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)

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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(A**v**) = 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

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