Question Number 91825 by john santu last updated on 03/May/20 $$\left(\mathrm{cos}\:{x}\right)\:\frac{{dy}}{{dx}}−{y}\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{cot}\:\left({x}\right) \\ $$ Commented by Tony Lin last updated on 03/May/20 $$\left({ycosx}\right)'={cotx} \\ $$$${ycosx}=\int{cotxdx}={ln}\mid{sinx}\mid+{c} \\…
Question Number 26240 by sorour87 last updated on 22/Dec/17 $${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \\ $$ Commented by prakash jain last updated on 23/Dec/17 $${y}^{\left(\mathrm{2}\right)} −{y}=\left(\mathrm{1}−{e}^{\mathrm{2}{x}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}}…
Question Number 91770 by john santu last updated on 03/May/20 $${y}''+\mathrm{3}{y}={x}^{\mathrm{3}} +\mathrm{3}{x} \\ $$ Commented by Joel578 last updated on 03/May/20 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{in}\:\mathrm{form} \\ $$$${y}_{{p}} \:=\:{Ax}^{\mathrm{3}}…
Question Number 91750 by jagoll last updated on 02/May/20 $$\left(\mathrm{2}{x}−{y}+\mathrm{1}\right){dx}\:=\:\left({x}−\mathrm{4}{y}+\mathrm{3}\right){dy} \\ $$ Commented by mr W last updated on 02/May/20 $${see}\:{Q}\mathrm{90770} \\ $$ Commented by…
Question Number 26198 by sorour87 last updated on 22/Dec/17 $$\left(\mathrm{2}{x}\sqrt{{x}}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dx}+\mathrm{2}{y}\sqrt{{x}}{dy}=\mathrm{0} \\ $$ Answered by ajfour last updated on 22/Dec/17 $${let}\:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\…
Question Number 157267 by cortano last updated on 21/Oct/21 $$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}={x}\: \\ $$ Commented by MathsFan last updated on 22/Oct/21 $${any}\:{caption}???? \\ $$ Commented by…
Question Number 26191 by sorour87 last updated on 21/Dec/17 $$\mathrm{2}\frac{{dy}}{{dx}}+{x}=\mathrm{4}\sqrt{{y}} \\ $$ Answered by mrW1 last updated on 22/Dec/17 $${y}={u}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{u}\frac{{du}}{{dx}} \\ $$$$\Rightarrow\mathrm{4}{u}\frac{{du}}{{dx}}+{x}=\mathrm{4}{u} \\…
Question Number 91678 by john santu last updated on 02/May/20 $$\left({y}+\mathrm{2}{px}\right)^{\mathrm{2}} \:=\:\mathrm{2}{px}^{\mathrm{2}} \\ $$$${solve}\:{by}\:{Clairaut}'{s}\:{method}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26127 by sorour87 last updated on 20/Dec/17 $${y}+\mathrm{2}{y}^{\mathrm{3}} {y}^{\left(\mathrm{1}\right)} =\left({x}+\mathrm{4}{y}\mathrm{ln}\:\left({y}\right)\right){y}^{\left(\mathrm{1}\right)} \\ $$ Answered by prakash jain last updated on 21/Dec/17 $${y}+\mathrm{2}{y}^{\mathrm{3}} \frac{{dy}}{{dx}}=\left({x}+\mathrm{4}{y}\mathrm{ln}\:{y}\right)\frac{{dy}}{{dx}} \\…
Question Number 91656 by jagoll last updated on 02/May/20 Commented by john santu last updated on 02/May/20 $$\frac{{dy}}{{dx}}\:=\:\frac{−\mathrm{3}{xy}−{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{xy}}\:=\:\frac{−\frac{\mathrm{3}{y}}{{x}}−\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\frac{{y}}{{x}}\right)} \\ $$$$\left[\:{y}\:=\:{px}\:\Rightarrow\:\frac{{dy}}{{dx}}\:=\:{p}\:+\:{x}\frac{{dp}}{{dx}}\:\right] \\ $$$${p}\:+{x}\:\frac{{dp}}{{dx}}\:=\:\frac{−\mathrm{3}{p}−{p}^{\mathrm{2}}…