Question Number 161241 by help last updated on 14/Dec/21
Answered by mr W last updated on 15/Dec/21
$${let}\:{y}={xt} \\ $$$${y}'={t}+{x}\frac{{dt}}{{dx}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}} \\ $$$${x}\frac{{dt}}{{dx}}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}}−{t}=−\frac{\mathrm{1}}{{t}} \\ $$$${tdt}=−\frac{{dx}}{{x}} \\ $$$$\frac{{t}^{\mathrm{2}} }{\mathrm{2}}=−\mathrm{ln}\:{Cx} \\ $$$$\frac{{y}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }=−\mathrm{ln}\:{Cx} \\ $$$$\Rightarrow{y}^{\mathrm{2}} =−\mathrm{2}{x}^{\mathrm{2}} \mathrm{ln}\:{Cx} \\ $$
Answered by yeti123 last updated on 15/Dec/21
$$\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} }{{xy}} \\ $$$$\frac{{dy}}{{dx}}\:−{x}^{−\mathrm{1}} {y}\:=\:−{xy}^{−\mathrm{1}} \:….\left(\mathrm{1}\right) \\ $$$$\mathrm{let}\:{z}\:=\:{y}^{\mathrm{2}} \:\Rightarrow\:\frac{{dz}}{{dx}}\:=\:\mathrm{2}{y}\frac{{dy}}{{dx}} \\ $$$$\left(\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}\:−\:\mathrm{2}{y}^{\mathrm{2}} {x}^{−\mathrm{1}} \:=\:−\mathrm{2}{x} \\ $$$$\frac{{dz}}{{dx}}\:−\:\mathrm{2}{zx}^{−\mathrm{1}} \:=\:−\mathrm{2}{x} \\ $$$${x}^{−\mathrm{2}} \frac{{dz}}{{dx}}\:−\:\mathrm{2}{zx}^{−\mathrm{3}} \:=\:−\mathrm{2}{x}^{−\mathrm{1}} \\ $$$$\frac{{d}}{{dx}}\left({zx}^{−\mathrm{2}} \right)\:=\:−\mathrm{2}{x}^{−\mathrm{1}} \\ $$$${zx}^{−\mathrm{2}} \:=\:−\mathrm{2ln}\left({x}\right)\:+\:{c} \\ $$$${z}\:=\:{x}^{\mathrm{2}} \left\{−\mathrm{2ln}\left({x}\right)\:+\:{c}\right\} \\ $$$${y}^{\mathrm{2}} \:=\:{x}^{\mathrm{2}} \left\{−\mathrm{2ln}\left({x}\right)\:+\:{c}\right\} \\ $$