Question Number 93312 by john santu last updated on 12/May/20 $$\left(\mathrm{y}−\mathrm{xy}^{\mathrm{2}} \right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \right)\mathrm{dy}\:=\:\mathrm{0}\: \\ $$ Answered by i jagooll last updated on 12/May/20 $$\mathrm{ydx}+\mathrm{xdy}\:−\left(\mathrm{xy}^{\mathrm{2}}…
Question Number 93283 by john santu last updated on 12/May/20 $$\frac{{dy}}{{dx}}\:=\:\mathrm{2}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:,\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 27696 by chernoaguero@gmail.com last updated on 12/Jan/18 $$ \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{ae}^{\mathrm{x}} +\mathrm{3e}^{\mathrm{2x}} −\mathrm{b}}{\mathrm{x}}\:=\mathrm{8} \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b} \\ $$ Commented by…
Question Number 93059 by Ar Brandon last updated on 10/May/20 $$\mathrm{y}^{'} +\mathrm{2xy}=\mathrm{cos}\:\mathrm{x}+\mathrm{2x}\:\mathrm{sin}\:\mathrm{x} \\ $$ Answered by TANMAY PANACEA … last updated on 10/May/20 $${dy}+\mathrm{2}{xydx}={cosxdx}+\mathrm{2}{xsinxdx} \\…
Question Number 93003 by i jagooll last updated on 10/May/20 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}−\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} \:\mathrm{dy}=\mathrm{0} \\ $$ Answered by TANMAY PANACEA … last updated on 10/May/20…
Question Number 92860 by i jagooll last updated on 09/May/20 $$\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{x}\right)\:\mathrm{dy}\:+\:\mathrm{y}\:\mathrm{dx}\:=\:\mathrm{0}\: \\ $$ Commented by john santu last updated on 09/May/20 $$\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\mathrm{sin}^{−\mathrm{1}}…
Question Number 92852 by i jagooll last updated on 09/May/20 $$\mathrm{y}''+\mathrm{2y}'+\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \mathrm{e}^{−\mathrm{x}} \mathrm{cos}\:\mathrm{x} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{particular}\:\mathrm{solution} \\ $$$$ \\ $$ Commented by i jagooll last updated…
Question Number 92838 by i jagooll last updated on 09/May/20 $$\left(\mathrm{x}+\mathrm{y}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{xy}+\mathrm{x}+\mathrm{1} \\ $$ Commented by john santu last updated on 09/May/20 $$\left(\mathrm{x}+\mathrm{y}\right)\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{x}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{x}+\mathrm{1} \\ $$$$\mathrm{set}\:\mathrm{x}+\mathrm{y}\:=\:\mathrm{v}\:\Rightarrow\mathrm{1}+\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{dv}}{\mathrm{dx}}…
Question Number 92808 by device4438043516@gmail.com last updated on 27/May/20 $$\mathrm{Can}\:\mathrm{you}\:\mathrm{prove}\:{it}: \\ $$$${t}=\Phi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158309 by cortano last updated on 02/Nov/21 $${y}''+{y}'−\mathrm{2}{y}=−\mathrm{18}{te}^{−\mathrm{2}{t}} \\ $$ Answered by TheSupreme last updated on 02/Nov/21 $$\lambda^{\mathrm{2}} +\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\lambda_{\mathrm{1},\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{9}}}{\mathrm{2}}=−\mathrm{2},\:\mathrm{1} \\…