Question

Solve the following differential equation ;
y' = 2y'x + 1/y

Answer 1

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.