diff --git a/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger - Opgaver.pdf b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger - Opgaver.pdf new file mode 100644 index 0000000..0a6a6e7 --- /dev/null +++ b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger - Opgaver.pdf @@ -0,0 +1,5346 @@ +%PDF-1.7 +% + +1 0 obj +<< + /Type /Pages + /Count 11 + /Kids [365 0 R 367 0 R 369 0 R 371 0 R 373 0 R 375 0 R 377 0 R 379 0 R 381 0 R 383 0 R 385 0 R] +>> +endobj + +2 0 obj +<< + /Type /Outlines + /First 3 0 R + /Last 18 0 R + /Count 9 +>> +endobj + +3 0 obj +<< + /Parent 2 0 R + /Next 4 0 R + /Title (Opgave 1) + /Dest 346 0 R +>> +endobj + +4 0 obj +<< + /Parent 2 0 R + /Next 5 0 R + /Prev 3 0 R + /Title (Opgave 2) + /Dest 347 0 R +>> +endobj + +5 0 obj +<< + /Parent 2 0 R + /Next 6 0 R + /Prev 4 0 R + /Title (Opgave 3) + /Dest 348 0 R +>> +endobj + +6 0 obj +<< + 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/Dest 356 0 R +>> +endobj + +15 0 obj +<< + /Parent 12 0 R + /Prev 14 0 R + /Title (Opgave 6c) + /Dest 357 0 R +>> +endobj + +16 0 obj +<< + /Parent 2 0 R + /Next 17 0 R + /Prev 12 0 R + /Title (Opgave 7) + /Dest 359 0 R +>> +endobj + +17 0 obj +<< + /Parent 2 0 R + /Next 18 0 R + /Prev 16 0 R + /Title (Opgave 8) + /Dest 360 0 R +>> +endobj + +18 0 obj +<< + /Parent 2 0 R + /Prev 17 0 R + /First 19 0 R + /Last 21 0 R + /Count -3 + /Title (Opgave 9) + /Dest 364 0 R +>> +endobj + +19 0 obj +<< + /Parent 18 0 R + /Next 20 0 R + /Title (Opgave 9a) + /Dest 361 0 R +>> +endobj + +20 0 obj +<< + /Parent 18 0 R + /Next 21 0 R + /Prev 19 0 R + /Title (Opgave 9b) + /Dest 362 0 R +>> +endobj + +21 0 obj +<< + /Parent 18 0 R + /Prev 20 0 R + /Title (Opgave 9c) + /Dest 363 0 R +>> +endobj + +22 0 obj +<< + /Type /StructTreeRoot + /RoleMap << + /Datetime /Span + /Terms /Part + /Title /P + /Strong /Span + /Em /Span + >> + /K [34 0 R] + /ParentTree << + /Nums [0 23 0 R 1 24 0 R 2 25 0 R 3 26 0 R 4 27 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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) $ +] diff --git a/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.pdf b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.pdf index cc08f9c..cc9e985 100644 --- a/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.pdf +++ b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.pdf @@ -2773,7 +2773,7 @@ endobj << /Type /Font /Subtype /Type0 - /BaseFont /PUPGNL+NewCMMath-Book-Identity-H + /BaseFont /YSBCVK+NewCMMath-Book-Identity-H /Encoding /Identity-H /DescendantFonts [255 0 R] /ToUnicode 258 0 R @@ -2784,7 +2784,7 @@ endobj << /Type /Font /Subtype /CIDFontType0 - /BaseFont /PUPGNL+NewCMMath-Book + /BaseFont /YSBCVK+NewCMMath-Book /CIDSystemInfo << /Registry (Adobe) /Ordering (Identity) @@ -2809,7 +2809,7 @@ endobj 257 0 obj << /Type /FontDescriptor - /FontName /PUPGNL+NewCMMath-Book + /FontName /YSBCVK+NewCMMath-Book /Flags 131076 /FontBBox [-1 -960 2003 1450] /ItalicAngle 0 @@ -2888,8 +2888,8 @@ endcodespacerange <0022> <0023> <2212> <0024> -<0025> <0031> -<0026> <0030> +<0025> <0030> +<0026> <0031> <0027> <0028> <0029> @@ -2985,40 +2985,61 @@ endobj stream x{tUv"A,A PB-Ť7ۖnY-˶,$SHo,B-K(Kذ<̝{>1c ed{5//޵ -3ӟ8g0k۾`;cw;)3gͥog0 674%/)uUJj$SRȲGz3sJr3s6y9Lsnٜ)#e -SSSHsSREs$sVlٴy\ɜɩ9ɐH_z)4`{|(Ч{yq{p^Ó>y y=|o_h~>_-/^Lyɗ^~hQ̢׽ U& ;&`,&Ux|rKq=kdJ7}]g&n#[vAbdPy-d2S(:yKQGN+RHA j@jRUgs2F!C#'=en&>eO =}-q_7 Ņ D>#(ezeaRre@~d<[6, q;V$=R_Ey,{gAWsZ6;݊EJf?>[}pAL(Q P|`^f虜lN9^l ״]~hh2k2_HJ EfrTpU`5y .%Ċ 8:lg=W>^r%g7 [ t`22ܿ*4j^TWXL?Ⓙ/>gնdj &iw:2H^zLPݸ]/-.T7͡KMuj>zBW h¤ 轀ig&35|= "qr@׉c)p u2)ȠDmȺrjSa5lڱ*nd D F?ajpAw ."|2w 6̀vRYASťK2 p=JNU^0@uMAyyMco8Fq|,ڋE#w~DuFZT77P0f<7WMO3lωC'Sou7j#%XqƍyExeፇ7˻qS7?+`~O^!&&6PRXeq*VUjmf~s#J]=uF!rgtƝ?|$ f7~397D5VD`TeI~m! yNYM&=jZ|pke!CHT$`&EZ,-ީ[8]ךw?OVRwzz嶻 Pm8Մ/߰h8 s1Q4sZc.G\|gdF;?x{'GatP3@g< -p#uxqǗoYC_e~ 6fҋۨU`#Z=mt۩K.S@y2:k`mdŧ75 ^ xuCC17Fy5ic077X1x`zXBaVA ݙm}_u] -`}ݖބބQ1@Hh$Ԏx5Ԏ\4tLDMfwyK_^-lKoKN̫KN^ƗA/IoMƊ^$hvOϿ);zȆrι OI{6J@i{5trЬ֙t&%PF -;8`>dp@ԻnS9"rhss2CJl)I^8T1v5?'3›\OSiA&?XYU_$QtRjJ@",VPg0YX@b#Ǐ uE֞>};[jf]!M-(3$('&e͉-hR^WP+E88TRï8(W\#x3vaqcnX(T;'f?fM_eHxii=}[CM`6VN0<+3xQTK!`~Yo7u&Su3 eɏυ,Noy5|O3c-O_ 5?COG0 LjN&A<3$)LHDY `ξCpZ35أPJӭhRz?Mj:gp+հ!wCL˓!ePؓ?ð/05U? |O=d^`'^$t7u:8E5 &Ia$Japma+q@0 n35?()3xU7?Co~>mas7^Q|p5A'4k$ k QR,Jغpis]x?y?g(i_^Fq8oH nģvC2+f}嵞R'MhڴY[ -(}Il7ҙt -5đ50aL&j ◀JjnAHe\"\"CN^gWmr^m:nKµ!~9b)BrU~_uh  ,XHPJaZDL`ˍZʣ7ur_.Vi;;:>!ob,-1 #IQ(Csm} ԁ["ڤ鋖K2@AVCe*rW \ė@F/uf& -X/ܺ~{h ,'/ ~(yt V6BP*kFO?ou @B(iu&x۳Zl[={}5hRfbmqgl=4 nc#a4>=i>u|A(5D-lwU8d8]ZJ\HttYY8eKTe"aJ&ye2\P!NT:jhN `4ĴMװwUM֊ww1b{f JC{UD@~ Pj RcYD%pkw:*MOnnk_|_~1@{XLb:TƩ9h TnHV:czNӆZ ᘖXz>,vj&$NMPl~k\ܘ8֞P{y#T2c2Y4UYk 2݈$Av ,w6"Jvj aZ%UiONܝ6|{<G(ܔ$8ǍXᝢxu6ۖe5 ^_ѪX j@lmj$d:qo. ˾dNDElPu_R-QR"|ɼRs o|G[[Vۗ|10 itn?<Ttcw(\*C9)l1^Cp; $vz=x$yhx{KR *eŰ(^p޸a =Èwk0@;V8>:1Arn&f`S Qho2H#ȫS `)Q#' e Ҳ:UWG#pAT15xzUHA\,/I:~&V7 %Y8Uws{#OH=S9PԽVn|uI<@Mf8:+hk3Of -RċUʽU0e5;BoWMev |఺nP"*<*ڗԸdh1-fmltTK2BaM JPm]k4|j[~MQmlz6ơ*J*iH1piZxfu?XԃԽT71,6=3M`#vzNu[5A9>c;;gDehʍ!AjnZͶ-=92ɵjd-Nʠ~Vwt3]G)ꦟy"̒yi'5#qǺ^JHvZ*s,o݀0P\z\Ov&5{:>X -Q,Gi |<0X}4uXq8;yKX9NkѭrЂZlRXM6ZzL~Ua%wq]y$@›NQ4-"[`qe9r&}QfJ({q ȲRTY ~_(e O Djy喨\R0Z~U2A :@Ѓ'q:?O$t>CSVK*βNW_iG6Ȭb[>A]@*P*58@fx_P#N90 fYjK)3-J2@͛b֩gwF1?å2pos9=ivH)X3}iPj֥A -tVTq s@fpq T#P@Ǯ՚PU9P$΃<14Q"(l{*|r VPV3,sͦ$cքtM2A@6@]eGx.y媱U -]=nεȋglG\gy䷄笀ZD!eYՇIePitFGp%L {PJ:~4rqI_5{7^*tH5{ ]{:}KޣT Vj .? b5b-ay6UTW@Y t u쭞Fm?,cüs9$ӳӳ a'$Ufy E]wqa)lmH(!W֏A.B -apg{ 8ۄu "詮\ΧI+#p;}߲Wؘhْ]KT};62;Id7l@rL>ի}~Qi!ȑ *yÛ-.!WuNT7h|j -|r /@ε8 -)a+|a؁Kns}ͺ =Cc&bn -)mhUv[GKOp;;t2Ro],fls|cE3mLoy>RF9mn`^Ov"_eggHl&-|u}cC=ƏwAԮD4Wh)eI5ҔDYu]*GB〸s5aq7NlOck{'xrRq$;P@P+><:ů$.˪)jon@XY#t6+޸Z)V@De@ #jE<7%kx[Ao/}K_^w'8+6sw\o!! 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TWE*\pW ]] 5 !}'. -:wKF +]|j}is<{ ٱQ,Qo?}CO,¦P\1!ɨptcDA#o QĄsgʌeм(y[$b$&Ɍ˂PxHyAT)RBOiL4R]+FDUKN)36 +YcHJe3 LSblPfirbaSbkk|αHgLXpS ~$9!EA$ïsIʈT zl.I)oN^k4MFHj*W7*v̖<Хϭ|F?U[nk!$BtҔt`n7 jӓ_2uHSXٽIS2Q 5Rn45 +G7 +Q8.]ޑF)@[`Riw6vHSl4%)Y@vcS45Ʀ<ijڍMyԴ;`A8%b) DJ:C +,[mj1&^۔ +Ђ5֦LhJ+k~V2F_F +FSn> +żJ:z][9\w\ѣuhJ`) _{kOë׿w˯}7_&}>/g?曇/S/ǿ|_}7oS&(gzQJLͿ=w[e/vtN=OOսt>/_{7{y{ץꠝ3~a3of endstream endobj @@ -3633,9 +3662,9 @@ l2h }A`Zx~%ő[y>B8Թy"pmM'uH(r{6?-Xkr0vPjbA0pen`_ "ȶ=aԀa7mt>GiŇx|g[oolPvY+C 0Bݥ1Rq͑BݥXH`d.3U`KPyIؠQ2 G:ڊx\C%QڦQG($.8q} #tWW)$#T@4$ͻe`{*.Ixi*.=pULcPgI^xA>k&<=B%OFH6TZđmȦ=B`2so2%[Cž6ArɔMfM>m2Z$ -]3i 0 AǺ ‘z@ V5K,+8Sڈo)g$(|P=9*<G)RSԺSS]^CIQĚDFEv&3*L<-Qq/dШVk +]3i 0 AǺ ‘z@ V5K,+8Sڈo)g$(|P=9*<G)RSԺSS]^CNQĚDFEv&3*L<-Qq/dШVk C)8\bQaER4UC:Q)?TJ,,l eqWa W*+ =յ9&9L%VuD t(o?RgyL%Gjet"gt:_*o?Rg9Lo7RgyL%.zĹQ# 3 J̛b D0&K+gt {5LHQ:@3#z=, G(lbX (>v' Q.FuxʦP.*Ьj6p1߁ vRJWyNcP.HMC5l#%+ Nȱ495ƒbO+0 k^k4V 9Xc9Ae?ǘfykZc9zY STkVn zK5T?#goqZI(WuH$S5OBv=u=X|?FCC\Y6m\ᏝG$Ri+ǡ-&ԂkGe`*Kz"4\zd\oKoGCO"h@V}pV"c%8TQ0R։u0~F3 -yA 2+QSD,W:F%Pɚ=# o =JΨj]Q(^RUs,B5^xAx9:B9x`.mZ^bY:!;_a9 ֦~~σO ؾp1?ǣW&0yҮ)ͧ0&Q#0hPևHa>Φa:/Pӫ ‘Oa:_2_>MA%Vuf"p54@} j[:xz?/n-fh\Ѷq^㊶{1> Gi0πU0ȏ4k:.畊+tqo]!5A+bӬ5 Uh4Zfpjha~8fy&E{vѭa~'BvUO.2xbgD?1JzJcĻ}`ۯ]jCkk_S›Oa2M2Gʗa:/РҭÑ:|,J##Mt^W#Nt"keP|K$ oV34DjD3h\#A A\նt~^*g[иm5`YmkbT}TӢaO7߫atKi؋Ju]9+W#ߺC>&k"VĦYkLi/6\.q*5L0 P[[{0O 챟\e>xT~ >E_ ClYa/;xnCbZq t?V]1%瑱Կ2xji+X_">I G>8Cp!l +@;00 sycrp0W#b5#AI#  t sx^PŷL#p] +"$Xb(˞A(Ũx5$67_ TkϸM,D͕86[G\:X`B8i~> endobj @@ -3935,7 +3968,7 @@ endobj /Subtype /XML >> stream -Typst 0.14.0en2025-12-02T10:51:16+00:002025-12-02T10:51:16+00:007application/pdf4gewiXm5GeEg+9RfFu3nAQ==4gewiXm5GeEg+9RfFu3nAQ==proof1.7 +Typst 0.14.0en2025-12-02T12:39:16+00:002025-12-02T12:39:16+00:007application/pdfVGItTyS6ZGlBct4+Pe4gmQ==VGItTyS6ZGlBct4+Pe4gmQ==proof1.7 endstream endobj @@ -4236,28 +4269,28 @@ xref 0000100950 00000 n 0000101679 00000 n 0000102007 00000 n -0000109474 00000 n -0000109802 00000 n -0000118723 00000 n -0000119051 00000 n -0000126515 00000 n -0000126843 00000 n -0000133565 00000 n -0000133893 00000 n -0000142237 00000 n -0000142565 00000 n -0000145129 00000 n -0000187379 00000 n -0000252687 00000 n -0000252804 00000 n -0000253890 00000 n +0000109458 00000 n +0000109786 00000 n +0000118706 00000 n +0000119034 00000 n +0000126504 00000 n +0000126832 00000 n +0000133554 00000 n +0000133882 00000 n +0000142226 00000 n +0000142554 00000 n +0000145118 00000 n +0000187368 00000 n +0000252676 00000 n +0000252793 00000 n +0000253879 00000 n trailer << /Size 292 /Root 291 0 R /Info 289 0 R - /ID [(4gewiXm5GeEg+9RfFu3nAQ==) (4gewiXm5GeEg+9RfFu3nAQ==)] + /ID [(VGItTyS6ZGlBct4+Pe4gmQ==) (VGItTyS6ZGlBct4+Pe4gmQ==)] >> startxref -254127 +254116 %%EOF \ No newline at end of file diff --git a/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.typ b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.typ index ba3e287..6c2bb4c 100644 --- a/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.typ +++ b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/Andenordens differentialligninger.typ @@ -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 $f(t)=g(t) dot e^abs((lambda_1 t))$. 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1a/Noter/Andenordens differentialligninger/opgaver.typ b/01001 - Matematik 1a/Noter/Andenordens differentialligninger/opgaver.typ new file mode 100644 index 0000000..002e877 --- /dev/null +++ b/01001 - Matematik 1a/Noter/Andenordens 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()[ + +]