solve the following differential equation?

Question

solve the following differential equation

( x2 + 3xy + y2 )dx - x2dy = 0

Answer 1

( x ² + 3xy + y ² ) = x ² dy/dx
dy/dx = 1 + 3 y/x + ( y/x ) ²
Let y = vx
dy/dx = v + x dv/dx
v + x dv/dx = 1 + 3v + v ²
x dv/dx = 1 + 2 v + v ²
x dv/dx = ( 1 + v ) ²
∫ dv / ( 1 + v ² ) = ∫ (1/x) dx
tan^(-1) v = log x + C
tan^(-1) (y / x) = log x + log k
tan^(-1) (y / x) = log (k x)

Answer 2

Edited
( x^2 + 3xy + y^2 )dx - x^2dy = 0
dy/dx = (x^2 + 3xy + y^2) / x^2
This is a homogeneous differential equation.
Let y = vx => dy/dx = v + xdv/dx
=> v + x dv/dx = (x^2 + 3vx^2 + v^2x^2) / x^2
=> x dv/dx = 1 + 3v + v^2 - v
=> dv / (v^2 + 2v + 1) = dx/x
=> dv (v + 1)^2 = dx/x
Integrating,
- 1 / (v + 1) = ln l x l + c
=> - 1 /(y/x + 1) = ln l x l + c

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Answer 3

( x2 + 3xy + y2 )dx - x2dy = 0
or, dy/dx = (x2 + 3xy +y2)/x2
putting, y= vx
we have, v + x(dv/dx) = 1 + 3v + v2
or, dv/(1 + v)2 = dx/x
Inte., - 1/(1+v) = logx + logc , c= constant
or, (x+y)logc + x = 0

Answer 4

x2+3xy+y2=x2(dy/dx)
1+(3/2)y+(y/x)=(dy/dx)
integrate on both the sideswith respect to x
x+(3/2)xy-y(1/xsquare)=y

Answer 5

You could try dividing by x^2 and substituting for v=y/x to make a separable equation, or multiply it by a factor to make it exact, but no guarantees if either would work. Good luck.