Question Number 227273 by Spillover last updated on 11/Jan/26
Answered by peace2 last updated on 12/Jan/26
$$\frac{\mathrm{1}}{{y}_{} }={z} \\ $$$$\Rightarrow\frac{{dz}}{{dx}}=−\frac{{dy}}{{dx}}.\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\Leftrightarrow−{x}\frac{{dz}}{{dx}}+{z}={ln}\left({x}\right) \\ $$$${z}\left({x}\right)={kx}\Rightarrow−{x}\left({k}'{x}\right)={ln}\left({x}\right)\Rightarrow{k}'=−\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}} } \\ $$$${k}=\frac{{ln}\left({x}\right)}{{x}}+\frac{\mathrm{1}}{{x}}+{c} \\ $$$${z}=\left({ln}\left({x}\right)+\mathrm{1}+{cx}\right);{y}=\frac{\mathrm{1}}{{ln}\left({x}\right)+\mathrm{1}+{cx}} \\ $$