Question Number 227220 by Spillover last updated on 06/Jan/26
$${Solve}\: \\ $$$$\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}} \\ $$$$\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$
Answered by breniam last updated on 09/Jan/26
$${y}'\left({x}\right)=\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \left({x}\right)}{{xy}\left({x}\right)} \\ $$$${y}'\left({x}\right)=\frac{{x}}{{y}\left({x}\right)}+\frac{{y}\left({x}\right)}{{x}} \\ $$$${z}\left({x}\right)=\frac{{y}\left({x}\right)}{{x}} \\ $$$${y}\left({x}\right)={xz}\left({x}\right) \\ $$$${y}'\left({x}\right)={z}\left({x}\right)+{xz}'\left({x}\right) \\ $$$${z}\left({x}\right)+{xz}'\left({x}\right)=\frac{\mathrm{1}}{{z}\left({x}\right)}+{z}\left({x}\right) \\ $$$${xz}'\left({x}\right)=\frac{\mathrm{1}}{{z}\left({x}\right)} \\ $$$${z}\left({x}\right){z}'\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$$$\int{z}\left({x}\right){z}'\left({x}\right)\mathrm{d}{x}=\left\{{t}={z}\left({x}\right)\right\}=\int{t}\mathrm{d}{t}=\frac{{t}^{\mathrm{2}} }{\mathrm{2}}=\frac{{z}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}}=\int\frac{\mathrm{1}}{{x}}\mathrm{d}{x}=\mathrm{ln}\mid{x}\mid+{A} \\ $$$${z}^{\mathrm{2}} \left({x}\right)=\mathrm{2ln}\mid{x}\mid+{A} \\ $$$${z}\left({x}\right)=\pm\sqrt{\mathrm{2ln}\mid{x}\mid+{A}} \\ $$$$\frac{{y}\left({x}\right)}{{x}}=\pm\sqrt{\mathrm{2ln}\mid{x}\mid+{A}} \\ $$$${y}\left({x}\right)=\pm{x}\sqrt{\mathrm{2ln}\mid{x}\mid+{A}} \\ $$$$ \\ $$
Answered by Spillover last updated on 10/Jan/26
