Please help me solve this problem. Transform the equation to new variables (u and v). How do you do this?

#2x (∂z)/(∂x) - y (∂z)/(∂y)=0#
when
#x= u^2v, and y=1/u#

1 Answer
Jun 28, 2018

For #z = z(x,y)#:

  • #2x \ bbz_x - y \ bbz_y=0 qquad square#

The substitutions:

#{(x= u^2v),(y=1/u):} qquad harr qquad {(u = 1/y),(v = xy^2):} to {(u_x = 0, u_ y = - 1/y^2),(v_x = y^2, v_y = 2xy):} #

The partials:

  • #bbz_x = z_u u_x + z_v v_x#

# = z_v y^2 = z_v/u^2#

  • #bbz_y = z_u u_y + z_v v_y #

# = - 1/y^2 z_u + 2xy z_v #

# = - u^2 z_u + 2uv z_v #

Plugging in:

#square implies 2 u^2 v z_v/u^2 - 1/u( - u^2 z_u + 2uv z_v ) = 0 #

# z_v cancel (( 2 v - 2v)) + u z_u = 0 #

#implies z(u,v) = z(v) = z(xy^2)#

Any function with the argument #xy^2# should satisfy this PDE