{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AMath350.F12.A7.Sol

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}