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


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


Separating variables:


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



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


Therefore, p=b/x³ where b=a³.

Back substitution:




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