MATHHX A
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2.1 Introduktion til differentialligninger
En differentialligning er en ligning, hvor den ubekendte er en funktion, og som indeholder funktionens differentialkvotient (deraf navnet – differentialligning). Funktionen betegner vi med \(y\) og differentialkvotienten
skrives dermed som \(y'\).
-
Eksempel 2.1.1
Betragt ligningen
\[y'=5.\]
Vi kan se at ligningen indeholder differentialkvotienten \(y'\), og derfor er det en differentialligning.
I ovenstående eksempel står der ikke \(y\) nogle steder i ligningen. Men det ændrer ikke på, at det er \(y\) der er den ubekendte. Man kan sige at \(y\) optræder i ligningen via sin differentialkvotient.
-
Eksempel 2.1.2
Betragt ligningen
\[y'\cdot x =y+2x^3.\]
Vi kan se at ligningen indeholder differentialkvotienten \(y'\), og derfor er det en differentialligning. Ligning indholder også funktionen \(y\) og variablen \(x\).
Øvelse 2.1.1
Bestem hvilke af følgende ligninger som er differentialligninger
-
a) \(y'=2\)
-
b) \(3x+y=y'\)
-
c) \(x+2y=y\)
-
d) \(y=3\)
En løsning til en differentialligning er en funktion, der gør ligningen sand, når man sætter funktionens forskrift ind i stedet for \(y\).
-
Eksempel 2.1.3
Funktionen \(f(x)=x^2\) er ikke en løsning til differentialligningen \(y'=5\). Det kan vi se, ved at sætte \(f\) ind i stedet for \(y\) i ligningen. I ligningen optræder
differentialkvotienten \(y'\), så vi finder først \(f'(x)\). Den giver \(f'(x)=2x\) og den kan vi nu sætte ind i ligningen:
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
y' &=5 && \text {(Vores oprindelige ligning)}\\ 2x &= 5 && \text {(Vi har indsat $f'(x)$ i stedet for $y'$)}\\
\end{align*}
Vi kan se at \(f\) ikke en løsning, da ligningen er falsk (der står ikke det samme på hver side af lighedstegnet).
Vi vil nu prøve om funktionen \(g(x)=5x+1\) er en løsning til differentialligningen. Vi finder først differentialkvotienten \(g'(x)=5\). Vi sætter nu ind i ligningen.
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
y' &=5 && \text {(Vores oprindelige ligning)}\\ 5 &= 5 && \text {(Vi har indsat $g'(x)$ i stedet for $y'$)}
\end{align*}
Vi kan se at \(g\) er løsning til differentialligningen, da ligningen er sand (der står det samme på begge sider).
Øvelse 2.1.2
Bestem hvilke af følgende funktioner som er løsning til differentialligningen \(y'=2x\)
-
a) \(f(x)=2\)
-
b) \(f(x)=x^2+1\)
-
c) \(f(x)=x^2-x\)
-
d) \(f(x)=x^2\)
Løsning 2.1.2
-
a) Ikke løsning
-
b) Er løsning
-
c) Ikke løsning
-
d) Er Løsning
-
Eksempel 2.1.4
Vi vil nu undersøge om funktionen \(f(x)=\textcolor {red}{x^3-3x}\) er en løsning til ligningen \(y'\cdot x =y+2x^3.\) Vi finder først
differentialkvotienten \(f'(x)=\textcolor {blue}{3x^2-3},\) hvorefter vi indsætter \(f\) og \(f'\) i ligningen:
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
y'\cdot x &=y+2x^3 && \text {(Oprindelige ligning)}\\ (\textcolor {blue}{3x^2-3})\cdot x &=\textcolor {red}{x^3-3x}+2x^3 && \text {(Indsat $f$ for $y$ og $f'$
for $y'$)}\\ 3x^3-3x &=3x^3-3x && \text {(Reduceret)}
\end{align*}
Vi kan se at ligningen er sand og vi kan dermed konkludere at \(f(x)=x^3-3x\) er løsning til ligningen \(y'\cdot x =y+2x^3.\)
Øvelse 2.1.3
Tjek om:
-
a) \(f(x)=x^2+3x\) er løsning til \(y'\cdot x = x^2+y\)
-
b) \(f(x)=\ln (x)\) er løsning til \(x\ln (x)=(y'+y)x-1\)
-
c) \(f(x)=x^3+3x^2-1\) er løsning til \(y'-y+x^3=6x\)
-
d) \(f(x)=e^{3x}\) er løsning til \(y^2-e^{6x}+y'=3y\)
-
Eksempel 2.1.5
Vi vil bestemme konstanten \(c\) således at funktion
\[f(x)=x^2+c\cdot x\]
er løsning til differentialligningen
\[y'- 2 =2x+1.\]
Vi regner først \(f':\)
\[f'(x)=2x+c.\]
Vi indsætter nu i ligningen og isolerer \(c\):
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
y'-2 &=2x+1 && \text {(Oprindelige ligning)}\\ 2x+c -2 &= 2x+1&& \text {(Indsat $f'$ for $y'$)}\\ c &=3 && \text {(Isoleret $c$)}
\end{align*}
Vi konkluderer, at hvis \(c=3\), så er \(f(x)=x^2+c\cdot x\) løsning til \(y'- 2 =2x+1\).
Øvelse 2.1.4
Lad \(f(x)=x^2+5x+k\).
Øvelse 2.1.5
Betragt ligningen \(2y'=-a(y-x)+2\)
Øvelse 2.1.6
Gør rede for at:
Der er flere måder at opskrive den samme ligning på. Følgende variationer er mulige:
-
• Den ubekendte kaldes \(f(x)\) i stedet for \(y\).
-
• Differentialkvotienten skrives som \(\frac {dy}{dx}\) i stedet for \(y'\) .
-
• Løsningen betegnes med \(y\) i stedet for \(f(x)\).
-
Eksempel 2.1.6
Differentialligningen \(y'\cdot x =y+2x^3\) kan skrives på følgende måder:
\(\seteqnumber{0}{2.}{0}\)
\begin{align*}
y'\cdot x &= y+2x^3\\ f'(x)\cdot x &= f(x)+2x^3\\ \frac {dy}{dx}\cdot x &= y+2x^3
\end{align*}
Løsningen (tjek selv at rent faktisk er en løsning) kan skrives som
\[f(x)=x^3-3x\quad \textrm {eller}\quad y=x^3-3x\]
Om differentialligningen, eller løsningen, står på den ene eller den anden måde har selvfølgelig ingen betydning for, hvordan man regner på ligningen.
Øvelse 2.1.7
Gør rede for at:
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a) \(f(x)=x^3+x^2\) er løsning til \(2\frac {dy}{dx}-6x^2=4x\)
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b) \(y=x^2\) er løsning til \(y'=2x\)
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c) \(f(x)=e^x\) er løsning til \(f(x)=f'(x)\)
Ekstra
Differentialligninger kan også indeholdte den dobbelt afledte. Så kaldes det en andenordens differentialligning. Den type ligninger møder man meget ofte i fysik (sikkert også i økonomi). De kan være svære at løse
Øvelse 2.1.8
Lad \(f(x)=e^{2x}-4e^{-3x}\)