MATHHX A

MATHHX A

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(numerical range)}\TextOrMath { }{\ }}\) \(\def \LWRsiunitxdecimal {.}\)

2.5 Differentialligninger i GeoGebra

Igennem dette afsnit vil jeg tage udgangspunkt i ligningen \(y'=x-y\)

Fuldstændige løsninger

Vi finder den fuldstændige løsning i et CAS-vindue med kommandoen BeregnODE.

(image)

Vi kan se at den fuldstændige løsning til ligningen \(y'=x-y\) er givet ved \(y=c\cdot e^{-x}+x-1\). GeoGebra kalder konstanten for \(c_1\), men vi kalder vi den bare for \(c\) som vi plejer.

Partikulærer løsninger

Vi finder en partikulær løsning i et CAS-vindue med kommandoen BeregnODE, men hvor vi tilføjer punktet som løsningen skal gå igennem.

(image)

Vi kan se at den partikulære løsning til ligningen \(y'=x-y\) gennem punktet \((1,2)\) er givet ved \(y=x+ 2\cdot e^{-x}\cdot e -1\).

Differentialligninger med andre variabelnavne

Vi vil få brug for at løse differentialligninger, hvor de variable ikke hedder \(x\) og \(y\). Det giver lidt problemer i GeoGebra, fordi hvordan skal GeoGebra kunne vide, hvilken variabel der er den uafhængige, og hvilken der er den afhængige? Problemet er dog ikke større end, at vi bare kan omskrive ligningen, så variablene hedder \(x\) og \(y\).

  • Eksempel 2.5.1
    Lad \(N(t)\) være en funktion og betragt ligningen

    \[\frac {dN}{dt}=-0{,}02(40-N)\]

    Vi vil nu bestemme den partikulære løsning som går gennem \((0,100)\).

    Vi omskriver først ligningen, så den uafhængige variabel hedder \(x\), den afhængige \(y\) og differentialkvotienten \(y'\):

    \[y'=-0{,}02(40-y)\]

    Vi kan nu finde den partikulære løsning på sædvanlig vis i GeoGebra:

    (image)

    Vi skal nu bare tage GeoGebras løsning og huske, hvad de oprindelige variabelnavne var. Det giver os:

    \[N(t)=60e^{-0{,}02x}+40\]

Der er dog også en anden og mere elegant måde at håndtere det på. Man kan nemlig fortælle GeoGebra, hvilken en variabel der hvad.

  • Eksempel 2.5.2
    Lad \(N(t)\) være en funktion og betragt ligningen

    \[\frac {dN}{dt}=-0{,}02(40-N)\]

    Vi vil nu bestemme den partikulære løsning som går gennem \((0,100)\). Vi skriver følgende i GeoGebra:

    (image)

    Det giver os altså løsningen:

    \[N(t)=60e^{-0{,}02x}+40\]

    Ville vi have den fuldstændige løsning, kunne vi bare have undladt at skrive punktet til sidst.

Øvelse 2.5.1

Betragt differentialligningen:

\[s'(t)=s(t)+2t\]

  • a) Bestem den fuldstændige løsning til ligningen.

  • b) Bestem den partikulære løsning til ligningen som går gennem (0,4).

Øvelse 2.5.1

  • a) \(s(t)=c\cdot e^t-2t-2\)

  • b) \(s(t)=6e^t-2t-2\)

Retningsfelter og linjeelementer

Vi bruger algebravinduet til at tegne retningsfelter. Vi bruger kommandoen RetningsFelt. Så retningsfeltet for ligningen \(y'=x-y\) kan bestemmes således:

(image)

Læg mærke til at at man ikke skal taste ligningen ind, men kun højresiden. Kommandoen kræver at ligningen har formen \(y'=f(x,y)\). Altså at man har \(y'\) på venstresiden og et udtryk som kun afhænger af \(x\) og \(y\) på højresiden.

Der er ikke nogen kommando til at beregne linjeelementer, men det er også så nemt at vi ikke behøver nogen kommando.

Øvelse 2.5.2

Betragt differentialligningen \(y'=x^2-y\). I GeoGebra skal du:

  • a) Bestemme den fuldstændige løsning til differentialligningen.

  • b) Bestemme den partikulære løsning \(f\) til differentialligningen som opfylder \(f(3)=2\).

  • c) Lave et retningsfelt for ligningen.

Øvelse 2.5.2

  • a) \(y = c \cdot e^{-x} + x^2 - 2x + 2\)

  • b) \(y = x^2 - 2x - 3e^{-x} \cdot e^3 + 2\)

  • c)

    (image)