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Question Number 85116 by niroj last updated on 19/Mar/20
$$\:\mathrm{Reduce}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{to}\:\mathrm{Clairaut}'\mathrm{s}\:\mathrm{form} \\ $$$$\:\mathrm{and}\:\mathrm{find}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:: \\ $$$$\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \boldsymbol{\mathrm{p}}^{\mathrm{2}} +\boldsymbol{\mathrm{yp}}\left(\mathrm{2}\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}\right)+\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\mathrm{0}\:\:\:\:\:\:\left({put}\:\boldsymbol{{y}}=\boldsymbol{{u}}\:{and}\:\boldsymbol{{xy}}=\boldsymbol{{v}}\right) \\ $$$$\: \\ $$
Commented by MJS last updated on 19/Mar/20
$$\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equation} \\ $$$$\mathrm{with}\:\mathrm{3}\:\mathrm{unknowns}? \\ $$$${p}=−\frac{\left(\mathrm{2}{x}+{y}\right){y}}{\mathrm{2}{x}^{\mathrm{2}} }\pm\frac{\sqrt{\left(\mathrm{4}{x}+{y}\right){y}^{\mathrm{3}} }}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${x}={y}\left(−\frac{\mathrm{1}}{{p}}\pm\frac{\mathrm{1}}{\sqrt{−{p}}}\right) \\ $$$${y}={x}\left(−\frac{{p}}{{p}+\mathrm{1}}\pm\frac{\sqrt{−{p}^{\mathrm{3}} }}{{p}+\mathrm{1}}\right) \\ $$