Question Number 14430 by tawa tawa last updated on 31/May/17 $$\mathrm{y}\:=\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{4}} \right)^{\mathrm{2}} \:,\:\:\mathrm{find}\:\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{x} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 14288 by tawa tawa last updated on 30/May/17 $$\mathrm{Solve}:\:\:\mathrm{y}'\:+\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{0} \\ $$ Answered by Tinkutara last updated on 30/May/17 $$\frac{{dy}}{{dx}}\:=\:−\:\frac{{y}}{{x}}\: \\ $$$$\frac{{dy}}{{y}}\:=\:−\:\frac{{dx}}{{x}} \\ $$$$\mathrm{ln}\:{y}\:=\:−\:\mathrm{ln}\:{x}\:+\:{C}…
Question Number 79761 by Najaah last updated on 28/Jan/20 $$\left({x}^{\mathrm{2}} +\mathrm{2}\right){y}''\:+\:{xy}\:=\:\mathrm{0} \\ $$ Commented by abdomathmax last updated on 28/Jan/20 $${solution}\:{developpable}\:{at}?{integr}\:{serie} \\ $$$${y}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}}…
Question Number 145300 by imjagoll last updated on 04/Jul/21 $$\:\:\:\:\:\:\:\mathrm{yy}'\:=\:\mathrm{x}\:\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \: \\ $$ Answered by puissant last updated on 04/Jul/21 $$\mathrm{y}'=\frac{\mathrm{x}}{\mathrm{y}}\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \:\:\mathrm{put}\:\:\:\mathrm{t}=\frac{\mathrm{x}}{\mathrm{y}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dt}}\:=\:\mathrm{te}^{\mathrm{t}} \Rightarrow\:\mathrm{dy}\:=\:\mathrm{te}^{\mathrm{t}}…
Question Number 79626 by M±th+et£s last updated on 26/Jan/20 $${find}\:{tbe}\:{general}\:{solution}\:{of}\:{the}\:{D}.{E} \\ $$$${y}^{''} −\mathrm{5}{y}^{'} +\mathrm{4}{y}={e}^{{t}} {cos}\left({t}\right) \\ $$ Answered by mind is power last updated on…
Question Number 13983 by wodus103 last updated on 26/May/17 Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/May/17 $${x}={y}=\mathrm{1} \\ $$ Commented by ajfour last updated on…
Question Number 13976 by tawa tawa last updated on 25/May/17 $$\mathrm{Solve}: \\ $$$$\mathrm{x}^{\mathrm{2}} \left(\mathrm{y}\:+\:\mathrm{1}\right)\:+\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{x}\:−\:\mathrm{1}\right)\mathrm{y}'\:=\:\mathrm{0} \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17 $${y}^{'}…
Question Number 79469 by M±th+et£s last updated on 25/Jan/20 $${Q}.{solve} \\ $$$$−\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }−{coth}\left({t}\right)\frac{{dy}}{{dt}}+\left(\mathrm{20}+\frac{\mathrm{4}}{{sinh}^{\mathrm{2}} \left({t}\right)}\right){y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 13649 by ajfour last updated on 22/May/17 $$\mathrm{2}{xyy}'+\left({x}−\mathrm{1}\right){y}^{\mathrm{2}} ={x}^{\mathrm{2}} {e}^{{x}} \\ $$ Answered by ajfour last updated on 22/May/17 $${let}\:{y}^{\mathrm{2}} ={tx}\:\:\:\:\:\Rightarrow\:\:\:\mathrm{2}{yy}'={t}+{x}\frac{{dt}}{{dx}} \\ $$$${substituting}…
Question Number 13626 by tawa tawa last updated on 21/May/17 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cosech}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\mathrm{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right) \\ $$ Answered by mrW1 last updated on 21/May/17 $${u}=\mathrm{cosech}^{−\mathrm{1}}…