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Question Number 128689 by bemath last updated on 09/Jan/21

(x^3 +y^3 ) dx = 3xy^2 dy

$$\:\left(\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \right)\:\mathrm{dx}\:=\:\mathrm{3xy}^{\mathrm{2}} \:\mathrm{dy}\: \\ $$

Answered by liberty last updated on 09/Jan/21

(dy/dx) = ((x^3 +y^3 )/(3xy^2 )) ⇒ (dy/dx) = (1/3)x^2 y^(−2) + (y/(3x)) let v = y^3 ⇒ (dv/dx) = 3y^2 (dy/dx) or (dy/dx) = (1/3)y^(−2) (dv/dx) then (1/3)y^(−2) (dv/dx) = (1/3)x^2 y^(−2) + (y/(3x)) ⇔ (dv/dx) = x^2 +(y^3 /x) ; (dv/dx) − (v/x) = x^2 put integrating factor μ = e^(∫−(dx/x)) = (1/x) we get v = ((∫ x^2 ((1/x)) dx + C)/(((1/x)))) v = x ((1/2)x^2 +C)⇒ y= (((1/2)x^3 +Cx))^(1/3) ⇔ y = (1/2) ((4x^3 +λx))^(1/3) .

$$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} }{\mathrm{3xy}^{\mathrm{2}} }\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \mathrm{y}^{−\mathrm{2}} \:+\:\frac{\mathrm{y}}{\mathrm{3x}} \\ $$$$\:\mathrm{let}\:\mathrm{v}\:=\:\mathrm{y}^{\mathrm{3}} \:\Rightarrow\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\mathrm{3y}^{\mathrm{2}} \:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{or}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{y}^{−\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\:\mathrm{then}\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{y}^{−\mathrm{2}} \:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x}^{\mathrm{2}} \mathrm{y}^{−\mathrm{2}} \:+\:\frac{\mathrm{y}}{\mathrm{3x}} \\ $$$$\:\Leftrightarrow\:\frac{\mathrm{dv}}{\mathrm{dx}}\:=\:\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}}\:;\:\frac{\mathrm{dv}}{\mathrm{dx}}\:−\:\frac{\mathrm{v}}{\mathrm{x}}\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{put}\:\mathrm{integrating}\:\mathrm{factor}\:\mu\:=\:\mathrm{e}^{\int−\frac{\mathrm{dx}}{\mathrm{x}}} \:=\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{v}\:=\:\frac{\int\:\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{x}}\right)\:\mathrm{dx}\:+\:\mathrm{C}}{\left(\frac{\mathrm{1}}{\mathrm{x}}\right)} \\ $$$$\:\mathrm{v}\:=\:\mathrm{x}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}} \:+\mathrm{C}\right)\Rightarrow\:\mathrm{y}=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{3}} +\mathrm{Cx}} \\ $$$$\Leftrightarrow\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sqrt[{\mathrm{3}}]{\mathrm{4x}^{\mathrm{3}} +\lambda\mathrm{x}}\:.\: \\ $$

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