Question Number 175734 by Engr_Jidda last updated on 05/Sep/22 $${form}\:{a}\:{differential}\:{equation}\:{from}\:{the}\:{following} \\ $$$$\left.\mathrm{1}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{ax}+\mathrm{1}=\mathrm{0}\:\:{a}={constant} \\ $$$$\left.\mathrm{2}\right)\:{y}={A}\varrho^{\mathrm{3}{x}} +{B}\varrho^{−\mathrm{2}{x}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 110178 by Rio Michael last updated on 27/Aug/20 $$\mathrm{Define}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{transformation}\:\mathrm{equation}\:\mathrm{and} \\ $$$$\mathrm{use}\:\mathrm{the}\:\mathrm{transformation}\:\mathrm{equation}\:\mathrm{transform}\:\frac{{dy}}{{dx}}\:\mathrm{and}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$$$\mathrm{hence}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{equation}\::\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{5}\:\frac{{dy}}{{dx}}\:+\:\mathrm{4}\:=\:{e}^{−{x}} \:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{laplace}\:\mathrm{transformation}\:\mathrm{equations}\:\mathrm{derived}\:\mathrm{above}. \\ $$ Answered…
Question Number 175670 by Engr_Jidda last updated on 04/Sep/22 $${Solve}\:{the}\:{differential}\:{equation} \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dx}−\left(\mathrm{1}+{x}^{\mathrm{2}} \right){xydy}=\mathrm{0} \\ $$ Answered by mahdipoor last updated on 04/Sep/22 $$\Rightarrow\left(\mathrm{1}+{y}^{\mathrm{2}} \right){dx}=\left({x}+{x}^{\mathrm{3}}…
Question Number 44537 by needhelp last updated on 01/Oct/18 Answered by $@ty@m last updated on 01/Oct/18 $${v}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{2}{as} \\ $$$$\left(\mathrm{7}.\mathrm{10}\right)^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} +\mathrm{2}.{a}.\left(\mathrm{35}.\mathrm{4}\right) \\ $$$$\mathrm{50}.\mathrm{41}=\mathrm{70}.\mathrm{8}{a}…
Question Number 175464 by nadovic last updated on 31/Aug/22 $$\:\mathrm{Find}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{the}\:\mathrm{following}\: \\ $$$$\mathrm{differential}\:\mathrm{equations} \\ $$$$\left({a}\right)\:\:\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\frac{{dy}}{{dx}}\:=\:\mathrm{6}{y} \\ $$$$\left(\mathrm{b}\right)\:\:\:\left(\mathrm{3}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dx}\:+\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dy}\:=\:\mathrm{0} \\ $$$$ \\…
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Question Number 175406 by princeDera last updated on 29/Aug/22 $$\frac{{d}^{\mathrm{2}} \boldsymbol{{y}}}{\boldsymbol{{dx}}^{\mathrm{2}} }\:−\mathrm{tan}\:\left(\boldsymbol{{x}}\right)\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}\:=\:\mathrm{0} \\ $$ Answered by Ar Brandon last updated on 29/Aug/22 $${y}''−\mathrm{tan}{x}\centerdot{y}'=\mathrm{0} \\ $$$$\Rightarrow\frac{{y}''}{{y}'}=\mathrm{tan}{x}…
Question Number 175041 by rexford last updated on 17/Aug/22 $${Solve}\:{the}\:{differential} \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}^{\mathrm{2}} −\mathrm{3}{xy}}{{x}+{y}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 43616 by Tawa1 last updated on 12/Sep/18 $$\mathrm{Solve}:\:\:\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''\:\:+\:\:\mathrm{xy}'\:\:−\:\mathrm{y}\:=\:\mathrm{x}\:,\:\:\:\:\:\:\:\:\mathrm{y}_{\mathrm{1}} \:=\:\mathrm{x} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 13/Sep/18 $${x}={e}^{{t}} \:\:\:\frac{{dx}}{{dt}}={e}^{{t}} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dt}}×\frac{{dt}}{{dx}}=\frac{{dy}}{{dt}}×\frac{\mathrm{1}}{{e}^{{t}}…
Question Number 43599 by rish@bh last updated on 12/Sep/18 $$\frac{{dy}}{{dx}}\left({x}^{\mathrm{2}} {y}^{\mathrm{2}} +{xy}\right)=\mathrm{1} \\ $$$$\mathrm{Solve}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\mathrm{Thanks} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com