Question Number 227249 by Spillover last updated on 10/Jan/26
Answered by breniam last updated on 10/Jan/26
$$\frac{{x}−{y}\left({x}\right)+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}} \\ $$$${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}'\left({x}\right)−\mathrm{1}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{z}\left({x}\right)−\mathrm{1}} \\ $$$${z}'\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{{z}\left({x}\right)−\mathrm{1}} \\ $$$${z}'\left({x}\right)\left({z}\left({x}\right)−\mathrm{1}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$$\int{z}'\left({x}\right)\left({z}\left({x}\right)−\mathrm{1}\right)\mathrm{d}{x}=\left\{{t}={z}\left({x}\right)\right\}=\int\left({t}−\mathrm{1}\right)\mathrm{d}{t}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−{t}=\frac{{z}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}−{z}\left({x}\right)=\int\left(\mathrm{2}{x}+\mathrm{1}\right)\mathrm{d}{x}={x}^{\mathrm{2}} +{x}+{A} \\ $$$$\frac{{z}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}−{z}\left({x}\right)−{x}^{\mathrm{2}} −{x}+{A}=\mathrm{0} \\ $$$$\:_{{z}} =\mathrm{1}−\mathrm{2}\left({x}^{\mathrm{2}} −{x}+{A}\right) \\ $$$${z}\left({x}\right)=\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{2}\left({x}^{\mathrm{2}} −{x}+{A}\right)}\vee{z}\left({x}\right)=\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{2}\left({x}^{\mathrm{2}} −{x}+{A}\right)} \\ $$$${y}\left({x}\right)=\mathrm{1}−{x}−\sqrt{\mathrm{1}−\mathrm{2}\left({x}^{\mathrm{2}} −{x}+{A}\right)}\vee{y}\left({x}\right)=\mathrm{1}−{x}+\sqrt{\mathrm{1}−\mathrm{2}\left({x}^{\mathrm{2}} −{x}+{A}\right)} \\ $$