Question Number 227250 by Spillover last updated on 10/Jan/26
Answered by breniam last updated on 10/Jan/26
$${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}\left({x}\right)\left({z}'\left({x}\right)−\mathrm{1}\right)=\mathrm{2}{x}−{z}\left({x}\right)+\mathrm{2} \\ $$$${z}'\left({x}\right){z}\left({x}\right)=\mathrm{2}\left({x}+\mathrm{1}\right) \\ $$$$\int{z}'\left({x}\right){z}\left({x}\right)\mathrm{d}{x}=\left\{{t}={z}\left({x}\right)\right\}=\int{t}\mathrm{d}{t}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}=\frac{{z}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}=\mathrm{2}\int\left({x}+\mathrm{1}\right)\mathrm{d}{x}={x}^{\mathrm{2}} +\mathrm{2}{x}+{A} \\ $$$$\frac{{z}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}={x}^{\mathrm{2}} +\mathrm{2}{x}+{A} \\ $$$${z}\left({x}\right)=\pm\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+{A}} \\ $$$${y}\left({x}\right)=\pm\sqrt{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+{A}}−{x} \\ $$