# Answer to Question #22473 in Differential Equations for Matthew Lind

Question #22473

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.

U_x + 3U_y + U =0.

Expert's answer

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)

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)

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