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Question Number 122006 by bemath last updated on 13/Nov/20

(dy/dx) = ((x−y)/(x+y)) ?

$$\:\frac{{dy}}{{dx}}\:=\:\frac{{x}−{y}}{{x}+{y}}\:? \\ $$

Answered by bobhans last updated on 13/Nov/20

let y = zx ⇒(dy/dx) = z + x (dz/dx) ⇔ z+x (dz/dx) = ((x−zx)/(x+zx)) = ((1−z)/(1+z)) ⇔ x (dz/dx) = ((1−z)/(1+z)) − z = ((1−2z−z^2 )/(1+z)) ⇔ (((1+z) dz)/(z^2 +2z−1)) = −(dx/x) ⇔ ∫ ((d(z^2 +2z−1))/(z^2 +2z−1)) = −2∫ (dx/x) ⇔ ℓn ∣z^2 +2z−1∣ = ℓn∣(C/x^2 ) ∣ ⇔ (z+1)^2 = ((C+2x^2 )/x^2 ) ⇔ (((y+x)/x))^2 = ((C+2x^2 )/x^2 ) ⇔ y+x = ± (√(C+2x^2 )) ⇔ y = −x ± (√(C+2x^2 )) . ⊝

$${let}\:{y}\:=\:{zx}\:\Rightarrow\frac{{dy}}{{dx}}\:=\:{z}\:+\:{x}\:\frac{{dz}}{{dx}} \\ $$$$\Leftrightarrow\:{z}+{x}\:\frac{{dz}}{{dx}}\:=\:\frac{{x}−{zx}}{{x}+{zx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}} \\ $$$$\Leftrightarrow\:{x}\:\frac{{dz}}{{dx}}\:=\:\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}}\:−\:{z}\:=\:\frac{\mathrm{1}−\mathrm{2}{z}−{z}^{\mathrm{2}} }{\mathrm{1}+{z}} \\ $$$$\:\Leftrightarrow\:\frac{\left(\mathrm{1}+{z}\right)\:{dz}}{{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}}\:=\:−\frac{{dx}}{{x}} \\ $$$$\Leftrightarrow\:\int\:\frac{{d}\left({z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}}\:=\:−\mathrm{2}\int\:\frac{{dx}}{{x}} \\ $$$$\Leftrightarrow\:\ell{n}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{1}\mid\:=\:\ell{n}\mid\frac{{C}}{{x}^{\mathrm{2}} }\:\mid \\ $$$$\Leftrightarrow\:\left({z}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{{C}+\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:\left(\frac{{y}+{x}}{{x}}\right)^{\mathrm{2}} =\:\frac{{C}+\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:{y}+{x}\:=\:\pm\:\sqrt{{C}+\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:{y}\:=\:−{x}\:\pm\:\sqrt{{C}+\mathrm{2}{x}^{\mathrm{2}} }\:.\:\circleddash \\ $$

Answered by TANMAY PANACEA last updated on 13/Nov/20

xdx−ydx=xdy+ydy xdy+ydx+ydy−xdx=0=dC d(xy)+d((y^2 /2))−d((x^2 /2))=dC xy+((y^2 −x^2 )/2)=C

$${xdx}−{ydx}={xdy}+{ydy} \\ $$$${xdy}+{ydx}+{ydy}−{xdx}=\mathrm{0}={dC} \\ $$$${d}\left({xy}\right)+{d}\left(\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\right)−{d}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)={dC} \\ $$$${xy}+\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{2}}={C} \\ $$

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