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Step-by-Step Solution
Step 1: Write down the given differential equation
We have the differential equation:
$$2x^2\,\frac{dy}{dx} = 2xy + y^2.$$
Step 2: Express the equation in a more convenient form
Divide both sides by $2x^2$:
$$\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}.$$
Next, divide both sides by $y^2$ (assuming $y\neq 0$):
$$\frac{1}{y^2}\,\frac{dy}{dx} = \frac{2xy + y^2}{2x^2\,y^2}
= \frac{2xy}{2x^2\,y^2} + \frac{y^2}{2x^2\,y^2}
= \frac{1}{xy} + \frac{1}{2x^2}.$$
Step 3: Introduce a substitution to simplify
Let
$$t = -\frac{1}{y}.$$
Then
$$\frac{dt}{dx} = \frac{1}{y^2}\,\frac{dy}{dx}.$$
So the left-hand side $\frac{1}{y^2}\,\frac{dy}{dx}$ is simply $\frac{dt}{dx}$. Hence, our equation becomes:
$$\frac{dt}{dx} = \frac{1}{xy} + \frac{1}{2x^2}.$$
Step 4: Rewrite the right-hand side in terms of t
Because $t = -\frac{1}{y}$, we have $\frac{1}{y} = -t$. Thus:
$$\frac{1}{xy} = \frac{1}{x} \left(\frac{1}{y}\right) = \frac{1}{x}(-t) = -\frac{t}{x}.$$
So the differential equation becomes:
$$\frac{dt}{dx} = -\frac{t}{x} + \frac{1}{2x^2}.$$
Step 5: Recognize this as a linear first-order differential equation
We can rewrite it as:
$$\frac{dt}{dx} + \frac{t}{x} = \frac{1}{2x^2}.$$
This is a standard linear differential equation in $t$.
Step 6: Find the integrating factor (I.F.)
The integrating factor is:
$$\text{I.F.} = e^{\int(1/x)\,dx} = e^{\ln x} = x.$$
Step 7: Multiply through by the integrating factor
Multiplying the entire differential equation by $x$ gives:
$$x \frac{dt}{dx} + t = x \left(\frac{1}{2x^2}\right) = \frac{1}{2x}.$$
Observe that the left side is the derivative of the product $x\,t$:
$$\frac{d}{dx}\bigl(x\,t\bigr) = \frac{1}{2x}.$$
Step 8: Integrate both sides
Integrate with respect to $x$:
$$x\,t = \int \frac{1}{2x}\,dx = \frac{1}{2}\ln x + C,$$
where $C$ is the constant of integration.
Step 9: Substitute back for t
Recall $t = -\frac{1}{y}$. Therefore,
$$x \left(-\frac{1}{y}\right) = \frac{1}{2}\ln x + C,$$
which simplifies to
$$-\frac{x}{y} = \frac{1}{2}\ln x + C.$$
Step 10: Apply the initial condition to find C
The curve passes through $(1,2)$, meaning $x=1$ and $y=2$ at that point. Substitute:
$$-\frac{1}{2} = \frac{1}{2}\ln(1) + C = 0 + C,$$
because $\ln(1)=0.$ Hence
$$C = -\frac{1}{2}.$$
Step 11: Write the particular solution
Substitute $C=-\frac{1}{2}$ into
$$-\frac{x}{y} = \frac{1}{2}\ln x + C,$$
giving
$$-\frac{x}{y} = \frac{1}{2}\ln x - \frac{1}{2}.$$
Multiply both sides by $-2$:
$$\frac{2x}{y} = -\ln x + 1.$$
Hence,
$$y = \frac{2x}{1 - \ln x}.$$
This function is $y=f(x)$.
Step 12: Find the required function value $f\bigl(\tfrac{1}{2}\bigr)$
Substitute $x=\tfrac{1}{2}$:
$$f\!\Bigl(\tfrac{1}{2}\Bigr)
= \frac{2\left(\tfrac{1}{2}\right)}{1 - \ln\!\bigl(\tfrac{1}{2}\bigr)}
= \frac{1}{1 - \left(-\ln 2\right)}
= \frac{1}{1 + \ln 2}.$$
Final Answer
$$f\!\Bigl(\tfrac{1}{2}\Bigr) = \frac{1}{1 + \ln 2}.$$
