Question Number 127589 by bramlexs22 last updated on 31/Dec/20 Answered by liberty last updated on 31/Dec/20 $$\left(\bullet\right)\:\mathrm{let}\:\mathrm{y}'=\mathrm{z}\:\Rightarrow\:\mathrm{xz}'\:+\:\mathrm{z}\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\:\frac{\mathrm{d}}{\mathrm{dx}}\:\left(\mathrm{xz}\right)\:=\:\mathrm{3x}^{\mathrm{2}} −\mathrm{x}\: \\ $$$$\:\:\:\mathrm{xz}\:=\:\mathrm{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:+\mathrm{C}_{\mathrm{1}}…
Question Number 127578 by bramlexs22 last updated on 31/Dec/20 $$\:\mathrm{y}\:''\:=\:\mathrm{5y}−\mathrm{10y}^{\mathrm{2}} \\ $$$$\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$ Answered by mr W last updated on 31/Dec/20 $${y}''=\frac{{dy}'}{{dx}}={y}'\frac{{dy}'}{{dy}} \\ $$$${y}'\frac{{dy}'}{{dy}}=\mathrm{5}{y}−\mathrm{10}{y}^{\mathrm{2}}…
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Question Number 127505 by bramlexs22 last updated on 30/Dec/20 $$\left(\mathrm{y}^{\mathrm{2}} +\mathrm{2x}^{\mathrm{2}} \mathrm{y}\right)\:\mathrm{dx}\:+\:\left(\mathrm{2x}^{\mathrm{3}} −\mathrm{xy}\right)\:\mathrm{dy}\:=\:\mathrm{0} \\ $$ Answered by liberty last updated on 30/Dec/20 $$\:\frac{{dy}}{{dx}}\:=\:\frac{{y}^{\mathrm{2}} +\mathrm{2}{x}^{\mathrm{2}} {y}}{{xy}−\mathrm{2}{x}^{\mathrm{3}}…
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Question Number 127369 by liberty last updated on 29/Dec/20 $$\:\left({x}−\mathrm{3}{y}+\mathrm{5}\right){dx}\:+\left(\mathrm{2}{x}+\mathrm{5}{y}+\mathrm{4}\right){dy}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 29/Dec/20 Terms of Service Privacy Policy Contact:…
Question Number 127354 by liberty last updated on 29/Dec/20 Answered by som(math1967) last updated on 29/Dec/20 $$\mathrm{1}.\:{let}\:{x}={X}+{h}\:\:\:{y}={Y}+{k} \\ $$$$\frac{{dx}}{{dX}}=\mathrm{1}\:\:\:\:\frac{{dy}}{{dY}}=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dY}}\:.\frac{{dY}}{{dX}}.\frac{{dX}}{{dx}}=\mathrm{1}.\frac{{dY}}{{dX}}.\mathrm{1}=\frac{{dY}}{{dX}} \\ $$$$\therefore\frac{{dY}}{{dX}}=\frac{{X}+{h}+{Y}+{k}−\mathrm{3}}{{X}+{h}−{Y}−{k}−\mathrm{1}} \\ $$$$\frac{{dY}}{{dX}}=\frac{{X}+{Y}+{h}+{k}−\mathrm{3}}{{X}−{Y}+{h}−{k}−\mathrm{1}}…
Question Number 127291 by bemath last updated on 28/Dec/20 $$\:\left({D}^{\mathrm{3}} −\mathrm{5}{D}^{\mathrm{2}} +\mathrm{7}{D}−\mathrm{3}\right){y}\:=\:{e}^{\mathrm{2}{x}} \:\mathrm{cosh}\:{x} \\ $$ Answered by liberty last updated on 28/Dec/20 $$\:{note}\:{that}\:{e}^{\mathrm{2}{x}} \:\mathrm{cosh}\:{x}\:=\:\frac{{e}^{\mathrm{3}{x}} +{e}^{{x}}…
Question Number 127287 by liberty last updated on 28/Dec/20 $$\:\:\left({x}+\mathrm{2}{y}−\mathrm{2}\right){dx}\:+\:\left(\mathrm{2}{x}+{y}+\mathrm{3}\right)\:{dy}\:=\:\mathrm{0} \\ $$ Answered by bemath last updated on 28/Dec/20 Commented by liberty last updated on…
Question Number 127260 by bramlexs22 last updated on 28/Dec/20 $$\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}\:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered by liberty last updated on 28/Dec/20 $${let}\:{y}\:=\:{xz}\:{for}\:{some}\:{z}={z}\left({x}\right) \\ $$$${y}'={xz}'+{z}\:\wedge\:{y}''={xz}''+\mathrm{2}{z}' \\…