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…
Question Number 129388 by MathSh last updated on 15/Jan/21 $${Solve}\:{the}\:{equation}: \\ $$$${y}''−\mathrm{2}{y}'+\mathrm{2}{y}={e}^{{x}} +{xcosx} \\ $$ Answered by mathmax by abdo last updated on 16/Jan/21 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}}…
Question Number 129334 by bramlexs22 last updated on 15/Jan/21 $$\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}\:+\:\mathrm{y}^{\mathrm{3}} \mathrm{cos}\:\mathrm{x}\: \\ $$ Answered by liberty last updated on 15/Jan/21 $$\:\mathrm{let}\:\mathrm{y}\:=\:\mathrm{w}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{w}^{−\mathrm{3}/\mathrm{2}} \:\frac{\mathrm{dw}}{\mathrm{dx}} \\ $$$$\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{xw}^{−\mathrm{3}/\mathrm{2}}…