2 apply second bc uy 1 y e 2y e y g1 e2y g1

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Unformatted text preview: ) e y G(1) = e2y ) G(1) = e3y . This is impossible, so the BVP has no solutions. This is impossible, so the BVP has no solutions. 2 5. (a) yuxx 4uxy + 2yuyy + exy ux x3u = xyex 2 xy 3 x 5. (a) yuxx 4uxy + 2yuyy + e u2 x u = xye 2 a = y, b = 4, c = 2y ) b x 4ac = 16 8y 2. a = y, b = 4, c = 2y ) b2 4ac = 16 8y . Equation is hyperbolic if 16 8y 2 > 0 Equation is hyperbolic if 16 8y 2 > 0 Equation is parabolic if 16 8y 2 = 0 ) y = ± 2 Equation is parabolic if 16 8y 2 = 0 ) y = ± 2 Equation is elliptic if 16 8y 2 < 0 Equation is elliptic if 16 8y 2 < 0 (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux a = 2, b = cos(x), c = (1 sin2(x)) a = 2, b = cos(x), c = (1 sin2(x)) ) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). ) b2 4ac = cos2 (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). uy = tan(x) uy = tan(x) AMATH 350 Assignment #7 Solutions - Fall 2012 Page 8 8. 2ux + uy + u = 0; u(x, 0) = cos x. This is a first-order linear PDE, and we cannot solve i...
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