Key Concepts and Formulas

• Bernoulli's Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)y^n$, where $n$ is a real number (except 0 or 1).
• Transformation for Bernoulli's Equation: Substitute $v = y^{1-n}$ to transform Bernoulli's equation into a linear differential equation.
• Integrating Factor (I.F.) for Linear Differential Equation: For $\frac{dv}{dx} + P'(x)v = Q'(x)$, the I.F. is $e^{\int P'(x)dx}$, and the solution is $v \cdot (\text{I.F.}) = \int Q'(x) \cdot (\text{I.F.}) dx + C$.

Step-by-Step Solution

Step 1: Identify and Rewrite the Differential Equation

We are given the differential equation $2x^2 dy = (2xy + y^2)dx$. Our first goal is to express this in the standard $\frac{dy}{dx}$ form.

Divide both sides by $2x^2 dx$: $\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}$

Simplify the right-hand side: $\frac{dy}{dx} = \frac{y}{x} + \frac{1}{2}\left(\frac{y}{x}\right)^2$

Rearrange to resemble Bernoulli's form: $\frac{dy}{dx} - \frac{1}{x}y = \frac{1}{2x^2}y^2$

Why: Rewriting the equation in this form allows us to recognize it as a Bernoulli differential equation. This identification is crucial for applying the correct solution method.

Step 2: Transform the Bernoulli Equation into a Linear Equation

The equation is now in the Bernoulli form $\frac{dy}{dx} + P(x)y = Q(x)y^n$ with $n=2$. We use the substitution $v = y^{1-n} = y^{-1} = \frac{1}{y}$.

Differentiate $v = \frac{1}{y}$ with respect to $x$: $\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$

Now, divide the original equation by $y^2$: $\frac{1}{y^2}\frac{dy}{dx} - \frac{1}{x}\frac{1}{y} = \frac{1}{2x^2}$

Substitute $v = \frac{1}{y}$ and $\frac{dv}{dx} = -\frac{1}{y^2}\frac{dy}{dx}$: $-\frac{dv}{dx} - \frac{1}{x}v = \frac{1}{2x^2}$

Multiply by -1: $\frac{dv}{dx} + \frac{1}{x}v = -\frac{1}{2x^2}$

Why: This transformation converts the non-linear Bernoulli equation into a linear first-order differential equation, which can be solved using an integrating factor.

Step 3: Solve the Linear Differential Equation

The equation is now in the form $\frac{dv}{dx} + P'(x)v = Q'(x)$, where $P'(x) = \frac{1}{x}$ and $Q'(x) = -\frac{1}{2x^2}$.

Calculate the integrating factor: $\text{I.F.} = e^{\int P'(x)dx} = e^{\int \frac{1}{x}dx} = e^{\ln x} = x$

Multiply the equation by the integrating factor: $x\frac{dv}{dx} + \frac{1}{x}xv = -\frac{1}{2x^2}x$ $x\frac{dv}{dx} + v = -\frac{1}{2x}$

Integrate both sides with respect to $x$: $\int \left(x\frac{dv}{dx} + v\right) dx = \int -\frac{1}{2x} dx$ $vx = -\frac{1}{2}\ln x + C$

Why: Multiplying by the integrating factor allows us to write the left-hand side as the derivative of a product, making integration straightforward.

Step 4: Substitute Back and Apply the Initial Condition

Substitute $v = \frac{1}{y}$: $\frac{x}{y} = -\frac{1}{2}\ln x + C$

Apply the initial condition $(1, 2)$: $\frac{1}{2} = -\frac{1}{2}\ln 1 + C$ Since $\ln 1 = 0$: $\frac{1}{2} = C$

Substitute $C = \frac{1}{2}$ back into the equation: $\frac{x}{y} = -\frac{1}{2}\ln x + \frac{1}{2}$ $\frac{x}{y} = \frac{1}{2}(1 - \ln x)$ $y = \frac{2x}{1 - \ln x}$

Why: Applying the initial condition allows us to find the particular solution that satisfies the given condition.

Step 5: Evaluate f(1/2)

Now we need to find $f\left(\frac{1}{2}\right)$: $f\left(\frac{1}{2}\right) = \frac{2\left(\frac{1}{2}\right)}{1 - \ln\left(\frac{1}{2}\right)} = \frac{1}{1 - \ln(2^{-1})} = \frac{1}{1 + \ln 2}$

Why: This is the final step to answer the question by finding the value of the function at the specified point.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs, especially when differentiating and substituting.
• Forgetting the Constant of Integration: Always include the constant of integration ($C$) after performing an indefinite integral.
• Incorrectly Identifying Bernoulli's Equation: Ensure the equation is truly in the correct form before applying the Bernoulli transformation.

Summary

We solved the given differential equation by recognizing it as a Bernoulli equation, transforming it into a linear first-order differential equation, solving the linear equation using an integrating factor, and applying the initial condition to find the particular solution. Finally, we evaluated the function at $x = \frac{1}{2}$ to obtain the answer.

Final Answer

The final answer is $\boxed{\frac{1}{1 + \ln_e 2}}$, which corresponds to option (B).