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