On substituting

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

\(\displaystyle{y}^{{t}}={\left({x}+{y}+{1}\right)}^{{2}}-{\left({x}+{y}-{1}\right)}^{{2}}\)

\(\displaystyle{y}^{{t}}-{4}{y}={4}{x}\)

On finding the integration factor

\(\displaystyle\mu{\left({x}\right)}={e}^{{-{{4}}}}{x}\)

Thus on writing the equation in the form

\(\displaystyle{\left(\mu{\left({x}\right)}{y}\right)}'=\mu{\left({x}\right)}{q}{\left({x}\right)}\)

\(\displaystyle{\left({e}^{{-{4}}}{y}\right)}'={4}{x}{e}^{{-{4}}}\)

\(\displaystyle{y}=-{x}-{\frac{{{1}}}{{{4}}}}+{c}_{{1}}{e}^{{{4}{x}}}\)

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

\(\displaystyle{y}^{{t}}={\left({x}+{y}+{1}\right)}^{{2}}-{\left({x}+{y}-{1}\right)}^{{2}}\)

\(\displaystyle{y}^{{t}}-{4}{y}={4}{x}\)

On finding the integration factor

\(\displaystyle\mu{\left({x}\right)}={e}^{{-{{4}}}}{x}\)

Thus on writing the equation in the form

\(\displaystyle{\left(\mu{\left({x}\right)}{y}\right)}'=\mu{\left({x}\right)}{q}{\left({x}\right)}\)

\(\displaystyle{\left({e}^{{-{4}}}{y}\right)}'={4}{x}{e}^{{-{4}}}\)

\(\displaystyle{y}=-{x}-{\frac{{{1}}}{{{4}}}}+{c}_{{1}}{e}^{{{4}{x}}}\)