4 cosxy x 1 4 x2 equating coe cients of like terms 1

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Unformatted text preview: solution (4+x2)M2(x) = 0, equation is (c) The general (x) = 1, of the nonhomogeneous ) M1 = 4 + x2 1 principle of superposition, and verify 2y 2y c) Form ) = general solution )e u(x, y the F (x)e + G(x using the 2 sin(xy ). Di erentiating gives 4+x 1 Thus p correct: that ituis(x, y ) = 4+x2 sin(xy ). x 2y 2y uy = 2 the nonhomogeneous (c) The general solution of F (x)e + 2G(x)e equationcos(xy ) 4 + x2 is 1 u(x, y ) = F (x)e 2y + G(x)e2y 4+x2 sin(xy ). Dix2 rentiating gives e uyy = 4F (x)e 2y + 4G(x)e2y + x sin(xy ) 2y 2y 4 + x2 cos(xy ) uy = 2F (x)e + 2G(x)e 2 Substituting into the PDE shows that u(x, y ) is 4 2 solution: a+ x x sin(xy ) 4 uyy 2y 4F (2x)e 2y + 4G(x)e2y + 2y = x 2y 4u = 4F (x)e +4G(x)e + sin(xy ) 4F (x)e 4 + x2(x)e2y + 4G sin(xy ) = sin(xy ). 4 + x2 4 + x2 Substituting into the PDE shows that u(x, y ) is a solution: 4. uyy + x2uy = 0 x2 4 2y 4u = 4F (x)e 2y +4G(x)e2y + sin(xy 4 = 4G(x)e2y + sin(xy ) = sin(xy ). @ 4 + x2 2 uyy ) 4uF (x)e 4 + x2 (a) Rewrite PDE as: ( uy + x u) = 0 @y 2 2 4. uyy 4F xxue = w.r.t. x)e2y + x x...
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This document was uploaded on 02/09/2014.

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