To solve the differential equation given by \( x \, dy - y \, dx = 2x^3y(x \, dy + y \, dx) \) with the initial condition \( y(1) = 1 \), we will follow these steps:
### Step 1: Rearranging the Equation
We start with the equation:
\[
x \, dy - y \, dx = 2x^3y(x \, dy + y \, dx)
\]
We can rearrange it to isolate the terms involving \( dy \) and \( dx \).
### Step 2: Dividing by \( x^2 \)
To simplify, we divide both sides by \( x^2 \):
\[
\frac{x \, dy - y \, dx}{x^2} = 2xy \left( \frac{x \, dy + y \, dx}{x^2} \right)
\]
This simplifies to:
\[
\frac{dy}{x} - \frac{y}{x^2} \, dx = 2y \left( \frac{dy}{x} + \frac{y}{x^2} \, dx \right)
\]
### Step 3: Recognizing the Form
The left-hand side can be recognized as the derivative of \( \frac{y}{x} \):
\[
d\left(\frac{y}{x}\right) = 2y \left( \frac{dy}{x} + \frac{y}{x^2} \, dx \right)
\]
This can be rewritten as:
\[
d\left(\frac{y}{x}\right) = 2y \, d\left(\frac{y}{x}\right)
\]
### Step 4: Integrating Both Sides
Integrating both sides gives us:
\[
\frac{y}{x} = 2 \cdot \frac{y^2}{2} + C
\]
This simplifies to:
\[
\frac{y}{x} = y^2 + C
\]
### Step 5: Applying the Initial Condition
Using the initial condition \( y(1) = 1 \):
\[
\frac{1}{1} = 1^2 + C \implies 1 = 1 + C \implies C = 0
\]
Thus, we have:
\[
\frac{y}{x} = y^2
\]
### Step 6: Solving for \( y \)
Rearranging gives us:
\[
y^2 - \frac{y}{x} = 0 \implies y(y - \frac{1}{x}) = 0
\]
This implies \( y = 0 \) or \( y = \frac{1}{x} \). Since \( y(1) = 1 \), we take:
\[
y = \frac{1}{x}
\]
### Step 7: Finding \( y(2) \) and \( y\left(\frac{1}{2}\right) \)
Now we can find:
\[
y(2) = \frac{1}{2}, \quad y\left(\frac{1}{2}\right) = \frac{1}{\frac{1}{2}} = 2
\]
### Step 8: Summing the Values
Finally, we compute:
\[
y(2) + y\left(\frac{1}{2}\right) = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}
\]
### Final Answer
Thus, the value of \( y(2) + y\left(\frac{1}{2}\right) \) is:
\[
\frac{5}{2}
\]
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