20d_04s_2s

20d_04s_2s - Math 20D Second Midterm Solutions May 26, 2004...

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Unformatted text preview: Math 20D Second Midterm Solutions May 26, 2004 1. (a) An integrating factor is exp y(x) = 1 cos x - tan x dx = cos x. Thus (y cos x) = cos x and sin x + C = tan x + C sec x. cos x cos x dx = (b) Since the characteristic equation for the homogeneous equation is 0 = r2 - 2r + 1 = (r - 1)2 , y = C1 et + C2 tet is the general solution to the homogeneous equation. By undetermined coefficients, a particular solution is y = At + B. Since y = A and y = 0, we have 4t = 0 - 2A + (At + B) = At + (B - 2A). Thus A = 4 and B = 8. The general solution to (b) is therefore y = C1 et + C2 tet + 4t + 8. (c) Rearrange, separate variables and integrate: dy = y dx 1 + x2 and so ln y = arctan x + C. You may leave the answer this way, with or without absolute values on y, or you may solve for y. 2. Since 4 + x2 = 0 for x = 2i, the radius of convergence of the series for Thus the best we can guarantee is |x| < 2. 1-x 4+x2 is 2. 3. Since y - 2y /t + y/t2 = et and W [t, t2 ] = t 2 et dt + t2 t2 t 1 t2 2t = t2 , a particular solution is et dt . t y = -t tet dt = -t t2 et dt + t2 4. By the table for Laplace transforms, sY (s) + y(0) + e-s Y (s) = 1/s and so Y (s) = 1 . s(s + e-s ) ...
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