This document contains solutions to 4 practice problems involving differential equations. The first problem involves solving the differential equation y''' - 6y'' = 3 - cos(x). The general solution is the sum of the complementary and particular solutions, where the complementary solution is C1e6x + C2e-6x + C3 and the particular solution is (1/7)sin(x) - (1/2)x. The second problem involves solving the differential equation y'v - y'' = 4x + 2xe-x. The general solution is a sum of exponential and polynomial terms, along with (-2/3)x3 and (-1/2)x2 - (

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