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 $$\lambda^{\mathrm{2}} +\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\lambda_{\mathrm{1},\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{9}}}{\mathrm{2}}=−\mathrm{2},\:\mathrm{1} \\…
Question Number 92702 by i jagooll last updated on 08/May/20 $${y}''+\mathrm{2}{y}'−\mathrm{3}{y}={e}^{{x}} +{e}^{\mathrm{2}{x}} \\ $$ Answered by niroj last updated on 08/May/20 $$\:\:\mathrm{y}^{''} +\mathrm{2y}^{'} −\mathrm{3y}=\:\mathrm{e}^{\mathrm{x}} +\:\mathrm{e}^{\mathrm{2x}}…
Question Number 27028 by sorour87 last updated on 01/Jan/18 $${y}^{\left(\mathrm{2}\right)} +\mathrm{4}{y}=\mathrm{sinh}\:{x}×\mathrm{sin}\:\mathrm{2}{x} \\ $$ Answered by prakash jain last updated on 02/Jan/18 $$\lambda^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\pm\mathrm{2}{i}…
Question Number 158087 by cortano last updated on 31/Oct/21 $$\left({y}^{\mathrm{2}} +\mathrm{4}{y}+\mathrm{8}\right){dx}+\sqrt{\mathrm{2}{x}+\mathrm{5}}\:\left({y}+\mathrm{3}\right){dy}=\mathrm{0}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 26983 by chernoaguero@gmail.com last updated on 31/Dec/17 $$\mathrm{Find}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\:\: \\ $$$$\:\:\:\:\mathrm{x}^{\mathrm{y}} =\mathrm{y}^{\mathrm{x}} \\ $$ Answered by mrW1 last updated on 31/Dec/17 $${y}\:\mathrm{ln}\:{x}={x}\:\mathrm{ln}\:{y} \\ $$$${y}'\:\mathrm{ln}\:{x}+\frac{{y}}{{x}}=\mathrm{ln}\:{y}+\frac{{x}}{{y}}\:{y}'…
Question Number 92468 by jagoll last updated on 07/May/20 $$\mathrm{y}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}}−\mathrm{9y}\:=\:\mathrm{3} \\ $$ Answered by mr W last updated on 07/May/20 $${u}=\frac{{dy}}{{dx}} \\ $$$${yu}\frac{{du}}{{dy}}−\mathrm{9}{y}=\mathrm{3} \\…
Question Number 92464 by jagoll last updated on 07/May/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{e}^{\mathrm{y}} }{\mathrm{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$ Commented by mr W last updated on 07/May/20 $${how}\:{do}\:{you}\:{get} \\ $$$$\frac{{dy}}{{e}^{{y}}…