Ejercicios john rangel
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1. This document contains solutions to several calculus exercises involving differential equations. 2. One exercise involves solving the differential equation y'' + 9y = 0, which has general solution y(x) = c1sin3x + c2cos3x. 3. Another exercise involves solving the non-homogeneous differential equation y'' + y' = 2e^2x/csc3x, yielding the general solution y(x) = c1 + c2e^-x + (1/5)e^2x(sinx - cosx).
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![2015
ALUMNO:
JOHN RANGEL
C.I: 25.753.613
[EJERCICIOS]
EJERCICIOSNUMERO 2 DE CALCULO 4](https://image.slidesharecdn.com/ejerciciosjohnrangel-150721021351-lva1-app6891/85/Ejercicios-john-rangel-1-320.jpg)
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- 1. 2015 ALUMNO: JOHN RANGEL C.I: 25.753.613 [EJERCICIOS] EJERCICIOSNUMERO 2 DE CALCULO 4
- 2. Ejercicio nº4 y 2.a 𝑦´´ + 9𝑦 = 1 4 𝑐𝑠𝑐3𝑥 𝑦´´ + 9𝑦 = 0 𝑅 = ±3𝑖 Por lo cual 𝑦ℎ( 𝑥) = 𝑐1 𝑠𝑒𝑛3𝑥 + 𝑐2 𝑐𝑜𝑠3𝑥 𝑓1( 𝑥) = 𝑠𝑒𝑛3𝑥 𝑓2( 𝑥) = 𝑐𝑜𝑠3𝑥 Y de la ecuación diferancial 𝑔( 𝑥) = 𝑐𝑠𝑒3𝑥 4 𝑤 = 𝑓1( 𝑥) 𝑓2( 𝑥) 𝑓1(𝑥) 𝑓2(𝑥) = 𝑠𝑒𝑛3𝑥 𝑐𝑜𝑠3𝑥 3𝑐𝑜𝑠3𝑥 −3𝑠𝑒𝑛3𝑥 = −3 𝑠𝑒𝑛2 3𝑥 − 3𝑐𝑜𝑠2 3𝑥 = −3( 𝑠𝑒𝑛2 3𝑥 + 𝑐𝑜𝑠2 3𝑥) = −3 𝑤1 = 0 𝑓2(𝑥) 𝑔(𝑥) 𝑓2(𝑥) = 0 𝑐𝑜𝑠3𝑥 𝑐𝑠𝑐3 3 −𝑠𝑒𝑛3𝑥 𝑤2 = 𝑓1(𝑥) 0 𝑓1(𝑥) 𝑔(𝑥) = 𝑠𝑒𝑛3𝑥 0 3𝑐𝑜𝑠3𝑥 𝑐𝑠𝑐3𝑥 3 = 1 3 𝑙𝑛3𝑥 − csc 3𝑥
- 3. Ejercicio nº 3.a 𝑦´´ + 𝑦´ = 2𝑒2𝑥 𝑠𝑒𝑛𝑥 𝑦´´ + 𝑦´ = 0 𝑟2 + 𝑟 = 0 𝑟( 𝑟 + 1) = 0 𝑎𝑠𝑖 𝑦ℎ( 𝑥) = 𝐶1 𝑒0𝑥 + 𝐶2 𝑒̅ 𝑥 𝑦ℎ( 𝑥) = 𝐶1 + 𝐶2 𝑒̅ 𝑥 𝑦´´ + 𝑦´𝑝 = 𝑒2𝑥[ (3𝐴 − 4𝐵) 𝑠𝑒𝑛 𝑥 + (4𝐴 + 3𝐵) 𝑐𝑜𝑠𝑥] +𝑒2𝑥 [(2𝐴 − 𝐵) 𝑠𝑒𝑛𝑥 + (2𝐵 + 𝐴) 𝑐𝑜𝑠𝑥] = 2𝑒2𝑥 𝑠𝑒𝑛𝑥 𝑎𝑠𝑖: 5𝐴 − 5𝐵 = 2 5𝐴 + 5𝐵 = 0 10𝐴 = 2 𝐴 = 1 5 𝐵 = − 1 5 𝑐𝑜𝑛 𝑙𝑜 𝑐𝑢𝑎𝑙 𝑦𝑝( 𝑥) = 𝑒2𝑥 ( 1 5 𝑠𝑒𝑛 − 1 5 𝑐𝑜𝑠𝑥) = 1 5 𝑒2𝑥 ( 𝑠𝑒𝑛𝑥 − 𝑐𝑜𝑠𝑥) 𝑎𝑠𝑖 𝑙𝑎 𝑠𝑜𝑙𝑢𝑐𝑖𝑜𝑛 𝑡𝑜𝑡𝑎𝑙 𝑒𝑠 𝑦( 𝑥) = 𝑐1 + 𝑐2𝑒2𝑥 + 1/5𝑒2𝑥( 𝑠𝑒𝑛 𝑥 − 𝑐𝑜𝑠𝑥)
- 4. Ejercicio 3.b 𝑦´´ + 9𝑦 = 3𝑥 + 3𝑐𝑜𝑠𝑥 𝑦´´ + 9𝑦 = 0 𝑟2 + 9 = 0 𝑟 = ±3𝑖 𝑒𝑛 𝑦ℎ( 𝑥) 𝑐1 cos 3𝑥 + 𝑐2 𝑠𝑒𝑛𝑜3𝑥 𝑠𝑒𝑎 𝑦𝑝( 𝑥) = 𝑎𝑥 + 𝐵𝑐𝑜𝑠 + 𝐶𝑠𝑒𝑛𝑥 𝑦𝑝( 𝑥) = 𝐴 + 𝐵 𝑠𝑒𝑎 𝑥 + 𝐶 𝑐𝑜𝑠𝑥 𝑦𝑝´´( 𝑥) = −𝐵 𝑐𝑜𝑠𝑥 − 𝐶 𝑠𝑒𝑛𝑥 𝑦𝑝 + 𝑎𝑦𝑝 = −𝐵𝑐𝑜𝑠𝑥 − 𝐶𝑠𝑒𝑛𝑥 +9( 𝑎𝑥 + 𝐵𝑐𝑜𝑠𝑥 + 𝐶𝑠𝑒𝑛𝑥) −𝐵 𝑐𝑜𝑠𝑥 − 𝐶 𝑠𝑒𝑛𝑥 + 9 𝐴𝑥 + 9𝐵𝑐𝑜𝑠𝑥 = 3𝑥 + 3𝑐𝑜𝑠𝑥 9𝐴𝑥 + (8𝐵𝑐𝑜𝑠𝑥) + 𝐵𝐶𝑠𝑒𝑛𝑥 = 3𝑥 + 3𝑐𝑜𝑠𝑥 9𝐴 = 3 −→ 𝐴 = 3 9 = 1 3 𝐵𝐵 = 3 −→ 𝐵 = 3 8 𝐵𝑐 = 0 𝑎𝑠𝑖 𝑦𝑝( 𝑥) = 1 3 𝑥 + 3 8 𝑐𝑜𝑠𝑥 𝑠𝑜𝑙𝑢𝑐𝑖𝑜𝑛 𝑡𝑜𝑡𝑎𝑙 𝑦( 𝑥) = 𝑐1𝑐𝑜𝑠3𝑥 + 𝑐2 𝑠𝑒𝑛3𝑥 + 1 3 𝑥 + 3 8 𝑐𝑜𝑠𝑥
- 5. Ejercicio nº 2.b ( 𝑒2𝑦 − 𝑦𝑐𝑜𝑠𝑥𝑦) 𝑑𝑥 + (2 𝑥 𝑒2𝑦 − 𝑥𝑐𝑜𝑠𝑥𝑦 + 2𝑦) 𝑑𝑦 𝑀( 𝑥, 𝑦) = 𝑒2𝑦 − 𝑦 cos 𝑥𝑦 𝑀 𝑦 = 2𝑒2𝑦 − 𝑐𝑜𝑠𝑥𝑦 + 𝑥𝑦 𝑠𝑒𝑛𝑥𝑦 𝑛( 𝑥, 𝑦) = 2𝑥𝑒2𝑦 − 𝑥𝑐𝑜𝑠𝑥𝑦 + 2𝑦 𝑛 𝑥 2𝑒2𝑦 + 𝑥𝑦 𝑠𝑒𝑛 𝑥𝑦 − 𝑐𝑜𝑠𝑥𝑦 𝑙𝑢𝑒𝑔𝑜 𝑙𝑎 𝑒𝑐𝑢𝑎𝑐𝑖𝑜𝑛 𝑑𝑖𝑓𝑒𝑟𝑒𝑛𝑐𝑖𝑎𝑙 𝑒𝑠 𝑒𝑥𝑎𝑐𝑡𝑎 𝑓( 𝑥, 𝑦) = ∫ 𝑀( 𝑥, 𝑦) 𝑑𝑥 ∫( 𝑒2𝑦 − 𝑦𝑐𝑜𝑠𝑥𝑦) 𝑑𝑥 𝑥𝑒2𝑦 − 𝑠𝑒𝑛 𝑥𝑦 + 𝑔( 𝑦) 𝑦 𝑎𝑓 𝑎𝑦 𝑛 2𝑥𝑒2𝑦 − 𝑥𝑐𝑜𝑠𝑥𝑦 + 𝑔´( 𝑔) = 2𝑥𝑒2𝑦 − 𝑥𝑐𝑜𝑠𝑥𝑦 𝑔´( 𝑔) = 2𝑦 𝑐𝑜𝑛 𝑙𝑜 𝑐𝑢𝑎𝑙: 𝑔( 𝑦) = ∫ 2𝑦𝑑𝑦 = 2𝑦2 2 . 𝑐 𝑎𝑠𝑖 𝑓( 𝑥, 𝑦) = 𝑥𝑒2𝑦 − 𝑠𝑒𝑛 𝑥𝑦 + 𝑦2 + 𝑐 𝑠𝑜𝑙𝑢𝑐𝑖𝑜𝑛 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑒𝑠 𝑒2𝑦 − 𝑠𝑒𝑛𝑥𝑦 + 𝑦2 + 𝑐 = 0