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x-2-y-2-2x-dy-2y-dx-0-y-0-1-

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Question Number 127780 by bemath last updated on 02/Jan/21
(x^2 +y^2 +2x) dy−2y dx = 0 y(0) = 1
$$\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2x}\right)\:\mathrm{dy}−\mathrm{2y}\:\mathrm{dx}\:=\:\mathrm{0} \\ $$$$\:\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$
Answered by liberty last updated on 02/Jan/21
(dy/dx) = ((2y)/(x^2 +y^2 +2x)) ; ⇒ (dx/dy) = ((x^2 +y^2 +2x)/(2y)) let x = vy ⇒(dx/dy) = v + y (dv/dy) ⇒ v+ y (dv/dy) = ((v^2 y^2 +y^2 +2vy)/(2y)) ⇒v + y (dv/dy) = (1/2)v^2 y+(1/2)y+v ⇒ y (dv/dy) = (1/2)v^2 y + (1/2)y ⇒ (dv/dy) = (1/2)v^2 + (1/2) ⇒ ((2dv)/(v^2 +1)) = dy ⇒ 2v arctan (v)+C = y ⇒ 2((x/y)) arctan ((x/y))+C = y where y(0)=1⇒ 0+C = 1 ; C = 1 we get solution : y =2((x/y))arctan ((x/y))+1 or y^2 −y =2x arctan ((x/y)).
$$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{2y}}{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2x}}\:;\:\Rightarrow\:\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2x}}{\mathrm{2y}} \\ $$$$\mathrm{let}\:\mathrm{x}\:=\:\mathrm{vy}\:\Rightarrow\frac{\mathrm{dx}}{\mathrm{dy}}\:=\:\mathrm{v}\:+\:\mathrm{y}\:\frac{\mathrm{dv}}{\mathrm{dy}} \\ $$$$\Rightarrow\:\mathrm{v}+\:\mathrm{y}\:\frac{\mathrm{dv}}{\mathrm{dy}}\:=\:\frac{\mathrm{v}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2vy}}{\mathrm{2y}} \\ $$$$\Rightarrow\mathrm{v}\:+\:\mathrm{y}\:\frac{\mathrm{dv}}{\mathrm{dy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}^{\mathrm{2}} \mathrm{y}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}+\mathrm{v}\: \\ $$$$\Rightarrow\:\mathrm{y}\:\frac{\mathrm{dv}}{\mathrm{dy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}^{\mathrm{2}} \mathrm{y}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{y}\: \\ $$$$\Rightarrow\:\frac{\mathrm{dv}}{\mathrm{dy}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\frac{\mathrm{2dv}}{\mathrm{v}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{dy}\: \\ $$$$\Rightarrow\:\mathrm{2v}\:\mathrm{arctan}\:\left(\mathrm{v}\right)+\mathrm{C}\:=\:\mathrm{y}\: \\ $$$$\Rightarrow\:\mathrm{2}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\:\mathrm{arctan}\:\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{C}\:=\:\mathrm{y} \\ $$$$\mathrm{where}\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow\:\mathrm{0}+\mathrm{C}\:=\:\mathrm{1}\:;\:\mathrm{C}\:=\:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{solution}\::\:\mathrm{y}\:=\mathrm{2}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{arctan}\:\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{1} \\ $$$$\mathrm{or}\:\mathrm{y}^{\mathrm{2}} −\mathrm{y}\:=\mathrm{2x}\:\mathrm{arctan}\:\left(\frac{\mathrm{x}}{\mathrm{y}}\right). \\ $$

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