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Asignacion de Calculo4 Carlos gonzalez Saia E
- 1. UNIVERSIDAD FERMIN TORO VICERECTORADO ACADEMICO DECANATO DE INGENIERIA MATEMATICA IV ACTIVIDAD EJERCICIOS Carlos Gonzรกlez 24538470
- 2. 1.) Determine si la funciรณn es soluciรณn de la ecuaciรณn diferencial. a y ๏ฝ senx ๏จ x ๏ซ ctgx๏ฉ y ๏ซ y ๏ฝ ๏ญctgx ,, ) ln csc ; 1) ํฆ = ํ ํํ ํฅln (csc ํฅ + ํํํกํ ํฅ) ํฆยด = ํํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) + 1 csc ํฅ + ํํํกํ ํฅ (โ csc ํฅํํํกํ ํฅ) โ csc 2 ํฅ ํ ํํ ํฅ ํฆยด = ํํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) + 1 csc ํฅ + ํํํกํ ํฅ (โ csc ํฅ)(ํํํกํ ํฅ +csc ํฅ)ํ ํํ ํฅ ํฆยด = ํํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ csc ํฅ ํ ํํ ํฅ ํฆยด = ํํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ 1 ํฆยดยด = โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) + 1 csc ํฅ + ํํํกํ ํฅ (โ csc ํฅํํํกํ ํฅ โ csc 2 ํฅ) ํํํ ํฅ ํฆยดยด = โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) + 1 csc ํฅ + ํํํกํ ํฅ (โ csc ํฅ)(ํํํกํ ํฅ +csc ํฅ)ํํํ ํฅ ํฆยดยด = โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ csc ํฅ cos ํฅ ํฆยดยด = โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ 1 ํ ํํํฅ ํํํ ํฅ ํฆยดยด = โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ ํํํกํํฅ ํฆยดยด + ํฆ = โํํํกํ ํฅ โํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) โ ํํํกํํฅ + ํ ํํ ํฅ ln(csc ํฅ + ํํํกํ ํฅ) = โํํํกํ ํฅ โํํํกํ ํฅ = โํํํกํํฅ
- 3. 2) Resolver las siguientes ecuaciones diferenciales de primer orden de acuerdo al mรฉtodo correspondiente. a .) senx cos xy y tg x , 2 ๏ซ ๏ฝ ) ๏จ 2 y cos ๏ฉ ๏จ2 2 y cos 2 ๏ฉ 0 b e y xy dx xe x xy y dy ๏ญ ๏ซ ๏ญ ๏ซ ๏ฝ 2- a) ํ ํํํฅ cos ํฅ ํํฆ ํํฅ + ํฆ = ํกํ2ํฅ ํ ํํํฅ cos ํฅ ํํฆ ํํฅ = (ํกํ2ํฅ โ ํฆ) ํ ํํํฅ cos ํฅํํฆ = (ํกํ2ํฅ โ ํฆ)ํํฅ ํ ํํํฅ cos ํฅ ํํฆ + (ํฆโํกํ2ํฅ)ํํฅ = 0 ํ = ํฆโํกํ2ํฅ ํํฆ = 1 ํ = ํ ํํํฅ cos ํฅ ํํฅ = ํํํ ํฅ โ ํํํ ํฅ + ํ ํํํฅ โ (โํ ํํํฅ) = ํํํ 2ํฅ โ ํ ํํ2ํฅ ํํฆ โ ํํฅ ํ = 1 โ ํํํ 2ํฅ + ํ ํํ2ํฅ ํ ํํ ํฅ cos ํฅ = 2ํ ํํ2ํฅ ํ ํํ ํฅ cos ํฅ = 2ํ ํํํฅ cos ํฅ = 2ํกํํฅ ํ โซ 2ํกํํฅํํฅ = ํln(ํ ํํํฅ)2 = (ํ ํํํฅ)2 (ํ ํํํฅ)2ํ ํํํฅcos ํฅ ํํฆ + (ํ ํํํฅ)2(ํฆโํกํ2ํฅ)ํํฅ = 0 ํกํํฅ ํํฆ + (ํฆํ ํํ2ํฅ โ ํ ํํ2ํฅํกํ2ํฅ)ํํฅ = 0 ํกํํฅํํฆ + (ํฆํ ํํ2ํฅ โ ํ ํํ2ํฅํกํ2ํฅ)ํํฅ = 0 ํ = ํฆํ ํํ2ํฅ โ ํ ํํ2ํฅํกํ2ํฅ ํํฆ = ํ ํํ2ํฅ ํ = ํกํํฅ ํํฅ = sec 2 ํฅ
- 4. โซ ํํ = โซ ํํํฅ โซ ํํ = โซ(ํฆํ ํํ2ํฅ โ ํ ํํ2ํฅํกํ2ํฅ)ํํฅ ํ = ํฆํกํํฅ โ ํกํ3 3 + โ(ํฆ) ํํฆ = ํกํํฅ โ โโฒ(ํฆ) ํํฆ = ํ ํกํํฅ + โโฒ(ํฆ) = ํกํํฅ โซ ํโ(ํฆ) = โซ 0ํํฆ โ(ํฆ) = ํถ ํ = ํฆํกํํฅ + ํกํ3ํฅ 3 + ํ 2.b-) (ํ2ํฆ โ ํฆํํํ ํฅํฆ)ํํฅ + (2ํฅํ2ํฆ โ ํฅํํํ ํฅํฆ + 2ํฆ)ํํฆ = 0 ํ(ํฅ, ํฆ) = ํ2ํฆ โ ํฆํํํ ํฅํฆ ํ(ํฅ, ํฆ) = 2ํฅํ2ํฆ โ ํฅํํํ ํฅํฆ + 2ํฆ ํํฆ = 2ํ2ํฆ โ 1cos ํฅํฆ + (โํฆ)(โํ ํํ ํฅํฆ)ํฅ = 2ํ2ํฆ โ cos ํฅํฆ + (ํฅํฆํ ํํ ํฅํฆ) ํํฅ = 2ํ2ํฆ โ 1 cos ํฅํฆ + (โํฅ)(โํ ํํ ํฅํฆ)ํฆ = 2ํ2ํฆ โ cos ํฅํฆ + (ํฅํฆํ ํํ ํฅํฆ) ํํฆ = ํํฅ ํธํฅํํํกํ ํํ(ํฅ, ํฆ) ํํฅ = ํ(ํฅ, ํฆ) โซ ํํ(ํฅ, ํฆ) = โซ(ํ2ํฆ โ ํฆํํํ ํฅํฆ)ํํฅ = ํ(ํฅ, ํฆ) = ํฅํ2ํฆ โ ํฆํ ํํ ํฅํฆ ํฆ + โ(ํฆ) ํ(ํฅ, ํฆ) = ํฅํ2ํฆ โ ํ ํํ ํฅํฆ + โ(ํฆ)
- 5. ํํ(ํฅ, ํฆ) ํํฆ = 2ํฅํ2ํฆ โ ํฅํํํ ํฅํฆ + โยด(ํฆ) ํํ(ํฅ, ํฆ) ํํฆ = ํ(ํฅ, ํฆ) 2ํฅํ2ํฆ โ ํฅํํํ ํฅํฆ + โยด(ํฆ) = 2ํฅํ2ํฆ โ ํฅํํํ ํฅํฆ + 2ํฆ โยด(ํฆ) = 2ํฆ 2 โซ โ(ํฆ)ํํฆ โ(ํฆ) = ํฆ2 + ํถ ํ(ํฅ, ํฆ) = ํฅํ2ํฆ โ ํ ํํ ํฅํฆ + ํฆ2 + ํถ 3.) Resolver las ecuaciones diferenciales de orden N por coeficientes indeterminados. a .) y ,, ๏ซ y , ๏ฝ 2 e 2 x senx .) ,, 9 93 3cos b y ๏ซ y ๏ฝ x ๏ซ x 3 a) ํฆยดยด + ํฆยด = 2ํ2ํฅ ํ ํํ ํฅ Para la soluciรณn homogenea ํ2 + ํ = 0 ํ = 0 ํ ํ = โ1 ํฆโ = ํ1 + ํ2ํโํฅ ํท(ํท + 1)(ํท2 โ 4ํท + 5)ํฆ = 0 ํท(ํท + 1)(ํท โ 2 โ ํ)(ํท โ 2 + ํ)ํฆ = 0 ํฆ = ํ1 + ํ2ํโํฅ + ํดํ2ํฅ ํํํ ํฅ + ํตํ2ํฅ ํ ํํํฅ
- 6. Yh yp ํฆยดํ = ํด2ํ2ํฅ ํํํ ํฅ + (โํ ํํํฅ)ํดํ2ํฅ + ํต2ํ2ํฅ ํ ํํํฅ + ํํํ ํฅํตํ2ํฅ = (2ํด + ํต)ํ2ํฅ ํํํ ํฅ + (โํด + 2ํต)ํ2ํฅ ํ ํํํฅ ํฆยดยดํ = 2(2ํด + ํต)ํ2ํฅ ํํํ ํฅ + (2ํด + ํต)ํ2ํฅ (โํ ํํํฅ) + 2(โํด + 2ํต)ํ2ํฅ ํ ํํํฅ + (โํด + 2ํต)ํ2ํฅ ํํํ ํฅ = (3ํด + ํต)ํ2ํฅ ํํํ ํฅ + (โ4ํด + 3ํต)ํ2ํฅ ํ ํํํฅ ํฆยดยดํ + ํฆยดํ = (5ํด + 5ํต)ํ2ํฅ ํํํ ํฅ + (โ5ํด + 5ํต)ํ2ํฅ ํ ํํํฅ = 2ํ2ํฅ ํ ํํํฅ โ5ํด + 5ํต = 2 5ํด + 5ํต = 0 ํด = โ 1 5 ํฆ ํต = 1 5 ํฆ = ํ1 + ํ2ํโํฅ โ 1 5 ํ2ํฅ ํํํ ํฅ + 1 5 ํ2ํฅ ํ ํํํฅ 3 b ) ํฆยดยด + 9ํฆ = 93ํฅ + 3ํํํ ํฅ (ํท2 + 9)ํท2(ํท2 + 1)ํฆ = 0
- 7. ํฆ = ํถ1ํํํ 3ํฅ + 2ํถํ ํํ3ํฅ + ํถ3 + ํถ4ํฅ + ํถ5ํํํ ํฅ + ํถ6ํ ํํํฅ Yh yp ํฆยดํ = ํถ4 โ ํถ5ํ ํํํฅ + ํถ6ํํํ ํฅ ํฆยดยดํ = โํถ5ํํํ ํฅ โ ํถ6ํ ํํํฅ ํฆยดยดํ + 9ํฆํ = 93ํฅ + 3ํํํ ํฅ โํถ5ํํํ ํฅ โ ํถ6ํ ํํํฅ + 9(ํถ3 + ํถ4ํฅ + ํถ5ํํํ ํฅ + ํถ6ํ ํํํฅ) = 93ํฅ + 3ํํํ ํฅ 9ํถ3 + 9ํถ4ํฅ + 8ํถ5ํํํ ํฅ + 8ํถ6ํ ํํํฅ = 93ํฅ + 3ํํํ ํฅ 9ํถ3 = 0 ํถ3 = 0 9ํถ4 = 93 ํถ4 = 93 9 8ํถ5 = 3 ํถ5 = 3 8 8ํถ6 = 0 ํถ6 = 0 ํฆ = ํถ1ํํํ 3ํฅ + 2ํถํ ํํ3ํฅ + 0 + ํถ4ํฅ + ํถ5ํํํ ํฅ + 0ํ ํํํฅ ํฆ = ํถ1ํํํ 3ํฅ + ํถ2ํ ํํ3ํฅ + 93 9 ํฅ + 3 8 ํํํ ํฅ
- 8. 4.) Resuelva por variaciรณn de parรกmetros: 1 9 ,, ๏ซ ๏ฝ 4 y y Csc3x 4 a) ํฆยดยด + 9ํฆ = 1 4 csc 3ํฅ ํ2 + 9 = 0 ํฆโ = ํ1ํํํ 3ํฅ + ํ2ํ ํํ3ํฅ y1 y2 ํฆ1โฒ = โ3ํ ํํ3ํฅ ํฆ2โฒ = 3ํํํ 3ํฅ ํ(ํฅ) = 1 4 csc 3ํฅ ํค = | ํํํ 3ํฅ ํ ํํ3ํฅ โ3ํ ํํ3ํฅ 3ํํํ 3ํฅ | = 3ํํํ 23ํฅ + 3ํ ํํ23ํฅ = 3 ํค1 = | 0 ํ ํํ3ํฅ 1 4 csc 3ํฅ 3ํํํ 3ํฅ | = 0 โ 1 4 csc 3ํฅํ ํํ3ํฅ = โ 1 4 1 ํ ํํ3ํฅ ํ ํํ3ํฅ = โ 1 4 ํค2 = | ํํํ 3ํฅ 0 โ3ํ ํํ3ํฅ 1 4 csc 3ํฅ | = 1 4 csc 3ํฅํํํ 3ํฅ โ 0 = 1 4 1 ํ ํํ3ํฅ ํํํ 3ํฅ = 1 4 ํํกํ3ํฅ ํข1 โฒ = 1 4 3 โ = โ1 12 ํข1 = โซ โ1 12 ํํฅ = โ1 12 ํฅ + ํ
- 9. ํข2 โฒ = 1 4 ํํกํ3ํฅ 3 = 1 12 ํํํ 3ํฅ ํ ํํ3ํฅ ํข2 = โซ 1 12 ํํกํ3ํฅํํฅ 1 36 ํํ|ํ ํํ3ํฅ| + ํถ ํฆํ = ํฆ1 โ ํข1 + ํฆ2 โ ํข2 = ํํํ 3ํฅ โ (โ 1 12 ํฅ) + ํ ํํ3ํฅ โ 1 36 ํํ|ํ ํํ3ํฅ| ํฆ = ํ1ํํํ 3ํฅ + ํ2ํ ํํ3ํฅ โ 1 12 ํฅ โ ํํํ 3ํฅ + 1 36 . ํ ํํ3ํฅ โ ํํ|ํ ํํ3ํฅ|