MATH 3600 Final: MATH 3600 Iowa Final3600ANS
Document Summary
The wronskian of e tcos(t), e tsin(t) is w (e tcos(t), e tsin(t)) = e 2t. = e tcos(t)[e tcos(t) e tsin(t)] e tsin(t)[ e tsin(t) e tcos(t)] = e 2tcos2(t) e 2tcos(t)sin(t) + e 2tsin2(t) + e 2tcos(t)sin(t) Solve t dy dt + y = 8t2 ty + y = 8t2 is a rst order linear de. R [ty] = r 8t2 ty = 8. Longer method: t y = 8t y + 1. Multiply both sides by t: t = eln|t| = |t| ty + y = 8t2 and continue as above. Solve (3x4 + 2y)dx + (2x + 4y3)dy = 0. If x = 3x4 + 2y, then = r (3x4 + 2y)dx = 3. Y = 2x + h (y) = 2x + 4y3. 5 x5 + 2xy + y4 = c dx( 3. 5 x5 + 2xy + y4) = d d dx(c)
