Calculation:

Convert \(\displaystyle{4}{y}\text{}{4}{y}'+{y}={0}\) into characteristics equation.

\(\displaystyle{4}{D}^{{{2}}}+{4}{D}+{1}={0}\)

\(\displaystyle{\left({2}{D}\right)}^{{{2}}}+{2}\cdot{2}{D}\cdot{1}+{1}^{{{2}}}={0}\)

\(\displaystyle{\left({2}{D}+{1}\right)}^{{{2}}}={0}\)

\(\displaystyle{D}=-{\frac{{{1}}}{{{2}}}}\)

Thus, the solution of differential equation \(\displaystyle{4}{y}\text{}{4}{y}+{y}={0}\ \text{is}\ {y}={A}{e}^{{-{\frac{{{x}}}{{{2}}}}}}+{B}{x}{e}^{{-{\frac{{{x}}}{{{2}}}}}}\)

Convert \(\displaystyle{4}{y}\text{}{4}{y}'+{y}={0}\) into characteristics equation.

\(\displaystyle{4}{D}^{{{2}}}+{4}{D}+{1}={0}\)

\(\displaystyle{\left({2}{D}\right)}^{{{2}}}+{2}\cdot{2}{D}\cdot{1}+{1}^{{{2}}}={0}\)

\(\displaystyle{\left({2}{D}+{1}\right)}^{{{2}}}={0}\)

\(\displaystyle{D}=-{\frac{{{1}}}{{{2}}}}\)

Thus, the solution of differential equation \(\displaystyle{4}{y}\text{}{4}{y}+{y}={0}\ \text{is}\ {y}={A}{e}^{{-{\frac{{{x}}}{{{2}}}}}}+{B}{x}{e}^{{-{\frac{{{x}}}{{{2}}}}}}\)