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Question Number 103664 by byaw last updated on 16/Jul/20

Answered by bramlex last updated on 16/Jul/20

(x^2 +y^2 )dx = 2xy dy (dy/dx) = ((x^2 +y^2 )/(2xy)) ; set y = zx ⇒(dy/dx) = z + x (dz/dx) ⇔ z+x (dz/dx) = ((x^2 +z^2 x^2 )/(2x^2 z)) x (dz/dx) = ((1+z^2 )/(2z))−z =((1−z^2 )/(2z)) ((2z)/(1−z^2 )) = (dx/x) ⇒∫((d(1−z^2 ))/(1−z^2 ))=−ln∣x∣+c ln∣1−z^2 ∣ = ln∣(C/x)∣ ⇒1−z^2 = (C/x) 1−(C/x) = (y^2 /x^2 ) ⇒y^2 =x^2 −Cx ⊛

$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:=\:\mathrm{2xy}\:\mathrm{dy}\: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{2xy}}\:;\:\mathrm{set}\:\mathrm{y}\:=\:\mathrm{zx}\: \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{z}\:+\:\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}} \\ $$$$\Leftrightarrow\:\mathrm{z}+\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} }{\mathrm{2x}^{\mathrm{2}} \mathrm{z}} \\ $$$$\mathrm{x}\:\frac{\mathrm{dz}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }{\mathrm{2z}}−\mathrm{z}\:=\frac{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }{\mathrm{2z}} \\ $$$$\frac{\mathrm{2z}}{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:=\:\frac{\mathrm{dx}}{\mathrm{x}}\:\Rightarrow\int\frac{\mathrm{d}\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)}{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }=−\mathrm{ln}\mid\mathrm{x}\mid+\mathrm{c} \\ $$$$\mathrm{ln}\mid\mathrm{1}−\mathrm{z}^{\mathrm{2}} \mid\:=\:\mathrm{ln}\mid\frac{\mathrm{C}}{\mathrm{x}}\mid\:\Rightarrow\mathrm{1}−\mathrm{z}^{\mathrm{2}} =\:\frac{\mathrm{C}}{\mathrm{x}} \\ $$$$\mathrm{1}−\frac{\mathrm{C}}{\mathrm{x}}\:=\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} −\mathrm{Cx}\:\circledast \\ $$

Commented by byaw last updated on 16/Jul/20

Thank you. I am greatful.

$$\mathrm{Thank}\:\mathrm{you}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{greatful}. \\ $$

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