Question

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

Answer 1

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.