To solve the differential equation given by
\[ y = 2y' x + \frac{1}{y} \]
let's rewrite \( y' \) as \( \frac{dy}{dx} \). This gives us:
\[ y = 2 \frac{dy}{dx} x + \frac{1}{y}. \]
Now, multiply both sides by \( y \) to eliminate the fraction:
\[ y^2 = 2y' xy + 1. \]
Next, rearranging the terms gives:
\[ 2y' xy = y^2 - 1. \]
Now solve for \( y' \):
\[ y' = \frac{y^2 - 1}{2xy}. \]
This is a separable differential equation. We can separate the variables \( y \) and \( x \):
\[ \frac{2xy}{y^2 - 1} , dy = dx. \]
Next, let's integrate both sides. To do this, we first simplify the left side:
\[ \int \frac{2xy}{y^2 - 1} , dy = \int dx. \]
To integrate the left-hand side, we can use substitution. Let's set:
\[ u = y^2 - 1 \quad \Rightarrow \quad du = 2y , dy. \]
Thus, we have:
\[ \frac{2xy}{y^2 - 1} , dy = x \cdot du, \]
which transforms our integral into:
\[ \int x , du = \int dx. \]
Integrating both sides:
\[ xu = x(y^2 - 1) = x \cdot y^2 - x. \]
Thus we have:
\[ \int x , du = \frac{1}{2}x^2 + C, \]
where \( C \) is the constant of integration.
We need to express our solution in terms of \( y \), so we rewrite:
\[ x(y^2 - 1) = \frac{1}{2}x^2 + C. \]
Finally, we can solve for \( y^2 \):
\[ y^2 - 1 = \frac{\frac{1}{2}x^2 + C}{x}. \]
Simplifying gives:
\[ y^2 = \frac{1}{2}x + 1 + \frac{C}{x}. \]
Thus, our solution for \( y \) can be expressed as:
\[ y = \sqrt{\frac{1}{2}x + 1 + \frac{C}{x}}. \]
This solution can be verified by differentiating and substituting back into the original differential equation.
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