To solve the differential equation
\[ x(y'^2) + 2xy' - y = 0, \]
let's denote \( y' = \frac{dy}{dx} \).
Rearranging the equation, we have:
\[ x(y'^2) + 2xy' - y = 0 \implies x(y'^2) + 2xy' = y. \]
Dividing everything by \( x \) (assuming \( x \neq 0 \)), we get:
\[ y'^2 + 2y' = \frac{y}{x}. \]
Now let's rewrite this in a standard form:
\[ y'^2 + 2y' - \frac{y}{x} = 0. \]
This is a quadratic equation in \( y' \). We can solve it using the quadratic formula:
\[ y' = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 1 \), \( b = 2 \), and \( c = -\frac{y}{x} \). Plugging these values in:
\[ y' = \frac{-2 \pm \sqrt{2^2 - 4(1)(-\frac{y}{x})}}{2(1)} = \frac{-2 \pm \sqrt{4 + \frac{4y}{x}}}{2}. \]
Simplifying the expression:
\[ y' = -1 \pm \sqrt{1 + \frac{y}{x}}. \]
Now we have two cases based on the \( \pm \) sign:
Case 1: \( y' = -1 + \sqrt{1 + \frac{y}{x}} \)
This implies:
\[ \frac{dy}{dx} = -1 + \sqrt{1 + \frac{y}{x}}. \]
Rearranging gives us:
\[ \frac{dy}{dx} + 1 = \sqrt{1 + \frac{y}{x}}. \]
Case 2: \( y' = -1 - \sqrt{1 + \frac{y}{x}} \)
This implies:
\[ \frac{dy}{dx} = -1 - \sqrt{1 + \frac{y}{x}}. \]
Rearranging gives us:
\[ \frac{dy}{dx} + 1 = -\sqrt{1 + \frac{y}{x}}. \]
Both cases lead us to different relationships between \( y \) and \( x \). However, we can start solving either equation in standard techniques of separable or exact forms.
Since the equations can be complex, detailed integration would require transforming back and then finding an implicit solution since no straightforward separable or exact form arises directly from the generated differential equations.
Let’s proceed with the Case 1:
\[ \frac{dy}{dx} + 1 = \sqrt{1 + \frac{y}{x}}. \]
Now we will square both sides (careful of possible extraneous solutions):
\[ \left(\frac{dy}{dx} + 1\right)^2 = 1 + \frac{y}{x}. \]
Expanding the left side:
\[ \left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx} + 1 = 1 + \frac{y}{x}. \]
Cancelling and rearranging gives:
\[ \left(\frac{dy}{dx}\right)^2 + 2\frac{dy}{dx} = \frac{y}{x}. \]
Continuing from here would require specific methods under initial conditions or further information.
To summarize: You can solve explicitly based on the quadratic relationship, but you might use numerical simulation or graphical methodologies if numerical solutions are required. Further analysis of the situation could lead to solving either implicitly or through the substitution for more complex functions depending on actual conditions or applied constraints (initial conditions).
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