Question-114980

Question Number 114980 by srijachakraborty last updated on 22/Sep/20

Answered by Dwaipayan Shikari last updated on 22/Sep/20

(y^4 −2x^3 y)+(x^4 −2xy^3 )(dy/dx)=0 (dy/dx)=((2x^3 vx−v^4 x^4 )/(x^4 −2v^3 x^4 )) (y=vx v+x(dv/dx)=((2v−v^4 )/(1−2v^3 )) (dv/dx)=((2v−v^4 )/(1−2v^3 ))−v (dv/dx)=((v+v^4 )/(1−2v^3 )) ∫(((1−2v^3 )dv)/(v+v^4 ))=∫dx ∫((1+4v^3 )/(v+v^4 ))−((6v^2 )/((1+v^3 )))=x+C log(v+v^4 )−2log(1+v^3 )=x+C log(((v+v^4 )/((1+v^3 )^2 )))=x+C ((v+v^4 )/((1+v^3 )^2 ))=Ce^x (((y/x)+((y/x))^4 )/((x^3 +y^3 ))).x^6 =Ce^x ((x^3 y+y^4 )/((x^3 +y^3 ))).x^2 =Ce^x ((y+y^4 )/((1+y^3 )^2 ))=Ce (y/(1+y^3 ))=Ce (1/2)=Ce C=(1/(2e)) ((x^5 y+x^2 y^4 )/((x^3 +y^3 )^2 ))=(1/2)e^(x−1)

$$\left({y}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} {y}\right)+\left({x}^{\mathrm{4}} −\mathrm{2}{xy}^{\mathrm{3}} \right)\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}^{\mathrm{3}} {vx}−{v}^{\mathrm{4}} {x}^{\mathrm{4}} }{{x}^{\mathrm{4}} −\mathrm{2}{v}^{\mathrm{3}} {x}^{\mathrm{4}} }\:\:\:\:\:\:\:\:\left({y}={vx}\right. \\ $$$${v}+{x}\frac{{dv}}{{dx}}=\frac{\mathrm{2}{v}−{v}^{\mathrm{4}} }{\mathrm{1}−\mathrm{2}{v}^{\mathrm{3}} } \\ $$$$\frac{{dv}}{{dx}}=\frac{\mathrm{2}{v}−{v}^{\mathrm{4}} }{\mathrm{1}−\mathrm{2}{v}^{\mathrm{3}} }−{v} \\ $$$$\frac{{dv}}{{dx}}=\frac{{v}+{v}^{\mathrm{4}} }{\mathrm{1}−\mathrm{2}{v}^{\mathrm{3}} } \\ $$$$\int\frac{\left(\mathrm{1}−\mathrm{2}{v}^{\mathrm{3}} \right){dv}}{{v}+{v}^{\mathrm{4}} }=\int{dx} \\ $$$$\int\frac{\mathrm{1}+\mathrm{4}{v}^{\mathrm{3}} }{{v}+{v}^{\mathrm{4}} }−\frac{\mathrm{6}{v}^{\mathrm{2}} }{\left(\mathrm{1}+{v}^{\mathrm{3}} \right)}={x}+{C} \\ $$$${log}\left({v}+{v}^{\mathrm{4}} \right)−\mathrm{2}{log}\left(\mathrm{1}+{v}^{\mathrm{3}} \right)={x}+{C} \\ $$$${log}\left(\frac{{v}+{v}^{\mathrm{4}} }{\left(\mathrm{1}+{v}^{\mathrm{3}} \right)^{\mathrm{2}} }\right)={x}+{C} \\ $$$$\frac{{v}+{v}^{\mathrm{4}} }{\left(\mathrm{1}+{v}^{\mathrm{3}} \right)^{\mathrm{2}} }={Ce}^{{x}} \\ $$$$\frac{\frac{{y}}{{x}}+\left(\frac{{y}}{{x}}\right)^{\mathrm{4}} }{\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)}.{x}^{\mathrm{6}} ={Ce}^{{x}} \\ $$$$\frac{{x}^{\mathrm{3}} {y}+{y}^{\mathrm{4}} }{\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)}.{x}^{\mathrm{2}} ={Ce}^{{x}} \\ $$$$\frac{{y}+{y}^{\mathrm{4}} }{\left(\mathrm{1}+{y}^{\mathrm{3}} \right)^{\mathrm{2}} }={Ce} \\ $$$$\frac{{y}}{\mathrm{1}+{y}^{\mathrm{3}} }={Ce} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}={Ce} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{2}{e}} \\ $$$$\frac{{x}^{\mathrm{5}} {y}+{x}^{\mathrm{2}} {y}^{\mathrm{4}} }{\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}{e}^{{x}−\mathrm{1}} \\ $$

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