Answer to Question #22473 in Differential Equations for Matthew Lind

Question

Show that the following equation has a solution of the form u(x,y) = e^(ax + by), and find the constants a,b:

U_x + 3U_y + U =0.

Answer 1

U_x + 3U_y + U =0.
if u(x,y) = e^(ax + by) - a solution
then
(e^(ax + by))_x + 3(e^(ax + by))_y + e^(ax + by) =0.
(e^(ax + by))_x = d/dx (e^(ax + by)) = a*e^(ax +by)
(e^(ax + by))_y = d/dy (e^(ax + by)) = b*e^(ax + by)
a*e^(ax + by) + 3*b*e^(ax + by) + e^(ax + by) = 0
(a+3b+1)*e^(ax + by) = 0
a+3b+1 = 0
a = -(3b+1)
Answer: a = -(3b+1)