Question

Consider the partial differential equation, with the initial condition: 1 2yuz + 3x?uy = 9x?y?, u(x,0) = x3 + 1 Find the characteristic curves and the orthogonal trajectories and sketch both on the same graph. Find a solution of the partial differential equation with the given initial con- dition valid in the first quadrant of the (x, y)-plane. Is this solution unique? Explain.

Answer 1

solution :- (1) Given that 3x2 24ux + 3x ly = 9 x y z condition u(7,0) = +3+1 characteristic equation dy 2y arry from first & second fraction da dy 24 3x 3x dx = 2y dy Integrating Sexy dx = frydy on both sides 24r + 3x 3 x² - yr ac, From second & third fraction dy dz - 34r dy = d2 akryy 3 on both sides Integrating S syr dy = f de 343 + = 7 3 z = 434 C2

Orthogonal curves x 3_gr = ca 2-4 = 6 General solution C =0(c) z- y3 – 0 (x3 _ yr) z = y3 + $ (23_gr) uca, y) = y3 + Ø (x² - y²) Condition uca,0) = 0 +0 (23) u(1,0) = +3+1 0 083) = ! +3+1 Ø (x3 - y) = Hence required sdution 16,4) = 48+0003-y") ulx, y) = y3 + abym 23-441 O 2

I HOPE THIS ANSWER WILL BE HELPFUL TO YOU

PLEASE LIKE THIS ANSWER

THANK YOU