The same solution 4u be found would sin xy if we took

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Unformatted text preview: yy Find the homogeneous solution uh (x, y ): 4u = sin(xy ). a) (a) Associated homogeneous equation: uyy 4u = 0. Characteristic equation: 2 4 = 0 ) = 2, +2. 2y General solution: uh (x, y ) = F (x)e + G(x)e2y (F, G arbitrary functions). b) Find a particular solution up (x, y ): (b) Let up(x, y ) = M1 (x) sin(xy ) + M2 (x) cos(xy ). Derivatives: @ up = xM1(x) cos(xy ) xM2 (x) sin(xy ) @y (b) Let up(x, y ) = M1 (x) sin(xy ) + M2 (x) cos(xy ). Derivatives: @ 2 up 2 2 @ up = x M1 (x) sin(xy ) x M2 (x) cos(xy ) 2 @y = xM1(x) cos(xy ) xM2 (x) sin(xy ) @y Substitute into the PDE: @ 2 up 2 2 2 x M1 (x) sin(xyy 2 x= 2 (x) M1 (x) ) 4xy1(x)x Mxyx) cos(xy )) cos(xy ) = sin(xy ). ) 2M x cos(xy sin( M ) sin( 2 ( ) 4M2(x @ Equating coe cthe PDE: terms: Substitute into ients of like uyy uyy 2 1 x2M1 x) 1(x = sin( M 4 , M2 = 0. (4+x2()Msin()xy ) 1,x M2 (x) x2)M2() ) 4M0,(x)) xy )1 =M2(x) cos(xy ) = sin(xy ). (4+ cos(xy x = 1 4 + x2 Equating coe cients of like terms: 1 Thus up(x, y ) = 4+x2 sin(xy ). 1 , M2 = 0. (4+x2)M1...
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