Amath 350 page 5 assignment 7 solutions fall 2013 4

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Unformatted text preview: 2 1 , A= 5 0 10 50 , F= . (b) Equilibrium solution is given by y(t) = k, t 2 IR where k is a constant 2-vector satisfying Ak + F = 0. Thus 0 k = A 1 ( F) = 1 1 5 10 50 2 5 = 50 22 . (c) First find general solution of associated homogeneous equation: y0 = Ay. 2 5 Characteristic equation is: det = 0 ) 2 2 + 5 = 0. 1 Thus the eigenvalues are: 1 ± 2i. 1 2i 5 v1 0 Eigenvector for = 1 + 2i: = ) v2 = 1 52i v1 . 1 1 2i v2 0 5 5 0 Choose v = = +i = u + iw. The eigenvector for ¯ = 1 2i 1 2i 1 2 ¯ is v = u iw. From Theorem 4.22 in the Course Notes, the general solution of the homogeneous equation is 5 1 yh (t) = c1 et cos(2t) 0 2 sin(2t) + c2 et sin(2t) 5 1 + cos(2t) 0 2 . A particular solution is the equilibrium solution found above, i.e. yp (t) = [50, 22]T . Thus the general solution of the DE is et [5c1 cos(2t) + 5c2 sin(2t)] + 50 e [(c1 2c2 ) cos(2t) + (2c1 + c2 ) sin(2t)] + 22 y ( t) = t . 5. Solve the following boundary value problem. ux y u = xey , u(1, y ) = cos(y ). (4) Treating y as a constant, this can be viewed as a first order ODE in x, with R integrating factor e ydx = e xy . Incorporating this, we have e xy ux =) @ e @x u= ey xe y Integrate with respect to x: e xy xy ye xy u = xey e xy u = xey e xy 4 ey e y2 xy xy . + F (y ) . Solve for u to obtain the general solution: u (x, y ) = xe y y ey + exy F (y ) . y2 Apply the boundary conditions: u (1, y ) = cos (y ) =) cos (y ) = ey y ey + ey F (y ) y2 1 1 + 2 + e y cos (y ) . yy Hence the solution to the given BVP is ✓ ◆ xe y ey 1 1 xy y u (x, y ) = +e + + e cos y . y y2 y y2 =) F (y ) = The domain of existence is {(x, y ) 2 R2 |y > 0} or {(x, y ) 2 R2 |y < 0}. 12 (⇠ ⌘ )3/2 + ⌘ (⇠ 2 ⌘ )1/2 + F (⌘ ) AMATH 350 Assignment 3 Solutions - Fall 2013 #7 Page 6 1 Thus the solution of the original PDE is u(x, y ) = x3 + (y 2 x2)x + F (y 2 3 6. Consider the PDE the same as what was found in part (b). The same solution x2), which is p uyy 4 0, sin x = (5) would be found if we took x <u =i.e., (xy ) . ⇠ 2 ⌘ . = 3. uyy Find th...
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