Question Number 127943 by liberty last updated on 03/Jan/21 $$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{x}\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{y}−\mathrm{x}\right)\:=\:\mathrm{1}\: \\ $$ Answered by bramlexs22 last updated on 03/Jan/21 Terms of Service Privacy Policy…
Question Number 127783 by bemath last updated on 02/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:−\frac{\mathrm{y}}{\mathrm{x}}\:=\:\frac{\mathrm{y}^{\mathrm{3}} .\mathrm{e}^{\mathrm{x}} }{\mathrm{x}^{\mathrm{2}} }\: \\ $$ Answered by liberty last updated on 02/Jan/21 $$\:\mathrm{Bernoulli}\:\mathrm{diff}\:\mathrm{equation}\:. \\ $$$$\:\mathrm{let}\:\mathrm{v}\:=\:\mathrm{y}^{−\mathrm{2}}…
Question Number 127780 by bemath last updated on 02/Jan/21 $$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2x}\right)\:\mathrm{dy}−\mathrm{2y}\:\mathrm{dx}\:=\:\mathrm{0} \\ $$$$\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by liberty last updated on 02/Jan/21 $$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}}…
Question Number 127723 by bemath last updated on 01/Jan/21 $$\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \right)\:\mathrm{dx}\:−\mathrm{3xy}\:\mathrm{dy}\:=\:\mathrm{0}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127589 by bramlexs22 last updated on 31/Dec/20 Answered by liberty last updated on 31/Dec/20 $$\left(\bullet\right)\:\mathrm{let}\:\mathrm{y}'=\mathrm{z}\:\Rightarrow\:\mathrm{xz}'\:+\:\mathrm{z}\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{xz}\right)\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\mathrm{xz}\:=\:\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:+\mathrm{C}_{\mathrm{1}}…
Question Number 127578 by bramlexs22 last updated on 31/Dec/20 $$\:\mathrm{y}\:''\:=\:\mathrm{5y}−\mathrm{10y}^{\mathrm{2}} \\ $$$$\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by mr W last updated on 31/Dec/20 $${y}''=\frac{{dy}'}{{dx}}={y}'\frac{{dy}'}{{dy}} \\ $$$${y}'\frac{{dy}'}{{dy}}=\mathrm{5}{y}−\mathrm{10}{y}^{\mathrm{2}}…
Question Number 127520 by liberty last updated on 30/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127505 by bramlexs22 last updated on 30/Dec/20 $$\left(\mathrm{y}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}\right)\:\mathrm{dx}\:+\:\left(\mathrm{2x}^{\mathrm{3}} −\mathrm{xy}\right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$ Answered by liberty last updated on 30/Dec/20 $$\:\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} {y}}{{xy}−\mathrm{2}{x}^{\mathrm{3}}…
Question Number 127426 by sdfg last updated on 29/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127369 by liberty last updated on 29/Dec/20 $$\:\left({x}−\mathrm{3}{y}+\mathrm{5}\right){dx}\:+\left(\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{4}\right){dy}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 29/Dec/20 Terms of Service Privacy Policy Contact:…