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AMath350.F12.A7.Sol

# Thus the general solution of the de is et 5c1 cos2t

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Unformatted text preview: + 22 y ( t) = t AMATH 350 . Assignment 6 Solutions c S.A. Campbell 2010 5. 1. ux y u = xey . Fixing y , this can be treated as a ﬁrst order linear ODE w.r.t. x, with integration factor: e y dx = e xy . e xy ux ye xy u = xey e xy @ (e xy u) = xey e xy @x Integrate w.r.t. x holding y ﬁxed. Use int. by parts on rhs. ey ey xy xe xy e + F (y ) e xy u = y y2 4 xey y ey + exy F (y ), F an arbitrary function. y2 ey ey + ey F (y ) Apply B.C.: u(1, y ) = cos(y ) ) cos(y ) = y y2 1 1 ) F (y ) = + 2 + e y cos(y ). yy 1 1 xey ey + exy + 2 + e y cos(y ) . Solution of BVP: u(x, y ) = 2 y y yy Domain of existence: {(x, y ) IR2 | y > 0} or {(x, y ) IR2 | y < 0}. General solution: u(x, y ) = 2. yux + xuy = y 3 , x > 0, y > 0. (a) Characteristic equation: dy x = . Solving gives dx y Z So the characteristics are y 2 y dy = Z x dx y2 x2 = +C 2 2 x2 = k . Sketch: = (⇠ ⌘) + ⌘ (⇠ ⌘) + F (⌘ ) Assignment #7 Solutions - Fall 2012 1 Page 6 Thus the solution of the original PDE is u(x, y ) = x3 + (y 2 x2)x + F (y 2 3 x2), which is the same as what was found in part (b). The same solution 4u = be found would sin (xy ) . if we took x < 0, i.e., x = ⇠ 2 ⌘. AMATH 350 6. uyy 3 3. u...
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