Differential equation
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(x²+2y²)+(4xy-y²)(dy/dx)=0.

Let y=vx then dy/dx=v+xdv/dx and:

(x²+2v²x²)+(4vx²-v²x²)(dy/dx)=0,

(1+2v²)+(4v-v²)(v+xdv/dx)=0 (dividing through by x²≠0),

xdv/dx=-(1+2v²)/(4v-v²)-v=(v³-6v²-1)/(4v-v²).

Separating variables:

(v(v-4)/(v³-6v²-1))dv=-dx/x.

Let p=v³-6v²-1, then dp=3v²-12v=[3v(v-4)]dv,

v(v-4)dv=dp/3.

So:

dp/(3p)=-dx/x and, integrating: ⅓ln(p)=ln(a/x) where a is constant of integration.

ln(p)=3ln(a/x)=ln(a/x)³.

Therefore, p=b/x³ where b=a³.

Back substitution:

v³-6v²-1=b/x³,

(y/x)³-6(y/x)²-1=b/x³,

y³-6xy²-x³=b.

by Top Rated User (796k points)

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