To solve the differential equation
\[ y' = 2y'x + \frac{1}{y}, \]
we start by rewriting it in a standard form. First, let's express \(y'\) as a function of \(y\) and \(x\). We can rearrange the equation as follows:
\[ y' - 2yx' = \frac{1}{y}. \]
This is fairly complicated given the \(y'\) terms, so let's try a separation of variables approach.
Rearranging the differential equation gives us:
\[ y' (1 - 2x) = \frac{1}{y}. \]
Assuming that \(1 - 2x \neq 0\), we can separate variables:
\[ y' = \frac{1}{y(1-2x)}. \]
Now write \(y' = \frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{y(1 - 2x)}. \]
Now we can separate the variables:
\[ y , dy = \frac{1}{(1-2x)} , dx. \]
Now integrate both sides:
\[ \int y , dy = \int \frac{1}{1-2x} , dx. \]
The integrals are:
\[ \frac{y^2}{2} = -\frac{1}{2} \ln |1 - 2x| + C, \]
where \(C\) is a constant of integration.
Multiplying both sides by 2 gives us:
\[ y^2 = -\ln |1 - 2x| + D, \]
where \(D = 2C\).
So the general solution to the differential equation can be expressed as:
\[ y^2 = D - \ln |1 - 2x|. \]
This represents the relationship between \(y\) and \(x\) implicit in the original differential equation.
To express it in a more recognizable form, we can also write:
\[ y = \sqrt{D - \ln |1 - 2x|}. \]
This is the general solution to the given differential equation.
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