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Question Number 129777 by stelor last updated on 18/Jan/21

(x^2 −xy−y^2 )dx+x^2 dy = 0 hello... please help me.

$$\left({x}^{\mathrm{2}} −{xy}−{y}^{\mathrm{2}} \right){dx}+{x}^{\mathrm{2}} {dy}\:=\:\mathrm{0} \\ $$$${hello}...\:{please}\:{help}\:{me}. \\ $$

Answered by Ar Brandon last updated on 18/Jan/21

(x^2 −xy−y^2 )dx=−x^2 dy (dy/dx)=((y/x)+(y^2 /x^2 )−1) Let y=vx ⇒ (dy/dx)=v+x(dv/dx) ⇒v+x(dv/dx)=(v+v^2 −1) ⇒(dv/(v^2 −1))=−(dx/x) ⇒Arctanh(v)=ln(x)+C ⇒v=tanh(ln(x)+C) ⇒y=xtanh(ln(x)+C)

$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{xy}−\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}=−\mathrm{x}^{\mathrm{2}} \mathrm{dy} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\mathrm{y}}{\mathrm{x}}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$\mathrm{Let}\:\mathrm{y}=\mathrm{vx}\:\Rightarrow\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} \\ $$$$\Rightarrow\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\left(\mathrm{v}+\mathrm{v}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\Rightarrow\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{1}}=−\frac{\mathrm{dx}}{\mathrm{x}} \\ $$$$\Rightarrow\mathrm{Arctanh}\left(\mathrm{v}\right)=\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C} \\ $$$$\Rightarrow\mathrm{v}=\mathrm{tanh}\left(\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}\right) \\ $$$$\Rightarrow\mathrm{y}=\mathrm{xtanh}\left(\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{C}\right) \\ $$

Commented by stelor last updated on 18/Jan/21

thank... cool .

$${thank}...\:{cool}\:\:\:. \\ $$

Answered by bramlexs22 last updated on 19/Jan/21

y=vx ⇒ dy = x dv+v dx (x^2 −x^2 v−x^2 v^2 )dx+x^2 (x dv+v dx)=0 (1−v−v^2 )dx+x dv+vdx=0 (1−v^2 )dx+x dv=0 (dx/x) + (dv/(1−v^2 )) = 0 ln ∣x∣ −(1/2) ∫ (1/(1+v))−(1/(1−v)) dv=C ln ∣x∣−(1/2)(ln ∣1+x∣+ln ∣1−v∣)=C ln ∣x∣−(1/2)ln ∣((x^2 −y^2 )/x^2 )∣=C ln ∣C_1 x^4 ∣ = ln ∣x^2 −y^2 ∣ y^2 =x^2 −C_1 x^4 . y=±x(√(1−C_1 x^2 )) .

$$\mathrm{y}=\mathrm{vx}\:\Rightarrow\:\mathrm{dy}\:=\:\mathrm{x}\:\mathrm{dv}+\mathrm{v}\:\mathrm{dx} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} \mathrm{v}−\mathrm{x}^{\mathrm{2}} \mathrm{v}^{\mathrm{2}} \right)\mathrm{dx}+\mathrm{x}^{\mathrm{2}} \left(\mathrm{x}\:\mathrm{dv}+\mathrm{v}\:\mathrm{dx}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}−\mathrm{v}−\mathrm{v}^{\mathrm{2}} \right)\mathrm{dx}+\mathrm{x}\:\mathrm{dv}+\mathrm{vdx}=\mathrm{0} \\ $$$$\:\left(\mathrm{1}−\mathrm{v}^{\mathrm{2}} \right)\mathrm{dx}+\mathrm{x}\:\mathrm{dv}=\mathrm{0} \\ $$$$\:\frac{\mathrm{dx}}{\mathrm{x}}\:+\:\frac{\mathrm{dv}}{\mathrm{1}−\mathrm{v}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$$\:\mathrm{ln}\:\mid\mathrm{x}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{v}}−\frac{\mathrm{1}}{\mathrm{1}−\mathrm{v}}\:\mathrm{dv}=\mathrm{C} \\ $$$$\:\mathrm{ln}\:\mid\mathrm{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{ln}\:\mid\mathrm{1}+\mathrm{x}\mid+\mathrm{ln}\:\mid\mathrm{1}−\mathrm{v}\mid\right)=\mathrm{C} \\ $$$$\:\mathrm{ln}\:\mid\mathrm{x}\mid−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\mid=\mathrm{C} \\ $$$$\mathrm{ln}\:\mid\mathrm{C}_{\mathrm{1}} \mathrm{x}^{\mathrm{4}} \mid\:=\:\mathrm{ln}\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \mid\: \\ $$$$\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} −\mathrm{C}_{\mathrm{1}} \mathrm{x}^{\mathrm{4}} \:. \\ $$$$\:\mathrm{y}=\pm\mathrm{x}\sqrt{\mathrm{1}−\mathrm{C}_{\mathrm{1}} \mathrm{x}^{\mathrm{2}} }\:. \\ $$$$\: \\ $$

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