Let \(\displaystyle{u}={y}-\lambda,\ {v}={u}'={y}'\). then we can rewrite the differential equation as

\(\displaystyle{1}+{v}^{{2}}-{v}'{u}={0}\to{u}{\frac{{{d}{v}}}{{{\left.{d}{x}\right.}}}}={v}'{u}={1}+{v}^{{2}},\ {\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}={u}'={v}\)

which implies

\(\displaystyle{u}{\frac{{{d}{v}}}{{{d}{u}}}}={\frac{{{1}+{v}^{{2}}}}{{{v}}}}\to{\frac{{{2}{v}{d}{v}}}{{{1}+{v}^{{2}}}}}={\frac{{{2}{d}{u}}}{{{u}}}}\)

on integration, we get

\(\displaystyle{\left({1}+{v}^{{2}}\right)}={c}{u}^{{2}}\to{v}=\pm\sqrt{{{1}-{c}{u}^{{2}}}}\to{\frac{{{d}{u}}}{{\sqrt{{{1}-{c}{u}^{{2}}}}}}}=\pm{\left.{d}{x}\right.}\)

now the integration will depend on the sign of c: if \(\displaystyle{c}{>}{0}\), you get an inverse sine, \(\displaystyle{{\sin}^{{-{1}}}}\) and if \(\displaystyle{c}{<}{0}\), then \(\displaystyle{\ln{}}\)