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x-tan-y-x-y-sec-2-y-x-dx-x-sec-2-y-x-dy-0-




Question Number 105738 by bobhans last updated on 31/Jul/20
(x tan ((y/x))−y sec^2 ((y/x))) dx−x sec^2 ((y/x))dy=0
$$\left({x}\:\mathrm{tan}\:\left(\frac{{y}}{{x}}\right)−{y}\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{{y}}{{x}}\right)\right)\:{dx}−{x}\:\mathrm{sec}\:^{\mathrm{2}} \left(\frac{{y}}{{x}}\right){dy}=\mathrm{0} \\ $$
Answered by john santu last updated on 01/Aug/20
set (y/x) = ϑ ⇒y=ϑx  (dy/dx) = ϑ + x (dϑ/dx)  (x tan ϑ−ϑx sec^2 ϑ)dx = x sec^2 ϑdy  (dy/dx) = ((xtan ϑ−vx sec^2  ϑ)/(x sec^2 ϑ))  ϑ+x (dϑ/dx) = sin ϑ cos ϑ−ϑ   x (dϑ/dx) = (1/2)sin 2ϑ−2ϑ   ((2 dϑ)/(sin 2ϑ−4ϑ)) = (dx/x)
$${set}\:\frac{{y}}{{x}}\:=\:\vartheta\:\Rightarrow{y}=\vartheta{x} \\ $$$$\frac{{dy}}{{dx}}\:=\:\vartheta\:+\:{x}\:\frac{{d}\vartheta}{{dx}} \\ $$$$\left({x}\:\mathrm{tan}\:\vartheta−\vartheta{x}\:\mathrm{sec}\:^{\mathrm{2}} \vartheta\right){dx}\:=\:{x}\:\mathrm{sec}\:^{\mathrm{2}} \vartheta{dy} \\ $$$$\frac{{dy}}{{dx}}\:=\:\frac{{x}\mathrm{tan}\:\vartheta−{vx}\:\mathrm{sec}^{\mathrm{2}} \:\vartheta}{{x}\:\mathrm{sec}\:^{\mathrm{2}} \vartheta} \\ $$$$\vartheta+{x}\:\frac{{d}\vartheta}{{dx}}\:=\:\mathrm{sin}\:\vartheta\:\mathrm{cos}\:\vartheta−\vartheta\: \\ $$$${x}\:\frac{{d}\vartheta}{{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\vartheta−\mathrm{2}\vartheta\: \\ $$$$\frac{\mathrm{2}\:{d}\vartheta}{\mathrm{sin}\:\mathrm{2}\vartheta−\mathrm{4}\vartheta}\:=\:\frac{{dx}}{{x}} \\ $$

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