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AMath350.F13.A7.Sol

# Consider the pde the same as what was found in part b

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Unformatted text preview: e complementary function: 4u = sin(xy ). a) (a) Associated homogeneous equation: uyy 4u = 0. Characteristic equation: 2 4 = 0 ) = 2, +2. General solution: uh (x, y ) = F (x)e 2y + G(x)e2y (F, G arbitrary functions). b) Find a particular solution: (b) Let up(x, y ) = M1 (x) sin(xy ) + M2 (x) cos(xy ). Derivatives: @ up (b) Let up(x, y ) = M1 (x=sin(xy ) x) cos(xy ) xM2 (x) sin(xy ) ) xM1( + M2 (x) cos(xy ). Derivatives: @y @ up @ 2up = xM1(x) cos(xy ) xM2 (x) sin(xy ) 2 @ y = x M1 (x) sin(xy ) x M2 (x) cos(xy ) @2 2 y @ up 2 2 = Substitute into the PDE: x M1 (x) sin(xy ) x M2 (x) cos(xy ) @ y2 2 x2M1 (x) into the PDE: Substitutesin(xy ) x M2 (x) cos(xy ) 4M1(x) sin(xy ) 4M2(x) cos(xy ) = sin(xy ). uyy uyy Equating ) sin(xy ) x2Mlike)terms: ) 4M1(x) sin(xy ) 4M2(x) cos(xy ) = sin(xy ). x2M1 (x coe cients of 2 (x cos(xy 1 Equating coe ) cients of (4+x2)M2(x) = 0, ) M1 = like terms: , M2 = 0. (4+x2)M1(x = 1, 4 + x2 1 , M2 = 0. (4+x2)M1 ) = 1,1 (4+ ) 2 Thus up(x, y(x) = 4+x2 sin(xyx. )M2(x) = 0, ) M1 = 4 + x2 (c) The general solution1 of the nonhomogeneous equation is c) Form the x, y ) = 2solution, and. verify that it is correct: Thus up( general sin(xy ) 1 u(x, y ) = F (x)e y 4+xG(x)e2y 4+x2 sin(xy ). Di erentiating gives +2 (c) The general solution of the nonhomogeneous equation is x 1 u(x, y ) = F (xue 2y + G(x(x2y 2y4+x2 G(xxy2y Di erentiating ) ) y = 2F )e )e + 2sin( )e ). cos(xy gives 4 + x2 2x = 2F )e 2 2y 4G G)e )e2y x sin(xy ) uuy = 4F (x(x)ey + + 2(x(x2y + 4 +2x2 cos(xy ) yy 4 +2 x x 2y 2y sin(xy uyy = PDE e + 4G( u e + Substituting into the 4F (x)shows thatx)(x, y ) is a solution:) 4 + x2 2 Substituting (x)e y + x sin(xy ) 4 u x, is 4 solution: 4 sin(xy ) = sin(xy ). 4u = 4F (x)e 2y +4Ginto 2the PDE2 shows thatF ((x)ey )2y a G(x)e2y + 4 +2 x 4 + x2 x 4 4u uyy4+ (x)ey 2y +4G(x)e2y + sin(xy ) 4uF (x)e 2y 4G(x)e2y + 4= sin(xy ) = sin(xy ). 4. = F x2u = 0 4 + x2 uyy 4 + x2 @ 2 4. (a) 4F x2uye = y0 G (x) e2y uy + x2u) =(0 ) 4F (x) e 2y 4G (x) e2y + 4 sin (xy ) uyy Rewrite PDE as: + (x) 2 +4 (+ x sin xy @ y 4 + x2 4 + x2 In...
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