Izpit Matematika IV 27. avgust 2008
Reˇ sitve
1. naloga
f(t) = (1−t)(u0(t)−u1(t)) = (1−t)u0(t) + (t−1)u1(t) F(s) =1s − s12
+s12 − 1se−s= 1s − s12
1−e−s
2. naloga
y=C1x+C3x3+C5x5+· · ·
C1+ 3C3x2+ 5C5x4+· · ·= 1 + (C1x+C3x3+C5x5+· · ·)2 C1+ 3C3x2+ 5C5x4+· · ·= 1 +C12x2 + 2C1C3x4+· · · C1 = 1
3C3 =C12 → C3 = 13 5C5 = 2C1C3 → C5 = 152 y= x+13x3+ 152x5+· · ·
3. naloga
Ker v diferencialni enaˇcbi nastopajo odvodi samo po spremenljivki x, jo lahko reˇsujemo z metodami za navadne diferencialne enaˇcbe za neznanko u(x) in so pridelane konstante odvisne od spremenljivke y.
u00 −u = 0 je linearna dif. enaˇcba s konstantnimi koeficienti. Reˇsimo karakteristiˇcno enaˇcbo in na osnovi korenov zapiˇsemo reˇsitev.
r2−1 = 0 , r1,2 =±1 u=C1(y) coshx+C2(y) sinhx x= 0 → C1(y) = f(y) ux =C1(y) sinhx+C2(y) coshx x= 0 → C2(y) = g(y)
u(x, y) = f(y) coshx+g(y) sinhx
4. naloga
24y−(x22y0)0 = 0 24y−2x2y0 −x22y00= 0 x2y00+ 2xy0 −12y= 0
To je Eulerjevadif. enaˇcba in njene reˇsitve iˇsˇcemo v obliki potenc y =xr x2r(r−1)xr−2+ 2xrxr−1−12xr= 0
[r(r−1) + 2r−12]xr = 0 r2+r−12 = 0
(r−3)(r+ 4) = 0 r1 = 3 , r2 =−4 y=Ax3+Bx−4
A+B = 1 , 8A+B/16 = 8 B = 0 , A= 1
y= x3
5. naloga
P =P(ZZZ) +P( ˇCCˇC) =ˇ 104 · 39 · 28 +106 · 59 · 48 = 15