T a = 20

T(0) = 40

T(L) = 200

h’ = 0,01

L = 10

h = 1

CON LOS SIGUIENTES DATOS:

RESOLVER LA SIGUIENTE E.D.O. DE

SEGUNDO ORDEN

2. Definir un Valor Inicial para z(0)y se resuelven las dos ecuaciones simultáneamente

utilizando por ejemplo RK 4

1. Escribir la E.D.O. de segundo orden como dos E.D.O. de Primer orden:

0'2

2

TThdx

Tda

aTThdx

dz'z

dx

dT

hkZhkThxfk

hkZhkThxfk

hkZhkThxfk

ZTxfk

iii

iii

iii

iii

323141

222131

121121

11

,,

2

1,

2

1,

2

1

2

1,

2

1,

2

1

,,

hkkkkTT ii 413121111 226

1hkkkkZZ ii 423222121 22

6

1

hkZhkThxfk

hkZhkThxfk

hkZhkThxfk

ZTxfk

iii

iii

iii

iii

323142

222132

121122

12

,,

2

1,

2

1,

2

1

2

1,

2

1,

2

1

,,

h = 1

z(0) = 10i x T i k 11 k 21 k 31 k 41 Z i k 12 k 22 k 32 k 42

0 0 40 10 10,1 10,125 10,2505 10 0,2 0,25 0,2505 0,30125

1 1 50,11675 10,250375 10,4009588 10,4265847 10,60354729 10,250375 0,3011675 0,35241938 0,35317229 0,40543335

2 2 60,53491819 10,603339 10,8060136 10,832522 11,06271828 10,60333903 0,40534918 0,45836588 0,45937925 0,5136744

3 3 71,35877294 11,0624247 11,3192185 11,3468746 11,63260849 11,06242467 0,51358773 0,56889985 0,57018382 0,62705648

4 4 82,69664285 11,6322266 11,9457098 11,9747904 12,31892157 11,6322266 0,62696643 0,68512756 0,68669498 0,74671433

5 5 94,66200094 12,3184476 12,6917576 12,7225537 13,12852637 12,31844757 0,74662001 0,80821225 0,8100788 0,87384555

6 6 107,3746004 13,1279555 13,5648285 13,5976484 14,06952566 13,12795551 0,873746 0,93938578 0,94157015 1,00972249

7 7 120,9616728 14,0688522 14,5736606 14,6088327 15,15133727 14,06885223 1,00961673 1,07996099 1,08248503 1,15570506

8 8 135,5592022 15,1505545 15,7283505 15,7662269 16,38478831 15,15055454 1,15559202 1,23134479 1,23423377 1,31325429

9 9 151,3132852 16,3838884 17,0404549 17,0814146 17,78222357 16,38388845 1,31313285 1,39505229 1,39833513 1,483947

10 10 168,3815937 17,7811976 18,5231055 18,5675585 19,35762903 17,78119756 1,48381594 1,57272192 1,57643146 1,66949152

h = 1

z(0) = 20i x T k 11 k 21 k 31 k 41 Z i k 12 k 22 k 32 k 42

0 0 40 20 20,1 20,15 20,3005 20 0,2 0,3 0,3005 0,4015

1 1 60,13341667 20,3004167 20,5010838 20,5518348 20,80425625 20,30041667 0,40133417 0,50283625 0,50383959 0,60685251

2 2 80,66850167 20,8040064 21,1073489 21,1593589 21,51622815 20,80400639 0,60668502 0,71070505 0,71222176 0,81827861

3 3 101,8107767 21,5158093 21,9248631 21,9786527 22,44354135 21,51580927 0,81810777 0,92568681 0,92773208 1,03789429

4 4 123,7718404 22,4429492 22,9618084 23,0179158 23,59547669 22,44294924 1,0377184 1,14993315 1,15252745 1,26789756

5 5 146,7714861 23,5947054 24,2285629 24,2875496 24,98356311 23,59470543 1,26771486 1,38568839 1,38885768 1,51059036

6 6 171,0399017 24,982605 25,7378045 25,800261 26,62169303 24,98260499 1,51039902 1,63531204 1,63908804 1,76840163

7 7 196,8199732 26,6205385 27,5046383 27,5711897 28,52626138 26,62053846 1,76819973 1,90130242 1,90572292 2,04391163

8 8 224,3697159 28,5248988 29,5467474 29,6180596 30,7163297 28,5248988 2,04369716 2,18632165 2,1914309 2,33987776

9 9 253,9648563 30,7147455 31,8845698 31,9613566 33,21381688 30,71474547 2,33964856 2,49322229 2,49907141 2,65926213

10 10 285,9015922 33,2119952 34,5415031 34,6245331 36,04371859 33,21199516 2,65901592 2,8250759 2,83172344 3,00526125

3. Como T(10) = 168,3815 es menor al valor dado por las condiciones iniciales, hay que

definir otro Valor Inicial para z(0)y se resuelven las dos ecuaciones simultáneamente

utilizando por ejemplo RK 4

4. Como T(10) = 285,9015 es mayor al valor dado por las condiciones iniciales, se debe

interpolar para encontrar el valor de z(0) que dé T(10) = 200

200

12,69047

h = 1

z(0) = 12,69047i x T k 11 k 21 k 31 k 41 Z i k 12 k 22 k 32 k 42

0 0 40 12,6904703 12,7904703 12,8221965 12,95442263 12,69047 0,2 0,26345235 0,26395235 0,32822196

1 1 52,8117044 12,9543088 13,1183674 13,1507531 13,34801772 12,95430884 0,32811704 0,39288859 0,39370888 0,45962458

2 2 65,95179899 13,3477983 13,5775573 13,6109268 13,87520404 13,34779827 0,45951799 0,52625698 0,52740578 0,59562726

3 3 79,55179405 13,8748767 14,1726357 14,2073229 14,54125785 13,87487673 0,59551794 0,66489232 0,66638112 0,73759117

4 4 93,74780268 14,5408194 14,9095584 14,9459105 15,35284521 14,54081939 0,73747803 0,81018212 0,81202582 0,88693713

5 5 108,6819031 15,3522912 15,7957008 15,8340815 16,31808877 15,35229124 0,88681903 0,96358049 0,96579753 1,04515985

6 6 124,5035605 16,3174137 16,8399315 16,8807251 17,44664898 16,31741372 1,0450356 1,12662267 1,12923526 1,21384286

7 7 141,3711231 17,4458461 18,0527017 18,0963163 18,74982085 17,44584611 1,21371123 1,30094046 1,30397474 1,39467439

8 8 159,453407 18,7488821 19,4461492 19,4930214 20,24064693 18,74888212 1,39453407 1,48827848 1,49176482 1,58946428

9 9 178,9313853 20,2395629 21,0342199 21,0848188 21,93404789 20,23956294 1,58931385 1,69051167 1,69448495 1,80016204

10 10 200 21,9328078 22,8328078 22,8876398 23,84697183 21,9328078 1,8 1,90966404 1,91416404 2,0288764

X ANALITICA0 40

1 52,81170471

2 65,95179998

3 79,55179631

4 93,74780704

5 108,6819106

6 124,5035725

7 141,3711414

8 159,4534334

9 178,9314222

10 200,0000501

5. Reemplazamos el valor obtenido en la interpolación en z(0) para obtener la solución

de la ecuación

Valor Buscado

INTERPOLANDO

40

60

80

100

120

140

160

180

200

0 1 2 3 4 5 6 7 8 9 10

ANALITICA

T

69047.213815.1682003815.16885.90152

1020100z