Question Number 130830 by Ar Brandon last updated on 29/Jan/21 $$\mathrm{x}^{\mathrm{2}} \mathrm{y}''−\mathrm{xy}'+\mathrm{y}=\mathrm{0} \\ $$ Answered by bemath last updated on 29/Jan/21 $$\mathrm{Cauchy}−\mathrm{Euler}\:\mathrm{diff}\:\mathrm{eq} \\ $$$$\mathrm{let}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{m}} \:\rightarrow\begin{cases}{\mathrm{y}'=\mathrm{mx}^{\mathrm{m}−\mathrm{1}}…
Question Number 65219 by Rio Michael last updated on 26/Jul/19 Commented by Rio Michael last updated on 26/Jul/19 $${please}\:{check}\:{my}\:{solution}\:{and}\:{correct}\:{me}\:{for}\:{that}\:{question} \\ $$$${you}\:{can}\:{give}\:{new}\:{methods}\:{to}\:{solve}\:{please}. \\ $$ Answered by…
Question Number 64973 by Tawa1 last updated on 23/Jul/19 Answered by mr W last updated on 23/Jul/19 $$\left(\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}={u} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}={u}\frac{{du}}{{dy}} \\…
Question Number 130301 by sarahvalencia last updated on 24/Jan/21 Answered by benjo_mathlover last updated on 24/Jan/21 $$\left(\mathrm{2}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} }\:\Rightarrow\:\int\:\mathrm{y}^{\mathrm{2}} \mathrm{dy}−\int\mathrm{x}^{\mathrm{2}} \mathrm{dx}=\mathrm{C} \\ $$$$\:\mathrm{y}^{\mathrm{3}} −\mathrm{x}^{\mathrm{3}} \:=\:\mathrm{3C}\:;\:\mathrm{y}^{\mathrm{3}}…
Question Number 130040 by liberty last updated on 22/Jan/21 $$\:\mathrm{e}^{\frac{\mathrm{t}}{\mathrm{y}}} \left(\mathrm{y}−\mathrm{t}\right)\:\frac{\mathrm{dy}}{\mathrm{dt}}\:+\:\mathrm{y}\left(\mathrm{1}+\mathrm{e}^{\frac{\mathrm{t}}{\mathrm{y}}} \:\right)\:=\:\mathrm{0}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129949 by bemath last updated on 21/Jan/21 $$\mathrm{Given}\:\mathrm{8}\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:\right)\mathrm{dy}\:=\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}}}\:,\:\mathrm{x}>\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{y}\left(\mathrm{0}\right)=\sqrt{\mathrm{7}}\:.\:\mathrm{Find}\:\mathrm{y}\left(\mathrm{256}\right). \\ $$ Answered by liberty last updated on 21/Jan/21 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{8}\sqrt{\mathrm{x}}\:\left(\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\right)\left(\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}}\right)} \\ $$$$\:\mathrm{let}\:\sqrt{\mathrm{4}+\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:}\:=\:\mathrm{z}\:;\:\sqrt{\mathrm{9}+\sqrt{\mathrm{x}}}\:=\:\mathrm{z}^{\mathrm{2}} −\mathrm{4}…
Question Number 129914 by MrJoe last updated on 20/Jan/21 Commented by MrJoe last updated on 20/Jan/21 Find x(t) Answered by Dwaipayan Shikari last updated on 20/Jan/21…
Question Number 129777 by stelor last updated on 18/Jan/21 $$\left({x}^{\mathrm{2}} −{xy}−{y}^{\mathrm{2}} \right){dx}+{x}^{\mathrm{2}} {dy}\:=\:\mathrm{0} \\ $$$${hello}…\:{please}\:{help}\:{me}. \\ $$ Answered by Ar Brandon last updated on 18/Jan/21…
Question Number 129760 by sdfg last updated on 18/Jan/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129656 by Ahmed1hamouda last updated on 17/Jan/21 Commented by Ahmed1hamouda last updated on 17/Jan/21 $${solve}\:{the}\:{differen}\mathrm{tial}\:\mathrm{equation} \\ $$ Answered by liberty last updated on…