Reˇ sitve kolokvija MATEMATIKE IV
5.4.2007 skupina A
1. naloga.
x00+ 4x0+ 13x = 0 s2X−3s+ 4(sX −3) + 13X = 0
(s2 + 4s+ 13)X = 3s+ 12 X = 3(s+2)+6(s+2)2+9
x(t) = e−2t(3 cos 3t+ 2 sin 3t)
2. naloga.
y0 =u+xu02x=u+ 2x2u0
y00 =u02x+ 4xu0+ 2x2u002x= 6xu0+ 4x3u00
6x3u0 + 4x5u00−xu−2x3u0+ (4x3−3)xu = 0 4x4u00+ 4x2u0+ (4x4−4)u = 0 z2u00+zu0+ (z2−1)u = 0
u = J1(z) y = xJ1(x2)
3. naloga.
p
Z
0
(qx−1)x dx =qp3 3 −p2
2 −→ qp3 = 12
p
Z
0
(15√
x−q)x dx = 152
p
p5 5 −qp2
2 −→ 6√ p= 2q
zmnoˇzi oboje 2√
p3 = 14 p = 14 q = 6
skupina B
1. naloga.
x00+ 6x0+ 13x = 0 s2X−2s+ 6(sX −2) + 13X = 0
(s2 + 6s+ 13)X = 2s+ 12 X = 2(s+3)+6(s+3)2+4
x(t) = e−3t(2 cos 2t+ 3 sin 2t)
2. naloga.
y0 =u+xu0(−x12) =u− ux0
y00 =u0(−x12)− u
00(−x12)x−u0
x2 = ux003
u00
x2 −u+ ux0 +xu2 = 0 z2u00+zu0+ (z2−1)u = 0
u = J1(z) y = xJ1(1x)
3. naloga.
b
Z
0
(x−a)x dx = b3 3 −ab2
2
b
Z
0
( 4
√x−a)x dx = 42√ b3 3 −ab2
2
izenaˇci oboje
b3
3 = 83√ b3
√
b3 = 8 b = 4 a = 83
