Question #25830

Is the function given by u(x,y) = x^2+y^2 a solution of the pde yu_x - xu_y = 0? Why or why not?

Expert's answer

yu_x - xu_y = 0

u(x,y) = x^2+y^2

At first we need to calculate u_x and u_y and then substitute obtained expressions into the initial equation:

u_x = d/dx[ x^2+y^2] = 2x

u_y = d/dy[ x^2+y^2] = 2y

y*u_x = y*2x = 2xy

x*u_y = x*2y = 2xy

yu_x - xu_y = 2xy - 2xy = 0

Therefore the given function is a solution of our pde.

Answer:& the function given by u(x,y) = x^2+y^2 is a solution of the pde yu_x - xu_y = 0

u(x,y) = x^2+y^2

At first we need to calculate u_x and u_y and then substitute obtained expressions into the initial equation:

u_x = d/dx[ x^2+y^2] = 2x

u_y = d/dy[ x^2+y^2] = 2y

y*u_x = y*2x = 2xy

x*u_y = x*2y = 2xy

yu_x - xu_y = 2xy - 2xy = 0

Therefore the given function is a solution of our pde.

Answer:& the function given by u(x,y) = x^2+y^2 is a solution of the pde yu_x - xu_y = 0

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