Problemario Ecuaciones Diferenciales
Enviado por Cesarx9 • 4 de Septiembre de 2021 • Prácticas o problemas • 1.883 Palabras (8 Páginas) • 91 Visitas
César Solorio - 200038
1.- (1 − 𝑥)𝑦′′ − 4𝑥𝑦′ + 5𝑦 = cos(𝑥) → 𝐸𝐷𝑂 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑜𝑟𝑑𝑒𝑛.
2. 𝑥 𝑑3𝑦 𝑑𝑦 4[pic 1][pic 2]
𝑑𝑥3 − (𝑑𝑥)
+ 𝑦 = 0 → 𝐸𝐷𝑂 𝑛𝑜 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑡𝑒𝑟𝑐𝑒𝑟 𝑜𝑟𝑑𝑒𝑛
3. 𝑡5𝑦(4) − 𝑡3𝑦′′ + 6𝑦 = 0 → 𝐸𝐷𝑂 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑐𝑢𝑎𝑟𝑡𝑜 𝑜𝑟𝑑𝑒𝑛[pic 3]
4. 𝑑2𝑢 + 𝑑𝑢 + 𝑢 = cos(𝑟 + 𝑢) → 𝐸𝐷𝑂[pic 4][pic 5][pic 6][pic 7]
𝑑𝑒
𝑜𝑟𝑑𝑒𝑛
𝑑𝑟2 𝑑𝑟
[pic 8]
5. = √[pic 9][pic 10][pic 11]
[pic 12]
𝑑𝑦 2
[pic 13]
1 + ( )
𝑑𝑥
→ 𝐸𝐷𝑂
𝑑𝑒
𝑜𝑟𝑑𝑒𝑛
6. 𝑑2𝑅 = − 𝑘[pic 14][pic 15][pic 16][pic 17]
→ 𝐸𝐷𝑂
𝑑𝑒
𝑜𝑟𝑑𝑒𝑛
𝑑𝑡2
𝑅2
7. (𝑠𝑒𝑛𝜃)𝑦′′′ − (𝑐𝑜𝑠𝜃)𝑦′ = 2 → 𝐸𝐷𝑂 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑡𝑒𝑟𝑐𝑒𝑟 𝑜𝑟𝑑𝑒𝑛[pic 18][pic 19]
8. ẍ − (1 −
[pic 20]
ẋ ) ẋ + 𝑥 = 0 → 𝐸𝐷𝑂 𝑛𝑜 𝑙𝑖𝑛𝑒𝑎𝑙 𝑑𝑒 𝑠𝑒𝑔𝑢𝑛𝑑𝑜 𝑜𝑟𝑑𝑒𝑛[pic 21]
3
9. (𝑦2 − 1)𝑑𝑥 + 𝑥𝑑𝑦 = 0; 𝑒𝑛 𝑦; 𝑒𝑛 𝑥
Formula: 𝑎 (𝑥) 𝑑𝑦 + 𝑎
[pic 22]
(𝑥)𝑦 = 𝑔(𝑥)
1 𝑑𝑥 0
𝑦2 − 1 + 𝑥 𝑑𝑦 = 0[pic 23]
𝑑𝑥
𝑥 𝑑𝑦 + 𝑦2 = 1 →[pic 24][pic 25][pic 26]
𝑑𝑥
𝑒𝑛 𝑦
(𝑦2 − 1) 𝑑𝑦 + 𝑥 = 0 →[pic 27][pic 28][pic 29]
𝑑𝑥
𝑒𝑛 𝑥
10. 𝑢𝑑𝑣 + (𝑣 + 𝑢𝑣 + 𝑢𝑒𝑢)𝑑𝑢 = 0; 𝑒𝑛 𝑣; 𝑒𝑛 𝑢
𝑢 𝑑𝑣 + 𝑣 + 𝑢𝑣 + 𝑢𝑒𝑢 = 0[pic 30]
𝑑𝑢
𝑢 𝑑𝑣 + (1 + 𝑢)𝑣 = −𝑢𝑒𝑢 →[pic 31][pic 32][pic 33]
𝑑𝑢
en v
𝑢 + (𝑣 + 𝑢𝑣 + 𝑢𝑒𝑢) 𝑑𝑢 = 0 →[pic 34][pic 35]
𝑑𝑣
𝑒𝑛 𝑢
11.[pic 36][pic 37][pic 38]
2𝑦′ + 𝑦 = 0
𝑡𝑖𝑒𝑛𝑒 𝑑𝑜𝑚𝑖𝑛𝑖𝑜 (−∞, ∞)[pic 39]
𝑦′
= − 1
2[pic 40]
−𝑥
𝑒 2 𝑡𝑖𝑒𝑛𝑒 𝑑𝑜𝑚𝑖𝑛𝑖𝑜 (−∞, ∞)
Haciendo la sustitución:
1 𝑥 𝑥
[pic 41] [pic 42]
𝑥 𝑥
[pic 43] [pic 44]
2 (−
𝑒−2) + 𝑒−2 = 0 → −𝑒−2 + 𝑒−2 = 0
2[pic 45]
0 = 0
Por lo que 𝑦 = 𝑒[pic 46]
𝑥
[pic 47]
2 es una solución de la ecuación diferencial dada en el intervalo 𝐼 = (−∞, ∞)
12. 𝑑 𝑦 + 20𝑦 = 24; 𝑦 =[pic 48]
𝑑𝑡[pic 49]
6 − 6
5 5
𝑒−20𝑡
[pic 50]
...