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Diapositiva de Estudio: SolPrac2Am4.pptx
- 1. SOLUCIÓN : PRÁCTICA 2 ANÁLISIS MATEMÁTICO IV INTEGRANTES: -PERCY EDUARDO TAFUR CAVIDES -JORGE JESÚS VELÁSQUEZ CRUZ
- 2. EJERCICIO (1): 𝑦′′′ − 6′′ = 3 − 𝑐𝑜𝑠𝑥 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a) 𝑦ℎ → 𝑦′′′ − 6′′ = 0 Ecuación característica: 𝑟3 − 6𝑟 = 0 𝑟1 = 0 → 1 𝑟 𝑟2 − 6 = 0 𝑟2 = 6 → 𝑒 6 𝑟3 = − 6 → 𝑒− 6 → 𝑦ℎ = 𝐶1 + 𝐶2 𝑒 6𝑥 + 𝐶3𝑒− 6𝑥 b) 𝑦𝑝1 → 𝑓(𝑥) = 3 𝑚 = 0, 0 ϵ 𝑠𝑝 𝑠 = 1 𝑦𝑝1 = 𝐴𝑥 𝑦𝑝1′ = 𝐴 𝑦𝑝1′′ = 0 𝑦𝑝1′′′ = 0
- 3. → −6𝐴 = 3 → 𝐴 = 1 2 𝑦𝑝1 = − 1 2 𝑥 𝑓 𝑥 = −𝑐𝑜𝑠𝑥 𝑎 = 0, 𝑏 = 1, 𝑎 + 𝑏𝑖 ϵ 𝑠𝑝 𝑦𝑝2 = 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑒𝑛𝑥 𝑦𝑝2 ′ = −𝐴𝑠𝑒𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥 𝑦𝑝3 ′′ = −𝐴𝑐𝑜𝑠𝑥 − 𝐵𝑠𝑒𝑛𝑥 𝑦𝑝3 ′′′ = 𝐴𝑠𝑒𝑛𝑥 − 𝐵𝑐𝑜𝑠𝑥 𝐴𝑠𝑒𝑛𝑥 − 𝐵𝑐𝑜𝑠𝑥 − 6 −𝐴𝑠𝑒𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥 = −𝑐𝑜𝑠𝑥 7𝐴𝑠𝑒𝑛𝑥 − 7𝐵𝑐𝑜𝑠𝑥 = −𝑐𝑜𝑠𝑥 → 𝐴 = 0, 𝑏 = 1 7 → 𝑦𝑝2 = 1 7 𝑠𝑒𝑛𝑥 𝑦𝑔 = 𝐶1 + 𝐶2𝑒 6𝑥 + 𝐶3 − 6𝑥 − 1 2 𝑥 + 1 7 𝑠𝑒𝑛𝑥
- 4. EJERCICIO (2): 𝑦′𝑣 − 𝑦′′ = 4𝑥 + 2𝑥𝑒−𝑥 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a) yh → 𝑦′𝑣 − 𝑦′′ = 0 → 𝑟4 − 𝑟2 = 0 → 𝑟2 𝑟2 − 1 = 0 𝑟1 = 0 → 1 𝑟2 = 0 → 𝑥 𝑟3 = 1 → 𝑒𝑥 𝑟4 = −1 → 𝑒−𝑥 𝑦ℎ = 𝐶1 + 𝐶2𝑥 + 𝐶3𝑒𝑥 + 𝐶4𝑒−𝑥 b) 𝑦𝑝1 → 𝑓(𝑥) = 4𝑥 𝑚 = 1 0ϵ 𝑠𝑝 𝑠 = 2 𝑦𝑝1 = 𝑥2 𝐴𝑥 + 𝐵 𝑦𝑝1 = 𝐴𝑥3 + 𝐵𝑥2 𝑦′𝑝1 = 3𝐴𝑥2 + 𝐵𝑥 𝑦′′𝑝1 = 6𝐴𝑥 + 𝐵 𝑦′′′𝑝1 = 6𝐴 𝑦𝑝1 ′𝑣 = 0
- 5. − 6𝐴𝑥 − 𝐵 = 4𝑥 −6𝐴 − 𝐵 = 4𝑥 𝐴 = − 2 7 , 𝐵 = 0 𝑦𝑝1 = − 2 3 𝑥3 𝑦𝑝2 → 𝑓(𝑥) = 2𝑥𝑒−𝑥 m = 1, a = −1, a ϵ sp , s = 1 𝑦𝑝2 = 𝑥 𝐴𝑥 + 𝐵 𝑒−𝑥 𝑦𝑝2 = (𝐴𝑥2 + 𝐵𝑥)𝑒−𝑥 𝑦′𝑝2 = 2𝐴𝑥 + 𝐵 𝑒−𝑥 − 𝐴𝑥2 + 𝐵𝑥 𝑒−𝑥 𝑦′𝑝2 = −𝐴𝑥2 + 2𝐴𝑥 − 𝐵𝑥 + 𝐵 𝑒−𝑥 𝑦′′𝑝2 = −2𝐴𝑥 + 2𝐴 − 𝐵 𝑒−𝑥 − −𝐴𝑥2 + 2𝐴𝑥 − 𝐵𝑥 + 𝐵 𝑒−𝑥 𝑦′′𝑝2 = 𝐴𝑥2 − 4Ax + Bx + 2A − 2B 𝑒−𝑥 → 𝑦′′′𝑝2 = 2𝐴𝑥 − 4𝐴 + 𝐵 𝑒−𝑥 − (𝐴𝑥2 − 4𝐴𝑥 + 𝐵𝑥 + 2𝐴 − 2𝐵)𝑒−𝑥
- 6. 𝑦′′′𝑝2 = −𝐴𝑥2 + 6𝐴𝑥 − 𝐵𝑥 − 6𝐴 + 3𝐵 𝑒−𝑥 𝑦𝑝2 ′𝑣 = −2𝐴𝑥 + 6𝐴 − 𝐵 𝑒−𝑥 − (−𝐴𝑥2 + 6𝐴𝑥 − 𝐵𝑥 − 6𝐴 + 3𝐵)𝑒−𝑥 𝑦𝑝2 ′𝑣 = 𝐴𝑥2 − 8𝐴𝑥 + 𝐵𝑥 + 12𝐴 − 4𝐵 𝑒−𝑥 𝐴𝑥2 − 8𝐴𝑥 + 𝐵𝑥 + 12𝐴 − 4𝐵 𝑒−𝑥 − 𝐴𝑥2 − 4𝐴𝑥 + 𝐵𝑥 + 2𝐴 − 2𝐵 𝑒−𝑥 = 2𝑥𝑒−𝑥 −4𝐴𝑥 + 10𝐴 − 2𝐵 = 2𝑥 𝑎 = − 1 2 , 10𝐴 − 2𝐵 = 0 → 𝑏 = − 5 2 𝑦𝑝2 = − 1 2 𝑥2 − 5 2 𝑥 𝑒−𝑥 𝑦𝑔 = 𝐶1 + 𝐶2𝑥 + 𝐶3𝑒𝑥 + 𝐶4𝑒−𝑥 − 2 3 𝑥3 − ( 1 2 𝑥2 + 5 2 𝑥)𝑒−𝑥
- 7. EJERCICIO (3): 𝑦′′ + 3𝑦′ + 2𝑦 = 1 1 + 𝑒𝑥 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a) yh: 𝑦′′ + 3𝑦′ + 2𝑦 = 0 𝑟2 + 3𝑟 + 2 = 0 𝑟1 = −1 → 𝑒−𝑥 𝑟2 = −2 → 𝑒−2𝑥 𝑦ℎ = 𝐶1𝑒−𝑥 + 𝐶2𝑒−2𝑥 b) yp: 𝑦𝑝 = 𝑣1𝑦1 + 𝑣2𝑦2 𝑦1 = 𝑒−𝑥 𝑦2 = 𝑒−2𝑥 𝑤 = 𝑒−𝑥 𝑒−2𝑥 −𝑒−𝑥 −2𝑒−2𝑥 𝑤 = −2𝑒−3𝑥 − −𝑒−3𝑥 𝑤 = −𝑒−3𝑥
- 8. 𝑣1 = − 𝑒−2𝑥 1 1 + 𝑒𝑥 −𝑒−3𝑥 𝑑𝑥 = 𝑒𝑥 1 + 𝑒𝑥 𝑑𝑥 𝑢 = 1 + 𝑒𝑥 𝑑𝑢 = 𝑒𝑥𝑑𝑥 𝑣1 = 𝑑𝑢 𝑢 = ln |𝑢| = ln |1 + 𝑒𝑥| 𝑣2 = 𝑒−𝑥( 1 1 + 𝑒𝑥) −𝑒−32 𝑑𝑥 = − 𝑒2𝑥 1 + 𝑒𝑥 𝑑𝑥 = − 𝑒𝑥 − 𝑒𝑥 1 + 𝑒𝑥 𝑑𝑥 = − 𝑒2 𝑑𝑥 + 𝑒𝑥 1 + 𝑒𝑥 𝑑𝑥 𝑣2 = −𝑒𝑥 + ln |1 + 𝑒𝑥 | 𝑦𝑝 = ln |1 + 𝑒𝑥 | 𝑒−𝑥 + −𝑒𝑥 + ln 1 + 𝑒𝑥 𝑒−2𝑥 𝑦𝑔 = 𝐶1𝑒−𝑥 + 𝐶2𝑒−2𝑥 + ln 1 + 𝑒𝑥 𝑒−𝑥 − (𝑒𝑥 − ln |1 + 𝑒𝑥 |) 𝑒−2𝑥
- 9. EJERCICIO (4): 𝑦′′′ + 𝑦′ = 𝑡𝑎𝑛𝑥 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a) yh: 𝑦′′′ + 𝑦′ = 0 𝑟3 + 𝑟 = 0 𝑟(𝑟2 + 1) = 0 𝑟1 = 0 → 1 𝑟2 = 𝑖 → 𝑐𝑜𝑠𝑥 𝑟3 = −𝑖 → 𝑠𝑒𝑛𝑥 𝑦ℎ = 𝐶1 +𝐶2 𝑐𝑜𝑠𝑥 + 𝐶3𝑠𝑒𝑛𝑥 b)𝑦𝑝 = 𝑣1𝑦1 + 𝑣2𝑦2 + 𝑣3𝑦3
- 10. 𝑤 = 1 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 0 −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 0 −𝑐𝑜𝑠𝑥 −𝑠𝑒𝑛𝑥 𝑤 = 𝑠𝑒𝑛2𝑥 − (−𝑐𝑜𝑠2𝑥) = 1 𝑤1 = 0 𝑐𝑜𝑠𝑥 𝑠𝑒𝑛𝑥 0 −𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥 −𝑐𝑜𝑠𝑥 −𝑠𝑒𝑛𝑥 𝑤1 = 𝑡𝑎𝑛𝑥 𝑐𝑜𝑠2𝑥 + 𝑠𝑒𝑛2𝑥 = 𝑡𝑎𝑛𝑥 𝑤2 = 1 0 𝑠𝑒𝑛𝑥 0 0 𝑐𝑜𝑠𝑥 0 𝑡𝑎𝑛𝑥 −𝑠𝑒𝑛𝑥 𝑤2 = −𝑡𝑎𝑛𝑥𝑐𝑜𝑠𝑥 𝑤3 = 1 𝑐𝑜𝑠𝑥 0 0 −𝑠𝑒𝑛𝑥 0 0 −𝑐𝑜𝑠𝑥 𝑡𝑎𝑛𝑥 𝑤3 = −𝑠𝑒𝑛𝑥𝑡𝑎𝑛𝑥 → 𝑆𝑖: 𝑣1′ = 𝑤1 𝑤 = 𝑡𝑎𝑛𝑥 1 → 𝑣1 = 𝑡𝑎𝑛𝑥𝑑𝑥 = ln |𝑠𝑒𝑐𝑥|
- 11. 𝑣2 = − 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 𝑐𝑜𝑠𝑥𝑑𝑥 = 𝑐𝑜𝑠𝑥 𝑣3 = −𝑠𝑒𝑛𝑥𝑡𝑎𝑛𝑥𝑑𝑥 = − 𝑠𝑒𝑛𝑥 𝑠𝑒𝑛𝑥 𝑐𝑜𝑠𝑥 𝑑𝑥 = − 𝑠𝑒𝑛2𝑥𝑑𝑥 𝑐𝑜𝑠𝑥 = − 1 − 𝑐𝑜𝑠2𝑥 𝑑𝑥 𝑐𝑜𝑠𝑥 𝑣3 = − 𝑑𝑥 𝑐𝑜𝑠𝑥 + 𝑐𝑜𝑠𝑥𝑑𝑥 = − 𝑠𝑒𝑐𝑥𝑑𝑥 + 𝑐𝑜𝑠𝑥𝑑𝑥 = −ln |𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥| + 𝑠𝑒𝑛𝑥 𝑦𝑝 = ln 𝑠𝑒𝑐𝑥 + 𝑐𝑜𝑠2𝑥 − ln 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑛𝑥 𝑠𝑒𝑛𝑥 𝑦𝑔 = 𝐶1 + 𝐶2𝑐𝑜𝑠𝑥 + 𝐶3𝑠𝑒𝑛𝑥 + ln 𝑠𝑒𝑐𝑥 + 𝑐𝑜𝑠2 𝑥 − ln 𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥 − 𝑠𝑒𝑛𝑥 𝑠𝑒𝑛𝑥
- 12. EJERCICIO (5): 3𝑦′′ − 6𝑦′ + 6𝑦 = 𝑒𝑥 𝑠𝑒𝑐𝑥 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a)yh : 3𝑦′′ − 6𝑦′ + 6𝑦 = 0 3𝑟2 − 6𝑟 + 6 = 0 𝑟2 − 2𝑟 + 2 = 0 𝑟1 = 1 + 𝑖 → 𝑒𝑥 𝑐𝑜𝑠𝑥 𝑟2 = 1 − 𝑖 → 𝑒𝑥 𝑠𝑒𝑛𝑥 𝑦ℎ = 𝐶1𝑒𝑥 𝑐𝑜𝑠𝑥 + 𝐶2𝑒𝑥 𝑠𝑒𝑛𝑥 b)yp : 𝑦𝑝 = 𝑣1𝑦1 + 𝑣2𝑦2 𝑦1 = 𝑒𝑥𝑐𝑜𝑠𝑥 𝑦2 = 𝑒𝑥𝑠𝑒𝑛𝑥
- 13. 𝑤 = 𝑒𝑥𝑐𝑜𝑠𝑥 𝑒𝑥𝑠𝑒𝑛𝑥 𝑒𝑥 𝑐𝑜𝑠𝑥 − 𝑒𝑥 𝑠𝑒𝑛𝑥 𝑒𝑥 𝑠𝑒𝑛𝑥 + 𝑒𝑥 𝑐𝑜𝑠𝑥 𝑤 = 𝑒𝑥𝑐𝑜𝑠𝑥 𝑒𝑥𝑠𝑒𝑛𝑥 + 𝑒𝑥𝑐𝑜𝑠𝑥 − 𝑒𝑥𝑠𝑒𝑛𝑥 𝑒𝑥𝑐𝑜𝑠𝑥 − 𝑒𝑥𝑠𝑒𝑛𝑥 𝑤 = 𝑒2𝑥 𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑥 + 𝑒2𝑥 𝑐𝑜𝑠2 𝑥 − 𝑒2𝑥 𝑠𝑒𝑛𝑥𝑐𝑜𝑠𝑥 + 𝑒2𝑥 𝑠𝑒𝑛2 𝑥 𝑤 = 𝑒2𝑥 𝑣1 = − 𝑒𝑥𝑠𝑒𝑛𝑥 𝑒𝑥𝑠𝑒𝑐𝑥 𝑑𝑥 𝑒2𝑥 = − 𝑒2𝑥𝑠𝑒𝑛𝑥𝑠𝑒𝑐𝑥𝑑𝑥 𝑒2𝑥 = − 𝑡𝑎𝑛𝑥𝑑𝑥 = − ln 𝑠𝑒𝑐𝑥 = ln |𝑐𝑜𝑠𝑥| 𝑣2 = 𝑒𝑥 𝑐𝑜𝑠𝑥 𝑒𝑥 𝑠𝑒𝑐𝑥 𝑑𝑥 𝑒2𝑥 = 𝑒2𝑥 𝑐𝑜𝑠𝑥𝑠𝑒𝑐𝑥𝑑𝑥 𝑒2𝑥 = 𝑐𝑜𝑠𝑥 𝑑𝑥 𝑐𝑜𝑠𝑥 = 𝑑𝑥 = 𝑥 𝑦𝑝 = ln 𝑐𝑜𝑠𝑥 𝑒𝑥𝑐𝑜𝑠𝑥 + 𝑥𝑒𝑥𝑠𝑒𝑛𝑥 𝑦𝑝 = 𝐶1𝑒𝑥 𝑐𝑜𝑠𝑥 + 𝐶2𝑒𝑥 𝑠𝑒𝑛𝑥 + ln cosx 𝑒𝑥 𝑐𝑜𝑠𝑥 + 𝑥𝑒𝑥 𝑠𝑒𝑛𝑥
- 14. EJERCICIO (6): 4𝑦′′ − 4𝑦′ + 𝑦 = 𝑒 𝑥 2 1 − 𝑥2 𝑦𝑔 = 𝑦ℎ + 𝑦𝑝 a)Yh : 4𝑦′′ − 4𝑦′ + 𝑦 = 0 𝑟1 = 1 2 → 𝑒 1 2𝑥 𝑟2 = 1 2 → 𝑥𝑒 1 2𝑥 𝑦ℎ = 𝐶1𝑒 1 2𝑥 + 𝐶2𝑒 1 2𝑥 b)yp:𝑦𝑝 = 𝑣1𝑦1 + 𝑣2𝑦2 𝑦1 = 𝑒 1 2𝑥 𝑦2 = 𝑥𝑒 1 2𝑥
- 15. 𝑤 = 𝑒 1 2 𝑥 𝑥𝑒 1 2 𝑥 1 2 𝑒 1 2 𝑥 𝑒 1 2 𝑥 + 1 2 𝑥𝑒 1 2 𝑥 𝑤 = 𝑒 1 2 𝑥 𝑒 1 2 𝑥 + 1 2 𝑥𝑒 1 2 𝑥 − 1 2 𝑒 1 2 𝑥 𝑥𝑒 1 2 𝑥 = 𝑒𝑥 + 1 2 𝑥𝑒𝑥 − 1 2 𝑥𝑒𝑥 = 𝑒𝑥 𝑣1 = − 𝑥𝑒 1 2 𝑥 𝑒 1 2 𝑥 1 − 𝑥2 𝑒𝑥 𝑑𝑥 = − 𝑒𝑥 𝑥 1 − 𝑥2 𝑒𝑥 𝑑𝑥 = − 𝑥 1 − 𝑥2 1 2𝑑𝑥 = 1 3 (1 − 𝑥2 ) 3 2 𝑣2 = 𝑒 1 2 𝑥 𝑒 1 2 𝑥 1 − 𝑥2 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 1 − 𝑥2 𝑒𝑥 𝑑𝑥 = 1 − 𝑥2𝑑𝑥 = 𝑥 1 − 𝑥2 2 + 𝑎𝑟𝑐𝑠𝑒𝑛𝑥 2 𝑦𝑝 = 1 3 (1 − 𝑥2) 3 2 𝑒 1 2 𝑥 + 𝑥 1 − 𝑥2 2 + 𝑎𝑟𝑐𝑠𝑒𝑛𝑥 2 𝑥𝑒 1 2 𝑥 𝑦𝑔 = 𝐶1𝑒 1 2𝑥 + 𝐶2𝑥𝑒 1 2𝑥 + 1 3 (1 − 𝑥2 ) 3 2 𝑒 1 2𝑥 + 𝑥 1 − 𝑥2 2 + 𝑎𝑟𝑐𝑠𝑒𝑛𝑥 2 𝑥𝑒 1 2𝑥