ln 2})^2 \cdot 2 = \ln 2 + C$ $2 \ln 2 = \ln 2 + C$
• Solve for $C$: $C = 2 \ln 2 - \ln 2 = \ln 2$
• Substitute $C = \ln 2$ back into the general solution: $y^2 x = \ln x + \ln 2$ $y^2 x = \ln(2x)$

Step 7: Match the Given Form

We rewrite the particular solution to match the form $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$.

• We have $y^2 x = \ln(2x)$. Exponentiating both sides: $e^{y^2 x} = e^{\ln(2x)}$ $e^{y^2 x} = 2x$ $2x = \exp(y^2 x)$
• Rewrite $y^2 x$ as $x^1 y^2$: $2x = \exp(x^1 y^2)$

Step 8: Determine $\alpha, \beta, \gamma$

We compare $2x = \exp(x^1 y^2)$ with $\alpha x=\exp \left(x^{\beta} y^{\gamma}\right)$ to find the values of $\alpha, \beta,$ and $\gamma$.

• By comparison, we have: $\alpha = 2$ $\beta = 1$ $\gamma = 2$

Step 9: Calculate $\alpha + \beta - \gamma$

Finally, we compute the value of $\alpha + \beta - \gamma$.

• $\alpha + \beta - \gamma = 2 + 1 - 2 = 1$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs during rearrangement and substitution.
• Logarithm Properties: Remember to use logarithm properties correctly, especially when combining or separating logarithmic terms. Also, remember that $\int \frac{1}{x} dx = \ln|x| + C$, but since $x > 0$, we can write $\ln x + C$.
• Integrating Factor: Double-check the calculation of the integrating factor and its application in solving the linear differential equation.

Summary

We solved the given differential equation by rearranging it, applying a suitable substitution ($t = y^2$), solving the resulting linear differential equation using an integrating factor, applying the initial condition to find the particular solution, and then matching the solution to the given form to determine the values of $\alpha, \beta,$ and $\gamma$. Finally, we calculated $\alpha + \beta - \gamma = 1$.

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