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A4-format til udskrift. - Aarhus Universitet

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140 IV. DIFFERENTIALLIGNINGER<br />

Løsninger<br />

<br />

k<br />

x(t) = C1 cos(<br />

m t) + C2<br />

<br />

k<br />

sin(<br />

m t)<br />

1.5. Fjeder ☞ [S] 7.1 Modelling with differential equations<br />

Fjeder<br />

3<br />

Løsninger<br />

• t tid og x(t) udsving<br />

• x ′ hastighed<br />

x acceleration<br />

• x ′′ = − k<br />

m<br />

d2x k<br />

= −<br />

dt2 m x<br />

<br />

k<br />

x(t) = C1 cos(<br />

m t) + C2<br />

<br />

k<br />

sin(<br />

m t)<br />

1.6. Pendul ☞ [S] 7.1 Modelling with differential equations<br />

Pendul <strong>til</strong>nærmet<br />

3<br />

Løsninger<br />

• t tid og x(t) udsving<br />

• x ′ hastighed<br />

x acceleration<br />

• x ′′ = −k<br />

m<br />

d2x k<br />

= −<br />

dt2 m x<br />

<br />

k<br />

x(t) = C1 cos(<br />

m t) + C2<br />

<br />

k<br />

sin(<br />

m t)<br />

1.7. Differentialligning ☞ [S] 7.1 Modelling with differential equations<br />

Generel ligning<br />

4 y ′ = xy<br />

eller<br />

4<br />

Løsning<br />

dy<br />

= xy<br />

dx<br />

y = f(x)<br />

f ′ (x) = xf(x)<br />

1.8. Differentier funktion ☞ [S] 7.1 Modelling with differential equations<br />

Eksempel 1<br />

1 + cet<br />

y =<br />

1 − cet er løsning <strong>til</strong><br />

4 y ′ = 1<br />

2 (y2 − 1)<br />

Gør prøve<br />

y ′ = cet (1 − cet ) + (1 + cet )cet (1 − cet ) 2<br />

2ce<br />

=<br />

t<br />

(1 − cet ) 2

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