To solve the differential equation \( y = 2y'x + \frac{1}{y'} \), we start by rewriting it in a more standard form.
Let \( y' = \frac{dy}{dx} \). Then the equation becomes:
\[ y = 2\frac{dy}{dx} x + \frac{1}{\frac{dy}{dx}} \]
Let's denote \( p = \frac{dy}{dx} \). Substituting \( p \) into the equation gives:
\[ y = 2px + \frac{1}{p} \]
To eliminate \( p \), we can rearrange it. Multiply both sides by \( p \):
\[ yp = 2p^2x + 1 \]
Rearranging gives a quadratic in \( p \):
\[ 2p^2x - yp + 1 = 0 \]
To solve for \( p \) (i.e., \( \frac{dy}{dx} \)), we can apply the quadratic formula, where \( a = 2x \), \( b = -y \), and \( c = 1 \):
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the coefficients:
\[ p = \frac{y \pm \sqrt{(-y)^2 - 4 \cdot 2x \cdot 1}}{2 \cdot 2x} \]
This simplifies to:
\[ p = \frac{y \pm \sqrt{y^2 - 8x}}{4x} \]
This means:
\[ \frac{dy}{dx} = \frac{y \pm \sqrt{y^2 - 8x}}{4x} \]
This gives us a first-order differential equation, which is separable. However, the solution may depend on the choice of \( + \) or \( - \). Therefore, the simpler separated form requires some manipulation as:
- Choose either sign (let's consider the + sign first).
- Rewrite it as:
\[ dy = \left(\frac{y + \sqrt{y^2 - 8x}}{4x}\right) dx \]
At this point, further analytical solutions or numerical solutions can be pursued based on specific initial conditions or the context of the problem, as the separation leads towards more complex integration.
Would you like to proceed further with a specific target for the solution, such as a numerical example or particular initial conditions?