Matematik og modeller 2012: Facit til udvalgte opgaver fra Modul 3–5
Matematik og modeller 2012: Facit til udvalgte opgaver fra Modul 3–5
Matematik og modeller 2012: Facit til udvalgte opgaver fra Modul 3–5
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<strong>Matematik</strong> <strong>og</strong> <strong>modeller</strong> <strong>2012</strong>:<br />
<strong>Facit</strong> <strong>til</strong> <strong>udvalgte</strong> <strong>opgaver</strong> <strong>fra</strong> <strong>Modul</strong> <strong>3–5</strong><br />
DL Opgave 136<br />
For a > 0 <strong>og</strong> b > 0: y = 1<br />
a+b+1 xb+1 +cx −a (c ∈ R).<br />
DL,<br />
(<br />
Opgave 152<br />
x<br />
y)<br />
= e<br />
−t ( 2) −5 +3e<br />
6t ( 1)<br />
1<br />
DL,<br />
(<br />
Opgave 169<br />
x<br />
(<br />
y)<br />
=<br />
7t+9<br />
)<br />
13t +c1 e −t( )<br />
1<br />
1 +c2 e 3t( )<br />
1<br />
5<br />
DL,<br />
(<br />
Opgave 180<br />
3<br />
)<br />
1 , ustabil<br />
DL, Opgave 187<br />
Ustabil<br />
DL, Opgave 192<br />
Ustabil<br />
DL, Opgave 194<br />
(1) ( 1 3 , 1 3 )<br />
DL, Opgave 204<br />
( )<br />
(1) (M ∗ ,N ∗ ) = 30(2α−1) 30<br />
4α−1<br />
,<br />
4α−1<br />
for α ≥ 1 2 .<br />
(2)<br />
1<br />
4α−1<br />
(3) Stabil<br />
( ) 2−4α 1−2α<br />
−1 −2α<br />
DL, Opgave 205<br />
(3,5)<br />
DL, Opgave 206<br />
(30,20); stabil<br />
1
Opgave S.3.1<br />
y = 1 3 ln( 3<br />
2 e2x +c ) (c ∈ R).<br />
Opgave S.3.2<br />
y = − 8 12<br />
13<br />
cos(2x)+<br />
13 sin(2x)+ce−3x (c ∈ R).<br />
Opgave S.3.3<br />
Ligevægte: x ∗ = 0 (stabil for a < 1) <strong>og</strong> x ∗ = lna (stabil for a > 1).<br />
Opgave S.3.4<br />
(a) Ligevægte: N ∗ = 300 (ustabil) <strong>og</strong> N ∗ = 700 (stabil).<br />
(b) H < 50. Den mindste ligevægt er ustabil, mens den største er stabil.<br />
Opgave S.3.5<br />
(a) y = 1<br />
x 2 +c<br />
(c ∈ R).<br />
(b)<br />
(c)<br />
x 0.0 0.2 0.4 0.6<br />
y(x) 1.000 0.962 0.862 0.735<br />
y(x) [Euler] 1 1 0.92 0.785<br />
x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0<br />
y(x) 1.000 0.990 0.962 0.917 0.862 0.800 0.735 0.671 0.610 0.552 0.500<br />
y(x) [Euler] 1 1 0.98 0.942 0.888 0.825 0.757 0.688 0.622 0.560 0.504<br />
(d) h = 0.01: y(1) = 0.50036 vha. Eulers metode. Afvigelse 0.00036.<br />
h = 0.001: y(1) = 0.500035 vha. Eulers metode. Afvigelse 0.000035.<br />
(e)<br />
(f)<br />
x 0.0 0.2 0.4 0.6<br />
y(x) [Euler forbedret] 1 0.96 0.86 0.735<br />
x 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0<br />
y(x) [Euler forbedret] 1 0.99 0.961 0.917 0.862 0.8 0.736 0.672 0.61 0.553 0.501<br />
(g) h = 0.01: y(1) = 0.5000096 vha. Eulers forbedrede metode. Afvigelse 0.0000096.<br />
h = 0.001: y(1) = 0.5000001 vha. Eulers forbedrede metode. Afvigelse 0.0000001.<br />
(h) Eulerny(f,1000,0,1,2) = 0.199987. Eulerforbedretny(f,1000,0,1,2) = 0.2000003.<br />
2
Opgave S.4.1<br />
(a) Lad x(t) <strong>og</strong> y(t) betegne mængderne af stoffet i tank 1 hhv. tank 2 <strong>til</strong> tiden t.<br />
( ) ( )<br />
x<br />
′ −<br />
b c (x ( )<br />
V W −<br />
b c (x<br />
V W<br />
y ′ = = , idet b = a+c.<br />
y)<br />
y)<br />
b<br />
V<br />
− a+c<br />
W<br />
b<br />
V<br />
− b W<br />
(b) Fuldstændig løsning:<br />
( x<br />
y)<br />
= c1 e −0.2t( )<br />
1<br />
4 +c2 e −t( )<br />
3<br />
−4<br />
Partikulær løsning: c 1 = 15 <strong>og</strong> c 2 = 10, så<br />
x(t) = 15e −0.2t +30e −t <strong>og</strong> y(t) = 60e −0.2t −40e −t .<br />
(c 1 ,c 2 ∈ R).<br />
0 10 20 30 40 50<br />
x<br />
y<br />
0 5 10 15 20 25 30<br />
t<br />
(c)<br />
( x<br />
′<br />
y ′ )<br />
=<br />
(<br />
−<br />
b<br />
V<br />
b<br />
V<br />
c<br />
W<br />
− b W<br />
) (x<br />
y)<br />
+<br />
( ) ka<br />
.<br />
0<br />
(d) ( x ∗<br />
y ∗ )<br />
=<br />
( kV<br />
kW)<br />
.<br />
(e) Slutkoncentrationen er k i begge tanke (<strong>og</strong> afhænger altså kun af k).<br />
Opgave S.4.2<br />
(a) e −2t cost hhv. e −2t sint<br />
(b) v(t) = e −2t( cost+sint<br />
cost<br />
)<br />
+ie<br />
−2t ( −cost+sint)<br />
sint<br />
Opgave<br />
(<br />
S.4.3<br />
x<br />
y)<br />
= c1 e −2t( cost+sint) cost +c2 e −2t( −cost+sint)<br />
sint<br />
(c 1 ,c 2 ∈ R)<br />
3
Opgave S.5.1<br />
t 0.0 0.1 0.2<br />
x(t) [Euler] 5 6.6 9.3<br />
y(t) [Euler] −2 0.3 3.63<br />
x(t) 5 7.28 11.60<br />
y(t) −2 0.94 5.87<br />
Opgave S.5.2<br />
y<br />
(a)<br />
y<br />
(b) h = 0.1:<br />
t 0.0 0.5 1.0<br />
x(t) [Euler] 2 0 0<br />
y(t) [Euler] 1 1.5 0.75<br />
0.0 0.5 1.0 1.5 2.0<br />
x<br />
y<br />
0.2 0.4 0.6 0.8 1.0<br />
0 1 2 3 4<br />
t<br />
0.0 0.5 1.0 1.5 2.0<br />
x<br />
h = 0.01:<br />
0.0 0.5 1.0 1.5 2.0<br />
x<br />
y<br />
0.2 0.4 0.6 0.8 1.0<br />
0 1 2 3 4<br />
t<br />
0.0 0.5 1.0 1.5 2.0<br />
x<br />
Opgave S.5.3<br />
(a) 326 gram den 1. oktober<br />
(b) Strategi A: Høst 255 gram<br />
Strategi B: Høst 609 gram<br />
Opgave S.5.4<br />
V ′ = h 1 b KV +h 4 b GV −(f V +l A )V<br />
BC ′ = h 4 b GC − ( (<br />
f C g BC<br />
) )<br />
1 V +lB BC.<br />
4
Opgave S.5.5<br />
(a) Ligevægte: (0,0), (15,0), (0,15a) <strong>og</strong> (20−10a,20a −10).<br />
M ∗ > 0 <strong>og</strong> N ∗ > 0 for 0.5 < a < 2.<br />
(b) Med M ′ = f(M,N) <strong>og</strong> N ′ = g(M,N) fås<br />
( (<br />
f 1−<br />
2<br />
15<br />
D (M,N) =<br />
g)<br />
M − 1 30 N − 1<br />
30 M<br />
)<br />
− 1 30 N a− 1 30 M − 2<br />
15 N .<br />
( f<br />
D (M<br />
g)<br />
∗ ,N ∗ ) = 1 ( ) 2a−4 a−2<br />
.<br />
3 1−2a 2−4a<br />
(c) Ligevægten er stabil for alle 0.5 < a < 2.<br />
5