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

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Question Number 118588 by bramlexs22 last updated on 18/Oct/20
x(x+y) dy −x^2 dx = 0
$${x}\left({x}+{y}\right)\:{dy}\:−{x}^{\mathrm{2}} \:{dx}\:=\:\mathrm{0}\: \\ $$
Commented by Dwaipayan Shikari last updated on 18/Oct/20
x(x+y)(dy/dx)−x^2 =0 (dy/dx)(x+y)=x (dy/dx)=(x/(x+y)) y=vx⇒(dy/dx)=v+x(dv/dx) v+x(dv/dx)=(1/(1+v)) x(dv/dx)=(1/(1+v))−v ∫((1+v)/(1−v−v^2 ))dv=∫(dx/x) ∫((1+v)/(v^2 +v−1))=log((C/x)) ∫(1/(v^2 +v−1))+∫(v/(v^2 +v−1))=log((C/x)) (1/2)∫(1/((v+(1/2))^2 −(((√5)/2))^2 ))+∫((2v+1)/(v^2 +v−1))=log((C/x)) (1/(2(√5)))log(((v+(1/2)−((√5)/2))/(v+(((√5)+1)/2))))+(1/2)log(v^2 +v−1)=log((C/x)) (1/(2(√5)))log((((y/x)−(((√5)−1)/2))/((y/x)+(((√5)+1)/2))))+(1/2)log((y^2 /x^2 )+(y/x)−1)=log((C/x)) (1/(2(√5)))log(((2y−(√5)x+x)/(2y+(√5)x+x)))+(1/2)log(y^2 −xy+x^2 )=log(Cx) ((((2y−(√5)x+x)/(2y+(√5)x+x)))^(1/(√5)) ).(y^2 −xy+x^2 )=C^2 x^2 (((C^2 x^2 )/((y^2 −xy+x^2 ))))^(√5) =((2y−(√5)x+x)/(2y+(√5)x+x)) ((2y+(√5)x+x)/(2y−(√5)x+x))=(((y^2 −xy+x^2 )/(C^2 x^2 )))^(√5)
$${x}\left({x}+{y}\right)\frac{{dy}}{{dx}}−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\left({x}+{y}\right)={x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{{dy}}{{dx}}=\frac{{x}}{{x}+{y}}\:\:\:\:\:\:\:\:{y}={vx}\Rightarrow\frac{{dy}}{{dx}}={v}+{x}\frac{{dv}}{{dx}} \\ $$$${v}+{x}\frac{{dv}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+{v}} \\ $$$${x}\frac{{dv}}{{dx}}=\frac{\mathrm{1}}{\mathrm{1}+{v}}−{v} \\ $$$$\int\frac{\mathrm{1}+{v}}{\mathrm{1}−{v}−{v}^{\mathrm{2}} }{dv}=\int\frac{{dx}}{{x}} \\ $$$$\int\frac{\mathrm{1}+{v}}{{v}^{\mathrm{2}} +{v}−\mathrm{1}}={log}\left(\frac{{C}}{{x}}\right) \\ $$$$\int\frac{\mathrm{1}}{{v}^{\mathrm{2}} +{v}−\mathrm{1}}+\int\frac{{v}}{{v}^{\mathrm{2}} +{v}−\mathrm{1}}={log}\left(\frac{{C}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({v}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} }+\int\frac{\mathrm{2}{v}+\mathrm{1}}{{v}^{\mathrm{2}} +{v}−\mathrm{1}}={log}\left(\frac{{C}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{log}\left(\frac{{v}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}{{v}+\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{log}\left({v}^{\mathrm{2}} +{v}−\mathrm{1}\right)={log}\left(\frac{{C}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{log}\left(\frac{\frac{{y}}{{x}}−\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}}{\frac{{y}}{{x}}+\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\frac{{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{{y}}{{x}}−\mathrm{1}\right)={log}\left(\frac{{C}}{{x}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}{log}\left(\frac{\mathrm{2}{y}−\sqrt{\mathrm{5}}{x}+{x}}{\mathrm{2}{y}+\sqrt{\mathrm{5}}{x}+{x}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{log}\left({y}^{\mathrm{2}} −{xy}+{x}^{\mathrm{2}} \right)={log}\left({Cx}\right) \\ $$$$\left(\sqrt[{\sqrt{\mathrm{5}}}]{\frac{\mathrm{2}{y}−\sqrt{\mathrm{5}}{x}+{x}}{\mathrm{2}{y}+\sqrt{\mathrm{5}}{x}+{x}}}\right).\left({y}^{\mathrm{2}} −{xy}+{x}^{\mathrm{2}} \right)={C}^{\mathrm{2}} {x}^{\mathrm{2}} \\ $$$$\left(\frac{{C}^{\mathrm{2}} {x}^{\mathrm{2}} }{\left({y}^{\mathrm{2}} −{xy}+{x}^{\mathrm{2}} \right)}\right)^{\sqrt{\mathrm{5}}} =\frac{\mathrm{2}{y}−\sqrt{\mathrm{5}}{x}+{x}}{\mathrm{2}{y}+\sqrt{\mathrm{5}}{x}+{x}} \\ $$$$\frac{\mathrm{2}{y}+\sqrt{\mathrm{5}}{x}+{x}}{\mathrm{2}{y}−\sqrt{\mathrm{5}}{x}+{x}}=\left(\frac{{y}^{\mathrm{2}} −{xy}+{x}^{\mathrm{2}} }{{C}^{\mathrm{2}} {x}^{\mathrm{2}} }\right)^{\sqrt{\mathrm{5}}} \\ $$
Answered by 1549442205PVT last updated on 19/Oct/20
x(x+y) dy −x^2 dx = 0 ⇔(x+y)dy−xdx=0(1) Put y=tx⇒dy=xdt+tdx.Then (1)⇔(x+tx)dy−xdx=0 ⇔(1+t)dy−dx=0 ⇔(1+t)(xdt+tdx)−dx=0 ⇔(1+t)xdt+(t^2 +t−1)dx=0 ⇔−(dx/x)=(((1+t)dt)/((t^2 +t−1)))⇔−ln∣x∣=(1/2)∫((d(t^2 +t−1))/(t^2 +t−1)) +(1/2)∫(dt/((t+(1/2))^2 −(((√5)/2))^2 )) ⇔lnC−ln∣x∣=(1/2)ln∣t^2 +t−1∣ +(1/2).(1/( (√5)))ln∣((t+1−(√5))/(t+1+(√5)))∣ ⇔ln(C/(∣x∣))=ln(√(t^2 +t−1))+(1/(2(√5)))ln∣((t+1−(√5))/(t+1+(√5)))⟨ ln(C/(∣x∣))=ln(√((y^2 /x^2 )+(y/x)−1))+(1/(2(√5)))ln∣((y+(1−(√5))x)/(y+(1+(√5))x))∣ (C/(∣x∣))=((√((y^2 /x^2 )+(y/x)−1)))(∣((y+(1−(√5))x)/(y+(1+(√5))x))∣)^(1/(2(√5)))
$${x}\left({x}+{y}\right)\:{dy}\:−{x}^{\mathrm{2}} \:{dx}\:=\:\mathrm{0}\: \\ $$$$\Leftrightarrow\left(\mathrm{x}+\mathrm{y}\right)\mathrm{dy}−\mathrm{xdx}=\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\mathrm{Put}\:\mathrm{y}=\mathrm{tx}\Rightarrow\mathrm{dy}=\mathrm{xdt}+\mathrm{tdx}.\mathrm{Then} \\ $$$$\left(\mathrm{1}\right)\Leftrightarrow\left(\mathrm{x}+\mathrm{tx}\right)\mathrm{dy}−\mathrm{xdx}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{1}+\mathrm{t}\right)\mathrm{dy}−\mathrm{dx}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{1}+\mathrm{t}\right)\left(\mathrm{xdt}+\mathrm{tdx}\right)−\mathrm{dx}=\mathrm{0} \\ $$$$\Leftrightarrow\left(\mathrm{1}+\mathrm{t}\right)\mathrm{xdt}+\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}\right)\mathrm{dx}=\mathrm{0} \\ $$$$\Leftrightarrow−\frac{\mathrm{dx}}{\mathrm{x}}=\frac{\left(\mathrm{1}+\mathrm{t}\right)\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}\right)}\Leftrightarrow−\mathrm{ln}\mid\mathrm{x}\mid=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{d}\left(\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\Leftrightarrow\mathrm{lnC}−\mathrm{ln}\mid\mathrm{x}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}\mid \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{t}+\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{t}+\mathrm{1}+\sqrt{\mathrm{5}}}\mid \\ $$$$\Leftrightarrow\mathrm{ln}\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\mathrm{ln}\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{t}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{t}+\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{t}+\mathrm{1}+\sqrt{\mathrm{5}}}\langle \\ $$$$\mathrm{ln}\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\mathrm{ln}\sqrt{\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}\mathrm{ln}\mid\frac{\mathrm{y}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\mathrm{x}}{\mathrm{y}+\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}}\mid \\ $$$$\frac{\mathrm{C}}{\mid\mathrm{x}\mid}=\left(\sqrt{\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{1}}\right)\left(\mid\frac{\mathrm{y}+\left(\mathrm{1}−\sqrt{\mathrm{5}}\right)\mathrm{x}}{\mathrm{y}+\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)\mathrm{x}}\mid\right)^{\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{5}}}} \\ $$

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