Question Number 121517 by liberty last updated on 09/Nov/20 $$\:\mathrm{y}''+\mathrm{4y}'+\mathrm{4y}\:=\:\left(\mathrm{x}+\mathrm{3}\right)\mathrm{e}^{−\mathrm{2x}} \\ $$$$\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2}\:\wedge\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{5} \\ $$ Answered by benjo_mathlover last updated on 09/Nov/20 Answered by TANMAY PANACEA…
Question Number 55939 by MJS last updated on 06/Mar/19 $$\frac{{y}'}{{y}}=\mathrm{1}−\mathrm{cot}\:{x} \\ $$$${y}=? \\ $$ Answered by kaivan.ahmadi last updated on 06/Mar/19 $$\frac{\mathrm{1}}{{y}}\:\frac{{dy}}{{dx}}=\mathrm{1}−{cotx}\Rightarrow \\ $$$$\frac{\mathrm{1}}{{y}}{dy}=\left(\mathrm{1}−{cotx}\right){dx}\Rightarrow \\…
Question Number 186999 by Mingma last updated on 12/Feb/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 121461 by bramlexs22 last updated on 08/Nov/20 Answered by Olaf last updated on 08/Nov/20 $$\left(\mathrm{3}{y}^{\mathrm{2}} +\mathrm{10}{xy}^{\mathrm{2}} \right){dx}+\left(\mathrm{6}{xy}−\mathrm{2}+\mathrm{10}{x}^{\mathrm{2}} {y}\right){dy}\:=\:\mathrm{0} \\ $$$$\mathrm{d}{f}\:=\:\frac{\partial{f}}{\partial{x}}\left({x},{y}\right){dx}+\frac{\partial{f}}{\partial{y}}\left({x},{y}\right){dy}\:=\:\mathrm{0} \\ $$$$ \\…
Question Number 121378 by liberty last updated on 07/Nov/20 $$\:\mathrm{cos}\:\mathrm{x}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{y}.\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2cos}\:^{\mathrm{3}} \mathrm{x}.\mathrm{sin}\:\mathrm{x}\: \\ $$ Answered by Ar Brandon last updated on 07/Nov/20 $$\mathrm{cosx}\centerdot\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\centerdot\mathrm{sinx}=\mathrm{2cos}^{\mathrm{3}} \mathrm{x}\centerdot\mathrm{sinx} \\ $$$$\mathrm{secx}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{ysecxtanx}=\mathrm{2cosxsinx}…
Question Number 121227 by benjo_mathlover last updated on 06/Nov/20 $$\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{e}^{−\mathrm{2x}} \: \\ $$ Answered by Lordose last updated on 06/Nov/20 $$\frac{\mathrm{dy}}{\mathrm{dx}}=\:\int\mathrm{e}^{−\mathrm{2x}} \mathrm{dx} \\…
Question Number 55597 by ajfour last updated on 27/Feb/19 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{6}{y}\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${Please}\:{solve}\:{the}\:{differential}\:{eq}. \\ $$ Answered by mr W last updated on 27/Feb/19…
Question Number 121028 by bramlexs22 last updated on 05/Nov/20 Answered by liberty last updated on 05/Nov/20 $$\mathrm{no}.\:\mathrm{this}\:\mathrm{not}\:\mathrm{separable}\:\mathrm{but}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{homogenous} \\ $$$$\mathrm{diff}\:\mathrm{equation}. \\ $$$$\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{2y}^{\mathrm{4}} }{\mathrm{xy}^{\mathrm{3}} }\:\:;\:\left[\:\mathrm{put}\:\mathrm{y}\:=\:\mathrm{zx}\:\right] \\…
Question Number 120905 by bemath last updated on 03/Nov/20 $$\:\:\mathrm{y}'\:=\:\mathrm{xy}^{\mathrm{2}} −\frac{\mathrm{y}}{\mathrm{x}} \\ $$ Answered by Bird last updated on 03/Nov/20 $$\Rightarrow\frac{{y}^{'} }{{y}^{\mathrm{2}} }={x}−\frac{\mathrm{1}}{{yx}}\:{let}\:\frac{\mathrm{1}}{{y}}={z}\:\Rightarrow{z}^{'} =−\frac{{y}^{'} }{{y}^{\mathrm{2}}…
Question Number 55351 by sayarthet92 last updated on 22/Feb/19 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com