Upload opgaver

2025-12-02 14:47:39 +01:00
parent 3df6ded18a
commit 41700ebeda
6 changed files with 9803 additions and 163 deletions
--- a/differentialligninger/Andenordens
+++ b/differentialligninger/Andenordens
--- a/differentialligninger/Andenordens
+++ b/differentialligninger/Andenordens
@@ -0,0 +1,403 @@
+#import("@local/dtu-template:0.5.1"):*
+
+
+#set math.vec(delim: "[")
+#set math.mat(delim: "[")
+
+
+#show: dtu-note.with(
+  title: "Andenordens lineære differentialligninger - OPGAVER",
+  date: datetime(year: 2025, month: 12, day: 02),
+  author: "Rasmus Rosendahl-kaa",
+  semester: "Spring 2025"
+)
+
+= Opgave 1
+Er følgende differentialligninger og systemer af differentialligninger
+homogene eller inhomogene?
+
+ $f'' (t) = f' (t) - 2 f (t)$.
+
+ $f' (t) - t dot.op f (t) - e^(- 3 t) = 0$.
+
+ $mat(delim: "[", f'_1 (t); f'_2 (t)) = mat(delim: "[", 4, 6; - 2, 7) dot.op mat(delim: "[", f_1 (t); f_2 (t)) + mat(delim: "[", 1; 1) .$
+
+ ${f'_1 (t) & = & f_2 (t)\
+  f'_2 (t) & = & f_1 (t) - f_2 (t)\
+  $
+
+#solution()[
+/ 1): Homogent da: $f''(t)-f'(t)+2f(t)=0$
+/ 2): Inhomogent. $q(t)=e^(-3t)$
+/ 3): Inhomogent da $q(t)=1$ (se Definition 13.2.1 i bogen)
+/ 4): Homogent da:
+$f'_2 (t) = f_1 (t) - f'_1 (t) <=> f''_1 (t) + f'_1 (t) - f_1 (t) = 0$
+]
+= Opgave 2
+Givet den homogene reelle differentialligning
+
+$ f''(t)+3f'(t)-4f(t)=0 $
+
+Find dens fuldstændige løsning.
+#note-box()[
+Find først rødderne i ligningens karakteristiske polynomium.
+Se Afsnit 13.3 fra lærebogen for flere oplysninger og eksempler.]
+#solution()[
+$
+P_(bold(A))(z) = z^2 + 3z -4\
+D = 3^2- 4 dot (-4) = 25\
+lambda_1 = (-3 + 5)/2 = 1, lambda_2 = (-3-5)/2 = -4
+$
+Finder egenrummene:
+$
+E_(1) = ker(mat(-1,1;4,-4)) = "span"(vec(1,1))\
+E_(-4) = ker(mat(4,1;4,1)) = "span"(vec(1,-4))
+$
+Fuldstændige løsning:
+$
+f(t)=c_1 dot e^(t) + c_2 dot e^(-4t), (c_1, c_2 in RR)
+$
+]
+
+= Opgave 3
+Givet følgende homogene reelle differentialligninger. Find deres
+fuldstændige løsning.
+
+ $ f'' (t) - 6 f' (t) + 9 f (t) = 0 . $
+
+ $ f'' (t) + 2 f' (t) + 5 f (t) = 0 . $
+
+#note-box()[
+Alt afhængig af om det karakteristiske polynomium har to reelle rødder, to ikke-reelle rødder eller en dobbeltrod, skal man bruge Scenarie 1, Scenarie 2, Scenarie 3 på sider 262 og 263 i Afsnit 13.3 fra lærebogen.
+]
+
+#solution()[
+Opstiller karakteristiske polynomium for dem begge
+$
+P_bold(A)^1 (z) = z^2 - 6z + 9\
+P_bold(A)^2 (z) = z^2 +2z + 5
+$
+Finder $D$:
+$
+D_1 = (-6)^2 - 4 dot 9 = 0\
+D_2 = 2^2 - 4 dot 5 = -16\
+$
+Finder rødder:
+$
+lambda_1^1 = lambda_2^1 = - (-6)/2 = 3\
+$
+For den første ligning er løsningen:
+$
+f_1 (t) = c_1 e^(3t) + c_2 t e^(3t)
+$
+For den anden:
+$
+lambda_1^2 = (-2 + i sqrt(|-16|))/2 = -1 + 2i\
+lambda_2^2 = overline(lambda_1^1) = -1 -2i
+$
+Løsning for den anden:
+$
+f_2 (t) = c_1 dot e^(-t) dot cos(2t) + c_2 dot e^(-t) dot sin(2t)
+$
+]
+= Opgave 4
+Givet den inhomogene differentialligning
+
+$ f'' (t) - 6 f' (t) + 9 f (t) = e^(2 t) . $
+== Opgave 4a
+Bestem et reelt tal $a$ således at funktionen $f (t) = a dot.op e^(2 t)$
+er en partikulær løsning til den givne differentialligning.
+#note-box()[
+Indsæt funktionen $f(t)=a dot e^(2t)$ i differentialligningen.  Hvilken ligning skal $a$ opfylde for at funktionen er en løsning til differentialligningen?
+]
+
+#solution()[
+Hvis $f(t)=a dot e^(2t)$ så:
+$
+f'(t) = 2 a e^(2t)\
+f''(t) = 4 a e^(2t)
+$
+Indsætter dem:
+$
+4 a e^(2t) - 6 dot (2 a e^(2t)) + 9 a e^(2t) &= e^(2t)\
+4 a e^(2t) - 12 a e^(2t) + 9 a e^(2t) &= e^(2t)\
+1 a e^(2t) &= e^(2t)\
+a=1
+$
+]
+== Opgave 4b
+Beskriv nu den fuldstændige løsning til den givne inhomogene
+differentialligning. Bemærk at den tilhørende homogene
+differentialligning allerede blev undersøgt i Opgave 3.
+#note-box()[
+Den fuldstændige løsning til den tilhørende homogene differentialligning $f''(t)-6f'(t)+9f(t)=0$ blev fundet i Opgave 3 til at være
+]
+
+#solution()[
+Homogene løsning fra før:
+$
+f_("hom") (t) = c_1 e^(3t) + c_2 t e^(3t)
+$
+
+$
+f(t) = f_("par")(t) + f_("hom")(t) = c_1 e^(3t) + c_2 t e^(3t) + e^(2t)
+$
+
+]
+= Opgave 5
+En lineær andenordens differentialligning med konstante koefficienter kan omskrives til et system af førsteordens differentialligninger (se Afsnit 13.3 fra lærebogen).
+I denne opgave betragtes et eksempel.
+== Opgave 5a
+Givet differentialligningen $f''(t)+2f'(t)+f(t)=cos(t)$.
+Omskriv denne differentialligning til et system af to førsteordensdifferentialligninger i funktionerne $f_1(t)$ og $f_2(t)$, hvor $f_1(t)=f(t)$ og $f_2(t)=f'(t)$.
+
+#solution()[
+Vi ved:
+$f_1 (t) = f(t), f_2 (t) = f'(t)$ og $f'_2 (t) = f''(t)$, så:
+$
+f'_2 (t) + 2 f_2 (t) + f_1 (t) = cos(t)
+$
+
+$
+vec(f'_1 (t), f'_2 (t)) = mat(0,1;-1,-2) dot vec(f_1 (t), f_2 (t)) + vec(0,cos(t))
+$
+]
+== Opgave 5b
+Tjek at Ligning (13.13) i Sætning 13.3.1 fra lærebogen ville have givet
+det samme system.
+
+#solution()[
+Hvis man kigger på Ligning (13.13) ser man, at $a_0 = 1$ og $a_1 = 2$, som kan indsættes for at få samme system.
+]
+= Opgave 6
+Givet den inhomogene differentialligning
+
+$ f'' (t) + 3 f' (t) - 4 f (t) = e^t . $
+== Opgave 6a
+Inspireret af Opgave 4, kunne man håbe at der findes et reelt tal $a$ således at $f (t) = a dot e^t$ er en partikulær løsning til den givne differentialligning. Vis at der faktisk ikke findes nogen funktion på formen $f (t) = a dot e^t$, som er en løsning. Hvad er problemet?
+
+#solution()[
+Når $f(t)= a e^t$ fås: $f'(t) = a e^(t)$ og $f''(t) = a e^(t)$. Indsættes de fås:
+$
+a e^t + 3a e^t - 4a e^t = e^t\
+0 a e^t = e^t
+$
+]
+== Opgave 6b
+Prøv nu at finde en partikulær løsning på formen
+$f (t) = a dot.op t dot.op e^t$. Tjek eventuelt først at
+$(t dot.op e^t)' = e^t + t dot.op e^t$ og
+$(t dot.op e^t)'' = 2 e^t + t dot.op e^t$.
+
+#solution()[
+Når $f(t)=a t e^t$ fås:
+$
+f'(t) = a e^t + a t e^t\
+f''(t) = 2 a e^t + a t e^t
+$
+Indsættes de fås:
+$
+2 a e^t + a t e^t + 3 dot (a e^t + a t e^t) - 4 dot (a t e^t) =\
+2 a e^t + a t e^t + 3 a e^t + 3 a t e^t - 4 a t e^t =\
+5 a e^t = e^t\
+5a = 1\
+a = 1/5
+$
+En partikulær løsning kunne være:
+$
+f(t) = 1/5 dot t dot e^t
+$
+]
+== Opgave 6c
+Beskriv nu den fuldstændige løsning til den givne inhomogene
+differentialligning. Bemærk at den tilhørende homogene
+differentialligning allerede blev undersøgt i Opgave 2.
+
+#note-box()[
+Den fuldstændige løsning til den tilhørende homogene differentialligning $f''(t)+3f'(t)-4f(t)=0$ blev fundet i Opgave 2 til at være $f(t)=c_1 dot e^t+c_2 dot e^(-4t), quad c_1,c_2 in RR.$
+]
+
+#solution()[
+Homogene løsning fra før:
+$
+f_("hom") (t) = c_1 dot e^(t) + c_2 dot e^(-4t)
+
+$
+
+$
+f(t) = f_("par")(t) + f_("hom")(t) = c_1 dot e^(t) + c_2 dot e^(-4t) + 1/5 t e^t
+$
+]
+= Opgave 7
+Det oplyses at den fuldstændige reelle løsning til en homogen lineær andenordens differentialligning med konstante reelle koefficienter er
+
+$ f(t)=c_1e^(-t)cos(2t)+c_2e^(-t) sin(2t), quad c_1,c_2 in RR $
+
+Opskriv differentialligningen.
+
+#note-box()[
+Fordi $0$ er en løsning (sæt $c_1=0$ og $c_2=0$ i den fuldstændige løsning), er differentialligningen man leder efter en homogen differentialligning. Differentialligningen kan derfor skrives på formen $f''(t)+a_1 f'(t)+a_0 f(t)=0$ for visse reelle tal $a_0,a_1$.
+]
+
+#note-box()[
+Hvilke rødder skal det karakteristiske polynomium $Z^2+a_1Z+a_0$ have?
+]
+
+#solution()[
+Vi ved:
+$
+f(t)=c_1 dot e^(alpha t) dot cos(beta t) + c_2 dot e^(alpha t) dot sin(beta t)\
+alpha = -a_1/2\
+beta = sqrt(|a_1^2 - 4a_0|)/2
+$
+Her har vi:
+$
+alpha = -1\
+beta = 2
+$
+Dvs:
+$
+alpha = -a_1/2 = -1\
+a_1 = 2
+$
+$
+beta = sqrt(|a_1^2 - 4a_0|)/2 = sqrt(|4 - 4a_0|)/2 = 2\
+sqrt(|4 - 4a_0|) = 4\
+|4-4a_0| = 16\
+"da vi ved D er negativ (pga cos og sin):"\
+4-4a_0 = -16\
+4a_0 = 20\
+a_0 = 5
+$
+
+Så differentialligningen er:
+$
+f''(t) + 2f'(t) + 5f(t) = 0
+$
+]
+= Opgave 8
+Det oplyses at den fuldstændige reelle løsning til en inhomogen lineær
+andenordens differentialligning er
+
+$ f (t) = c_1 e^(- t) cos(2 t) + c_2 e^(- t) sin(2 t) + 7 + 3 t + 5 e^t , quad c_1 , c_2 in RR. $
+
+Opskriv differentialligningen.
+#note-box()[
+Hvad er den fuldstændige løsning til den tilhørende homogene differentialligning? Kan den tilhørende homogene differentialligning bestemmes ved hjælp af svaret til Opgave 7?
+]
+
+#note-box()[
+Den tilhørende homogene differentialligning er ifølge Opgave 7 givet som $f''(t)+2f'(t)+5f(t)=0.$ Den inhomogene differentialligning der er efterspurgt i denne opgave kan derfor skrives på formen $f''(t)+2f'(t)+5f(t)=q(t).$
+]
+#note-box()[
+$f(t)=7+3t+5e^t$ er en partikulær løsning til den ønskede differentialligning. Hvad får man hvis man indsætter denne funktion i sidstnævnte differentialligningen fra forrige hint?
+]
+
+#solution()[
+Vi har lige fundet differentialligningen for dens tilhørende homogene ligning:
+$
+f''(t) + 2f'(t) + 5f(t) = 0
+$
+Så mangler vi at finde hvad $q(t)$ skal være. For at finde $q(t)$ kan vi indsætte $f(t)=7+3t+5e^t$ og se hvad resultatet bliver:
+Først:
+$
+f(t)=7+3t+5e^t\
+f'(t)=3+5e^t\
+f''(t)=5e^t
+$
+De indsættes:
+$
+5e^t + 2 dot (3 + 5e^t) + 5 dot (7 + 3t + 5e^t) =\
+5e^t + 6 + 10e^t + 35 + 15t + 25e^t =\
+40e^t+15t+41
+$
+Derfor må differentialligningen være:
+$
+f''(t) + 2f'(t) + 5f(t) = 40e^t+15t+41
+$
+]
+= Opgave 9
+Givet den homogene reelle differentialligning
+$ f'' (t) + 3 f' (t) - 4 f (t) = 0 . $
+
+Bemærk at differentialligningen er den samme som i Opgave 2. Målet med
+opgaven er at finde løsningen til differentialligningen som opfylder
+begyndelsesbetingelserne $f (0) = 1$ og $f' (0) = 2$.
+== Opgave 9a
+I Opgave 2 var resultatet at den givne differentialligning har den
+fuldstændige løsning
+
+$ f (t) = c_1 dot.op e^t + c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) . $
+
+Hvilken ligning skal $c_1$ og $c_2$ opfylde for at det gælder at $f(0) = 1$?
+
+#solution()[
+Den skal opfylde:
+$
+f(0)&=c_1 dot e^0 + c_2 dot e^(0) = 1\
+&= c_1 + c_2 = 1
+$
+]
+
+== Opgave 9b
+Hvilken ligning skal $c_1$ og $c_2$ opfylde for at det gælder at $f'(0)=2$?
+
+#note-box()[
+Ligesom i spørgsmål a, vides fra Opgave 2 at
+
+$ f(t)=c_1 dot e^t+c_2 dot e^(−4t), quad c_1,c_2 in RR. $
+
+Tag nu den afledte på begge sider af lighedstegnet og indsæt bagefter $t=0$.
+]
+
+#note-box()[
+Udfra den fuldstændige løsning
+
+$ f (t) = c_1 dot.op e^t + c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) $
+
+fås ved differentiation på begge sider af lighedstegnet at:
+
+$ f' (t) = c_1 dot.op e^t - 4 c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) . $
+
+Brug nu sidste del af forrige hint.
+]
+
+#solution()[
+Kan finde $f'(t)$ for den fuldstændige løsning:
+$
+f'(t) = c_1 dot e^t - 4 c_2 dot e^(-4t)
+$
+$
+f'(0) &= c_1 dot e^0 - 4 c_2 dot e^(0) = 2\
+&= c_1 - 4c_2 = 2
+$
+]
+
+== Opgave 9c
+Find nu den løsning $f (t)$ til differentialligningen som opfylder
+$f (0) = 1$ og $f' (0) = 2$.
+
+#note-box()[
+Bestem $c_1$ og $c_2$ ved at løse de to ligninger man fandt i spørgsmål a og b.
+]
+
+#solution()[
+To ligninger med to ubekendte. Isolerer $c_1$ i $c_1 - 4c_2 = 2$:
+$
+c_1 = 2 + 4c_2
+$
+Sætter ind i $c_1 + c_2 = 1$:
+$
+2 + 4c_2 + c_2 = 1\
+5c_2 = -1\
+c_2 = -1/5
+$
+Indsætter $c_2$ i $c_1 + c_2 = 1$:
+$
+c_1 - 1/5 = 1\
+c_1 = 6/5
+$
+
+Indsætter $c_1 "og" c_2$:
+$ f (t) = 6/5 dot e^t - 1/5 dot e^(- 4 t) $
+]
--- a/differentialligninger/Andenordens
+++ b/differentialligninger/Andenordens
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--- a/differentialligninger/Andenordens
+++ b/differentialligninger/Andenordens
@@ -42,7 +42,7 @@ $

 For nu: antag $q(t)=0$. Dvs vi er i det homogene tilfælde

-$bold(A)=mat(0,1,-a_0;-a_1)$
+$bold(A)=mat(0,1;-a_0,-a_1)$

 / 1): Undersøg $bold(A)$'s egenværdier.
 $ P_bold(A)(z)=det(mat(-z,1;-a_0,-_1-z))&=-z dot (-a_1-z) - 1 dot (-a_0)\
@@ -143,7 +143,7 @@ Husk $f(t)=g(t) dot e^abs((lambda_1 t))$.
 Derfor: Hvis $D=0$, så har $f''(t)+a_1 dot f'(t) + a_0 dot f(t)=0$ den fuldstændige løsning:

 $
-f(t)=c_1 dot t dot e^(lambda_1 t)+c_2 dot e^(lambda_1 t), (c_1, c_2 in RR)
+f(t)=c_1 dot e^(lambda_1 t)+c_2 dot t dot e^(lambda_1 t), (c_1, c_2 in RR)
 $


--- a/differentialligninger/opgaver.pdf
+++ b/differentialligninger/opgaver.pdf
--- a/differentialligninger/opgaver.typ
+++ b/differentialligninger/opgaver.typ
@@ -0,0 +1,226 @@
+#import("@local/dtu-template:0.5.1"):*
+
+
+#set math.vec(delim: "[")
+#set math.mat(delim: "[")
+
+
+#show: dtu-note.with(
+  title: "Andenordens lineære differentialligninger - OPGAVER",
+  date: datetime(year: 2025, month: 12, day: 02),
+  author: "Rasmus Rosendahl-kaa",
+  semester: "Spring 2025"
+)
+
+= Opgave 1
+Er følgende differentialligninger og systemer af differentialligninger
+homogene eller inhomogene?
+
+ $f'' (t) = f' (t) - 2 f (t)$.
+
+ $f' (t) - t dot.op f (t) - e^(- 3 t) = 0$.
+
+ $mat(delim: "[", f'_1 (t); f'_2 (t)) = mat(delim: "[", 4, 6; - 2, 7) dot.op mat(delim: "[", f_1 (t); f_2 (t)) + mat(delim: "[", 1; 1) .$
+
+ ${f'_1 (t) & = & f_2 (t)\
+  f'_2 (t) & = & f_1 (t) - f_2 (t)\
+  $
+
+#solution()[
+]
+= Opgave 2
+Givet den homogene reelle differentialligning
+
+$ f''(t)+3f'(t)-4f(t)=0 $
+Find dens fuldstændige løsning.
+#note-box()[
+Find først rødderne i ligningens karakteristiske polynomium.
+Se Afsnit 13.3 fra lærebogen for flere oplysninger og eksempler.]
+#solution()[
+
+]
+
+= Opgave 3
+Givet følgende homogene reelle differentialligninger. Find deres
+fuldstændige løsning.
+
+ $ f'' (t) - 6 f' (t) + 9 f (t) = 0 . $
+
+ $ f'' (t) + 2 f' (t) + 5 f (t) = 0 . $
+
+#note-box()[
+Alt afhængig af om det karakteristiske polynomium har to reelle rødder, to ikke-reelle rødder eller en dobbeltrod, skal man bruge Scenarie 1, Scenarie 2, Scenarie 3 på sider 262 og 263 i Afsnit 13.3 fra lærebogen.
+]
+
+#solution()[
+
+]
+= Opgave 4
+Givet den inhomogene differentialligning
+
+$ f'' (t) - 6 f' (t) + 9 f (t) = e^(2 t) . $
+== Opgave 4a
+Bestem et reelt tal $a$ således at funktionen $f (t) = a dot.op e^(2 t)$
+er en partikulær løsning til den givne differentialligning.
+#note-box()[
+Indsæt funktionen $f(t)=a dot e^(2t)$ i differentialligningen.  Hvilken ligning skal $a$ opfylde for at funktionen er en løsning til differentialligningen?
+]
+
+#solution()[
+
+]
+== Opgave 4b
+Beskriv nu den fuldstændige løsning til den givne inhomogene
+differentialligning. Bemærk at den tilhørende homogene
+differentialligning allerede blev undersøgt i Opgave 3.
+#note-box()[
+Den fuldstændige løsning til den tilhørende homogene differentialligning $f''(t)-6f'(t)+9f(t)=0$ blev fundet i Opgave 3 til at være
+]
+
+#solution()[
+
+]
+= Opgave 5
+En lineær andenordens differentialligning med konstante koefficienter kan omskrives til et system af førsteordens differentialligninger (se Afsnit 13.3 fra lærebogen).
+I denne opgave betragtes et eksempel.
+== Opgave 5a
+Givet differentialligningen $f''(t)+2f'(t)+f(t)=cos(t)$.
+Omskriv denne differentialligning til et system af to førsteordensdifferentialligninger i funktionerne $f_1(t)$ og $f_2(t)$, hvor $f_1(t)=f(t)$ og $f_2(t)=f'(t)$.
+
+#solution()[
+
+]
+== Opgave 5b
+Tjek at Ligning (13.13) i Sætning 13.3.1 fra lærebogen ville have givet
+det samme system.
+
+#solution()[
+
+]
+= Opgave 6
+Givet den inhomogene differentialligning
+
+$ f'' (t) + 3 f' (t) - 4 f (t) = e^t . $
+== Opgave 6a
+Inspireret af Opgave 4, kunne man håbe at der findes et reelt tal $a$ således at $f (t) = a dot e^t$ er en partikulær løsning til den givne differentialligning. Vis at der faktisk ikke findes nogen funktion på formen $f (t) = a dot e^t$, som er en løsning. Hvad er problemet?
+
+#solution()[
+
+]
+== Opgave 6b
+Prøv nu at finde en partikulær løsning på formen
+$f (t) = a dot.op t dot.op e^t$. Tjek eventuelt først at
+$(t dot.op e^t)' = e^t + t dot.op e^t$ og
+$(t dot.op e^t)'' = 2 e^t + t dot.op e^t$.
+
+#solution()[
+
+]
+== Opgave 6c
+Beskriv nu den fuldstændige løsning til den givne inhomogene
+differentialligning. Bemærk at den tilhørende homogene
+differentialligning allerede blev undersøgt i Opgave 2.
+
+#note-box()[
+Den fuldstændige løsning til den tilhørende homogene differentialligning $f''(t)+3f'(t)-4f(t)=0$ blev fundet i Opgave 2 til at være $f(t)=c_1 dot e^t+c_2 dot e^(-4t), quad c_1,c_2 in RR.$
+]
+
+#solution()[
+
+]
+= Opgave 7
+Det oplyses at den fuldstændige reelle løsning til en homogen lineær andenordens differentialligning med konstante reelle koefficienter er
+
+$ f(t)=c_1e^(-t)cos(2t)+c_2e^(-t) sin(2t), quad c_1,c_2 in RR $
+
+Opskriv differentialligningen.
+
+#note-box()[
+Fordi $0$ er en løsning (sæt $c_1=0$ og $c_2=0$ i den fuldstændige løsning), er differentialligningen man leder efter en homogen differentialligning. Differentialligningen kan derfor skrives på formen $f''(t)+a_1 f'(t)+a_0 f(t)=0$ for visse reelle tal $a_0,a_1$.
+]
+
+#note-box()[
+Hvilke rødder skal det karakteristiske polynomium $Z^2+a_1Z+a_0$ have?
+]
+
+#solution()[
+
+]
+= Opgave 8
+Det oplyses at den fuldstændige reelle løsning til en inhomogen lineær
+andenordens differentialligning er
+
+$ f (t) = c_1 e^(- t) cos(2 t) + c_2 e^(- t) sin(2 t) + 7 + 3 t + 5 e^t , quad c_1 , c_2 in RR. $
+
+Opskriv differentialligningen.
+#note-box()[
+Hvad er den fuldstændige løsning til den tilhørende homogene differentialligning? Kan den tilhørende homogene differentialligning bestemmes ved hjælp af svaret til Opgave 7?
+]
+
+#note-box()[
+Den tilhørende homogene differentialligning er ifølge Opgave 7 givet som $f''(t)+2f'(t)+5f(t)=0.$ Den inhomogene differentialligning der er efterspurgt i denne opgave kan derfor skrives på formen $f''(t)+2f'(t)+5f(t)=q(t).$
+]
+#note-box()[
+$f(t)=7+3t+5e^t$ er en partikulær løsning til den ønskede differentialligning. Hvad får man hvis man indsætter denne funktion i sidstnævnte differentialligningen fra forrige hint?
+]
+
+#solution()[
+
+]
+= Opgave 9
+Givet den homogene reelle differentialligning
+$ f'' (t) + 3 f' (t) - 4 f (t) = 0 . $
+
+Bemærk at differentialligningen er den samme som i Opgave 2. Målet med
+opgaven er at finde løsningen til differentialligningen som opfylder
+begyndelsesbetingelserne $f (0) = 1$ og $f' (0) = 2$.
+== Opgave 9a
+I Opgave 2 var resultatet at den givne differentialligning har den
+fuldstændige løsning
+
+$ f (t) = c_1 dot.op e^t + c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) . $
+
+Hvilken ligning skal $c_1$ og $c_2$ opfylde for at det gælder at $f(0) = 1$?
+
+#solution()[
+
+]
+
+== Opgave 9b
+Hvilken ligning skal $c_1$ og $c_2$ opfylde for at det gælder at $f'(0)=2$?
+
+#note-box()[
+Ligesom i spørgsmål a, vides fra Opgave 2 at
+
+$ f(t)=c_1 dot e^t+c_2 dot e^(−4t), quad c_1,c_2 in RR. $
+
+Tag nu den afledte på begge sider af lighedstegnet og indsæt bagefter $t=0$.
+]
+
+#note-box()[
+Udfra den fuldstændige løsning
+
+$ f (t) = c_1 dot.op e^t + c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) $
+
+fås ved differentiation på begge sider af lighedstegnet at:
+
+$ f' (t) = c_1 dot.op e^t - 4 c_2 dot.op e^(- 4 t) , quad c_1 , c_2 in bb(R) . $
+
+Brug nu sidste del af forrige hint.
+]
+
+#solution()[
+
+]
+
+== Opgave 9c
+Find nu den løsning $f (t)$ til differentialligningen som opfylder
+$f (0) = 1$ og $f' (0) = 2$.
+
+#note-box()[
+Bestem $c_1$ og $c_2$ ved at løse de to ligninger man fandt i spørgsmål a og b.
+]
+
+#solution()[
+
+]