Show that dy/dx=(4y-3x)/(2x-y) is homogeneous.
xv=y
v'x+v=y'
v'x+v=(4y-3x)/(2x-y)
Show that dy/dx=(4y-3x)/(2x-y) is homogeneous.
xv=y
v'x+v=y'
v'x+v=(4y-3x)/(2x-y)
dy/dx=(4y - 3x)/(2x - y)
(2x - y)dx - (4y - 3x)dy = 0
If you multiply x and y by a scalar, say A, you get
(2Ax - Ay)dx - (4Ay - 3Ax)dy = A(2x - y)dx - A(4y - 3x)dy
= A[(2x - y)dx - (4y - 3x)dy] = 0
Since this scales the whole left hand side by A, the equation is homogeneous.
In general:
A polynomial P(v) with all terms having degree n is homogeneous as P(Av) = A^{n}P(v) for all vectors v and scalars A.
A differential equation of the form dy/dx = M(x,y) / N(x,y) is homogeneous if M and N are homogeneous of the same degree.
Proof:
Let M and N be both homogeneous polynomials of degree K.
Rewrite M(x,y) dx - N(x,y) dy = 0
Now M(Ax,Ay) dx - N(Ax,Ay) dy = A^{K}M(x,y) dx - A^{K}N(x,y)dy = A^{K}[M(x,y) dx - N(x,y) dy].
(4y - 3x)dx + (2x - y)dy = 0
y = vx, dy = xdv + vdx
(4y - 3x)/(2x - y) = (4v - 3)/(2 - v)
λ = 1/x