dx-1-2xtan-y-dy-

Question Number 89748 by jagoll last updated on 19/Apr/20

$$\mathrm{dx}\:=\:\left(\mathrm{1}+\mathrm{2xtan}\:\mathrm{y}\right)\:\mathrm{dy}\: \\ $$

Commented by mr W last updated on 19/Apr/20

(dx/dy)−(2 tan y) x=1 −∫2 tan y dy=2∫((d (cos y))/(cos y))=2ln (cos y)=ln cos^2 y u(x)=e^(ln (cos^2 y)) =cos^2 y x=((∫cos^2 y dy+C)/(cos^2 y)) x=((∫(1−cos 2y) dy+2C)/(2 cos^2 y)) x=((y−((sin 2y)/2)+2C)/(2 cos^2 y)) ⇒x=((y+c)/(2 cos^2 y))−((tan y )/(2 ))

$$\frac{{dx}}{{dy}}−\left(\mathrm{2}\:\mathrm{tan}\:{y}\right)\:{x}=\mathrm{1} \\ $$$$−\int\mathrm{2}\:\mathrm{tan}\:{y}\:{dy}=\mathrm{2}\int\frac{{d}\:\left(\mathrm{cos}\:{y}\right)}{\mathrm{cos}\:{y}}=\mathrm{2ln}\:\left(\mathrm{cos}\:{y}\right)=\mathrm{ln}\:\mathrm{cos}^{\mathrm{2}} \:{y} \\ $$$${u}\left({x}\right)={e}^{\mathrm{ln}\:\left(\mathrm{cos}^{\mathrm{2}} \:{y}\right)} =\mathrm{cos}^{\mathrm{2}} \:{y} \\ $$$${x}=\frac{\int\mathrm{cos}^{\mathrm{2}} \:{y}\:{dy}+{C}}{\mathrm{cos}^{\mathrm{2}} \:{y}} \\ $$$${x}=\frac{\int\left(\mathrm{1}−\mathrm{cos}\:\:\mathrm{2}{y}\right)\:{dy}+\mathrm{2}{C}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{y}} \\ $$$${x}=\frac{{y}−\frac{\mathrm{sin}\:\mathrm{2}{y}}{\mathrm{2}}+\mathrm{2}{C}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{y}} \\ $$$$\Rightarrow{x}=\frac{{y}+{c}}{\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:{y}}−\frac{\mathrm{tan}\:{y}\:}{\mathrm{2}\:} \\ $$

Commented by niroj last updated on 19/Apr/20

dx= (1+2xtan y)dy (dx/dy)= 1+2xtany (dx/dy)−2xtany= 1 P=−2tany , Q=1 IF= e^(∫Pdy) = e^(−2∫tan y dy) = e^(−2log sec y) = e^(log sec^(−2) y) IF= sec^(−2) y x.IF= ∫IF.Q dy+C x.sec^(−2) y= ∫sec^(−2) y.1.dy+C (x/(sec^2 y))= ∫ (1/(sec^2 y))dy+C =∫ (1/(1+tan^2 y))dy+c x cos^2 y = tan^(−1) (tany)+C x = sec^2 y[ tan^(−1) (tany)+C]//.

$$\:\:\mathrm{dx}=\:\left(\mathrm{1}+\mathrm{2xtan}\:\mathrm{y}\right)\mathrm{dy} \\ $$$$\:\:\frac{\mathrm{dx}}{\mathrm{dy}}=\:\mathrm{1}+\mathrm{2xtany} \\ $$$$\:\:\:\frac{\mathrm{dx}}{\mathrm{dy}}−\mathrm{2xtany}=\:\mathrm{1}\:\:\: \\ $$$$\:\:\:\:\mathrm{P}=−\mathrm{2tany}\:\:,\:\mathrm{Q}=\mathrm{1} \\ $$$$\:\:\:\mathrm{IF}=\:\mathrm{e}^{\int\mathrm{Pdy}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{e}^{−\mathrm{2}\int\mathrm{tan}\:\mathrm{y}\:\mathrm{dy}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{e}^{−\mathrm{2log}\:\mathrm{sec}\:\mathrm{y}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{e}^{\mathrm{log}\:\mathrm{sec}^{−\mathrm{2}} \mathrm{y}} \\ $$$$\:\:\mathrm{IF}=\:\mathrm{sec}^{−\mathrm{2}} \mathrm{y} \\ $$$$\:\mathrm{x}.\mathrm{IF}=\:\int\mathrm{IF}.\mathrm{Q}\:\mathrm{dy}+\mathrm{C} \\ $$$$\:\:\mathrm{x}.\mathrm{sec}^{−\mathrm{2}} \mathrm{y}=\:\int\mathrm{sec}^{−\mathrm{2}} \mathrm{y}.\mathrm{1}.\mathrm{dy}+\mathrm{C} \\ $$$$\:\:\:\frac{\mathrm{x}}{\mathrm{sec}^{\mathrm{2}} \mathrm{y}}=\:\int\:\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} \mathrm{y}}\mathrm{dy}+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:=\int\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}}\mathrm{dy}+\mathrm{c} \\ $$$$\:\:\mathrm{x}\:\mathrm{cos}^{\mathrm{2}} \mathrm{y}\:\:\:\:\:\:=\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tany}\right)+\mathrm{C} \\ $$$$\:\mathrm{x}\:=\:\:\mathrm{sec}^{\mathrm{2}} \mathrm{y}\left[\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tany}\right)+\mathrm{C}\right]//. \\ $$$$\:\:\: \\ $$$$ \\ $$$$\:\:\: \\ $$$$ \\ $$

Commented by peter frank last updated on 19/Apr/20

$${thank}\:{you}\:{both} \\ $$

Commented by Coronavirus last updated on 27/Jun/20

need explanation please i don't understand anything

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