Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable. Select the values for A, B, and C dy/dx=(xy+y^2)/(3x^2+y^2) becomes integral (u^2+A)/(u^3+Bu^2-4u)

Answer:[tex]\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}[/tex]Step-by-step explanation:First step: I'm going to solve our substitution for y:[tex]u=\frac{y}{x}[/tex]Multiply both sides by x:[tex]ux=y[/tex]Second step: Differentiate the substitution:[tex]u'x+u=y'[/tex]Third step: Plug in first and second step into the given equation dy/dx=f(x,y):[tex]u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}[/tex][tex]u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}[/tex]We are going to simplify what we can.Every term in the fraction on the right hand side of equation contains a factor of [tex]x^2[/tex] so I'm going to divide top and bottom by [tex]x^2[/tex]:[tex]u'x+u=\frac{u+u^2}{3+u^2}[/tex]Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:Subtract u on both sides:[tex]u'x=\frac{u+u^2}{3+u^2}-u[/tex]Find a common denominator: Multiply second term on right hand side by [tex]\frac{3+u^2}{3+u^2}[/tex]:[tex]u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}[/tex]Combine fractions while also distributing u to terms in ( ):[tex]u'x=\frac{u+u^2-3u-u^3}{3+u^2}[/tex][tex]u'x=\frac{-u^3+u^2-2u}{3+u^2}[/tex]Third step: I'm going to separate the variables:Multiply both sides by the reciprocal of the right hand side fraction.[tex]u' \frac{3+u^2}{-u^3+u^2-2u}x=1[/tex]Divide both sides by x:[tex]\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}[/tex]Reorder the top a little of left hand side using the commutative property for addition:[tex]\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}[/tex]The expression on left hand side almost matches your expression but not quite so something seems a little off.