Question Number 227855 by Abdulwahab last updated on 22/Feb/26 Commented by mr W last updated on 22/Feb/26 $$\frac{{y}}{{x}}=\mathrm{tan}^{−\mathrm{1}} \frac{{x}}{{y}} \\ $$$$−\frac{{y}}{{x}^{\mathrm{2}} }+\frac{{y}'}{{x}}=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{x}}{{y}}\right)^{\mathrm{2}} }×\left(\frac{\mathrm{1}}{{y}}−\frac{{xy}'}{{y}^{\mathrm{2}} }\right) \\…
Question Number 227273 by Spillover last updated on 11/Jan/26 Answered by peace2 last updated on 12/Jan/26 $$\frac{\mathrm{1}}{{y}_{} }={z} \\ $$$$\Rightarrow\frac{{dz}}{{dx}}=−\frac{{dy}}{{dx}}.\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\Leftrightarrow−{x}\frac{{dz}}{{dx}}+{z}={ln}\left({x}\right) \\ $$$${z}\left({x}\right)={kx}\Rightarrow−{x}\left({k}'{x}\right)={ln}\left({x}\right)\Rightarrow{k}'=−\frac{{ln}\left({x}\right)}{{x}^{\mathrm{2}}…
Question Number 227272 by Spillover last updated on 11/Jan/26 Answered by breniam last updated on 12/Jan/26 $$\left({x}+\mathrm{1}\right){y}'\left({x}\right)−{y}\left({x}\right)={e}^{{x}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$${y}\left({x}\right)={a}\left({x}\right){b}\left({x}\right) \\ $$$${y}'\left({x}\right)={a}'\left({x}\right){b}\left({x}\right)+{a}\left({x}\right){b}'\left({x}\right) \\ $$$${a}\left({x}\right)\left(\left({x}+\mathrm{1}\right){b}'\left({x}\right)−{b}\left({x}\right)\right)+{a}'\left({x}\right){b}\left({x}\right)\left({x}+\mathrm{1}\right) \\…
Question Number 227248 by Spillover last updated on 10/Jan/26 Answered by Spillover last updated on 11/Jan/26 Answered by breniam last updated on 10/Jan/26 $${z}\left({x}\right)=\frac{\mathrm{2}{x}+{y}\left({x}\right)−\mathrm{1}}{{x}−\mathrm{2}}=\mathrm{2}+\frac{{y}\left({x}\right)+\mathrm{3}}{{x}−\mathrm{2}}\Rightarrow \\…
Question Number 227249 by Spillover last updated on 10/Jan/26 Answered by breniam last updated on 10/Jan/26 $$\frac{{x}−{y}\left({x}\right)+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{x}+{y}\left({x}\right)−\mathrm{1}} \\ $$$${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}'\left({x}\right)−\mathrm{1}=−\mathrm{1}+\frac{\mathrm{2}{x}+\mathrm{1}}{{z}\left({x}\right)−\mathrm{1}}…
Question Number 227250 by Spillover last updated on 10/Jan/26 Answered by breniam last updated on 10/Jan/26 $${z}\left({x}\right)={x}+{y}\left({x}\right) \\ $$$${y}\left({x}\right)={z}\left({x}\right)−{x} \\ $$$${y}'\left({x}\right)={z}'\left({x}\right)−\mathrm{1} \\ $$$${z}\left({x}\right)\left({z}'\left({x}\right)−\mathrm{1}\right)=\mathrm{2}{x}−{z}\left({x}\right)+\mathrm{2} \\ $$$${z}'\left({x}\right){z}\left({x}\right)=\mathrm{2}\left({x}+\mathrm{1}\right)…
Question Number 227221 by Spillover last updated on 06/Jan/26 $${Solve}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{xy}\right)\frac{{dy}}{{dx}}={xy}−{y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$ Answered by som(math1967) last updated on 07/Jan/26 $$\:\frac{{dy}}{{dx}}=\frac{{xy}−{y}^{\mathrm{2}}…
Question Number 227220 by Spillover last updated on 06/Jan/26 $${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…
Question Number 227222 by Spillover last updated on 06/Jan/26 $$\:\:\:\:{Solve} \\ $$$$\:\:\:\:\:\:\:{x}\frac{{dy}}{{dx}}+{y}={x}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$$$ \\ $$ Answered by som(math1967) last updated on 07/Jan/26…
Question Number 227219 by Spillover last updated on 06/Jan/26 $${Solve}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} +{xy}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}−{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\mathrm{6}/\mathrm{1}/\mathrm{2026} \\ $$$$ \\ $$ Answered by peace2…