Question Number 91812 by niroj last updated on 03/May/20
$$\:\mathrm{Solve}\:\mathrm{clairaut}'\mathrm{s}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{find} \\ $$$$\:\mathrm{general}\:\mathrm{and}\:\mathrm{singular}\:\mathrm{solution}: \\ $$$$\left(\mathrm{i}\right)\:\mathrm{y}=\mathrm{px}+\mathrm{p}^{\mathrm{n}} \\ $$$$\:\left(\mathrm{ii}\right)\:\left(\mathrm{y}+\mathrm{1}\right)\mathrm{p}−\mathrm{xp}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$
Commented by john santu last updated on 03/May/20
$$\left(\mathrm{1}\right){put}\:{xy}={Y}\:;\:{x}={X} \\ $$$${x}\frac{{dy}}{{dx}}+{y}\:=\:\frac{{dY}}{{dx}}\:=\:{P}\:;\:{dx}\:=\:{dX} \\ $$$${y}+{px}\:=\:{P}\:,\:{where}\:{p}=\frac{{dy}}{{dx}} \\ $$$$\Rightarrow{y}={px}+{p}^{{n}} \\ $$$${y}=\:{P}−{y}−{p}^{{n}} \Rightarrow\mathrm{2}{y}={P}−{p}^{{n}} \\ $$$$\frac{{Y}}{{x}}={P}−\left(\frac{{P}−{y}}{{x}}\right)^{{n}} \\ $$$${Y}={Px}−\frac{\left({P}−\frac{{Y}}{{x}}\right)^{{n}} }{{x}^{{n}−\mathrm{1}} } \\ $$$${continue}… \\ $$
Commented by john santu last updated on 03/May/20
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