MATH 251 Midterm: MATH 251 PSU s251Exam 2(fa02)
Document Summary
Answers: for y1 = y and y2 = y we have: (x2y y . 2 + xy2 + (x2 2)y1 = 0. 2 = 1 y x2 (x2 2)y1 1 x y2 y1(0) = y0, y2(0) = y . 0 y1(0) = y0, y2(0) = y . 0: y = ae2t + t(b sin(2t) + c cos(2t)) + d sin t + e cos t + f t2 + gt + h. 3. (b) f (t) = t2 + (6 t t2)u2(t) + (t 6)u6(t: it is critically damped, so 2 4km = 0. Therefore, = 4km = 12: l{y(t)} = s3 e2t , y(t) = e 4(t 3))e9u3(t) e2(t 3) + 2: y(t) = e 4t + u3(t)( e 4(t 3)); 2 e 4t + ( (s )2 + 9 3s 6 e3(t 3) . 9. (a) x(t) = c1(cid:18)1 (b) x(t) = c1et(cid:18) 1.
