\(\displaystyle{4}{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{4}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{y}={0}\)

Our task is to write our second order differential equation as a system of two first order differential equations. Let's introduce a substitution \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}\) and we get:

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}\)

\(\displaystyle{4}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}+{4}{v}+{y}={0}\)

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}\)

\(\displaystyle{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}=-{\frac{{{1}}}{{{4}}}}{y}-{v}\)

We got system of first order differential equation which solution is equivalent to solution our second order differential equation.

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={v}\)

\(\displaystyle{F}{\left({x},{v},{y}\right)}=-{\frac{{{1}}}{{{4}}}}{y}-{v}\)