{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AMath350.F13.A7.Sol

# apply second bc uy 1 y e e g1 e g1 e y this

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: le, so the BVP has no solutions. This is impossible, so the BVP has no solutions. 2 xy 3 x2 5. (a) yuxx 4uxy + 2yuyy + exy ux x33u = xyexx22 5. (a) yuxx 4uxy + 2yuyy + exy ux x33u = xyex 22 xy x 5. (a) yuxx 4uxy + 2yuyy + exy u2x x3u = xyex 2 5. (a) yuxx 4uxy + 2yuyy + exy u2x x u = xyex 2 yy xx a =xx b = xy 4,, c = 2y ) b x 4ac = 16 8y 2.. a = y, b = xy 4 c = yyy ) b2 4ac = 16 8y 2 y, 2 2 a = y, b = 4,, c = 2y ) b2 4ac = 16 8y 2.. a = y, b = 4 c = 2y ) b2 4ac = 16 8y 2 2 Equation is hyperbolic if 16 8y 22 > 0 Equation is hyperbolic if 16 8y22 > 0 p p Equation is hyperbolic if 16 82y2 > 0 y Equation is hyperbolic if 16 82 > 0 Equation is parabolic if 16 8y 2 = 0 ) y = ± p2 Equation is parabolic if 16 8y 2 = 0 ) y = ± p2 Equation is parabolic if 16 28y 2 = 0 ) y = ± 2 Equation is parabolic if 16 8 y 2 = 0 ) y = ± 2 Equation is elliptic if 16 8y 22 < 0 Equation is elliptic if 16 8y22 < 0 Equation is elliptic if 16 8y 2 < 0 Equation is elliptic if 16 8y < 0 2 2 (b) 2uxx + cos(x)uxy + (1 sin22(x))uyy + cos22(x)u + sin(xy )ux (b) 2uxx + cos(x)uxy + (1 sin22(x))uyy + cos22(x)u + sin(xy )ux (b) 2uxx + cos(x)uxy + (1 sin2(x))uyy + cos2(x)u + sin(xy )ux (b) 2uxx + cos(x)uxy + (1 sin (2 ))uyy + cos (x)u + sin(xy )ux x yy xx xy yy x xx x a = 2,, b = cos(xxy,, c = (1 sin 22(x)) a = 2 b = cos(x) c = (1 sin2(x)) ) 2 2(x)) a = 2,, b = cos(x),, c = (1 sin (x)) a = 2 b = cos(x) 2 = (1 sin c 2 2 2 ) b22 4ac = cos22(x) 8(1 sin22(x)) = 7(sin22(x) 1). ) b22 4ac = cos22(x) 8(1 sin22(x)) = 7(sin22(x) 1). 2 2 )b ) b 4ac = cos (x) 8(1 sin2 (x)) = 7(sin2 (x) 1). 4ac = cos (x) 8(1 sin (x)) = 7(sin (x) 1). uy = tan(x) uy = tan(x) uy = tan(x) uy = tan(x) y y AMATH 350 AMATH 350 8. Solve the PDE Assignment #7 Solutions - Fall 2013 Page 8 Assignment #7 Solutions - Winter 2013 Page 7 2 u x + uy + u = 0 8. subject to the= 0; u(condition uxx, 0) = cos x. 2ux + uy + u initial x, 0) = cos ( . This is a ﬁrst-order linear PDE, and we cannot solve it by pa...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online