(x^2+y^2)dx +3xydy=0

Question

Learning about homogeneous functions and this was the example given to practice this topic.

Answer 1

This is a long problem to solve.

Rearranging: dy/dx = - (x² + y²)/3xy  This is in fact homogeneous because the numerator degree = the denominator degree

 

Multiply top and bottom by 1/x², and distribute:

dy/dx = - (1 + y²/x²)/(3y/x)

 

Let v = y/x, so y = xv, (i)        and also dy/dx = -(v² + 1)/3v

and from (i): dy/dx = v + xdv/dx 

 

Therefore: v + xdv/dx = -(v² + 1)/3v

so  xdv/dx= -[4v/3 + 1/(3v)]

and xdv    = -[4v/3 + 1/(3v)]dx

so  -dv/[4v/3 + 1/(3v)]  =  1/x dx 

or [-3/(4v) - 3v]dv = 1/x dx 

integrating both sides gives us:

(-3/4) ln v - 3/2 v² = lnlxl + C

-3 ln v - 6 v²   = 4lnlxl + C   Now replace v by y/x

 

-3ln(y/x) - 6(y/x)² = 4lnlxl + C   

 or 3x²ln(y/x) - 6y² = 4lnlxl + C