用拉普拉斯变换求解无界弦的振动问题:[tex=15.429x6.929]fnpmC2J6JmQBLyo5NmGAzwWskHkYW3pO4L9W/DfMamEXcn6IwHXgIHEnwVmxmH9K+gDNbMAundikDLgmJ6bHatLJqeq2IXf7S3YAZPzd0k0IP9qZ1hpmMR/lK9+daYfAw0yCLIU50OsSiWxVJWLBcB63lM3YLPJC8ukVOixRcNxhxVTqv0pfVOEdsBazNchgeA36gql052p2D/C7BsuWLF4zDDjnVqylvbmZjDm319uT4eVIhBF05YzI3m0HZ0S1EvtuCbHlmVG6lWwDi3XjImF2xOBqn5tAOD11qYT/nkk=[/tex]其中[tex=2.071x1.357]+M91MMaKogfF4Ena+s6i7Q==[/tex]和[tex=2.0x1.357]eXqHnCRPYNHW2O3HwFxPDA==[/tex]是已知函数.
举一反三
- 用拉普拉斯变换求解无界弦的振动问题:[tex=15.429x6.929]fnpmC2J6JmQBLyo5NmGAzwWskHkYW3pO4L9W/DfMamEXcn6IwHXgIHEnwVmxmH9K+gDNbMAundikDLgmJ6bHatLJqeq2IXf7S3YAZPzd0k0IP9qZ1hpmMR/lK9+daYfAw0yCLIU50OsSiWxVJWLBcB63lM3YLPJC8ukVOixRcNxhxVTqv0pfVOEdsBazNchgeA36gql052p2D/C7BsuWLF4zDDjnVqylvbmZjDm319uT4eVIhBF05YzI3m0HZ0S1EvtuCbHlmVG6lWwDi3XjImF2xOBqn5tAOD11qYT/nkk=[/tex]其中[tex=2.071x1.357]+M91MMaKogfF4Ena+s6i7Q==[/tex]和[tex=2.0x1.357]eXqHnCRPYNHW2O3HwFxPDA==[/tex]是已知函数.
- 以下程序段实现的输出是()。for(i=0;i<;=9;i++)s[i]=i;for(i=9;i>;=0;i--)printf("%2d",s[i]);[/i][/i] A: 9 7 5 3 1 B: 1 3 5 7 9 C: 9 8 7 6 5 4 3 2 1 0 D: 0 1 2 3 4 5 6 7 8 9
- 【计算题】5 ×8= 6×4= 7×7= 9×5= 2×3= 9 ×2= 8×9= 7×8= 5×5= 4×3= 5+8= 6 ×6= 3×7= 4×8= 9×3= 1 ×2= 9×9= 6×8= 8×0= 4×7=
- set1 = {x for x in range(10)} print(set1) 以上代码的运行结果为? A: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} B: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10} C: {1, 2, 3, 4, 5, 6, 7, 8, 9} D: {1, 2, 3, 4, 5, 6, 7, 8, 9,10}
- 设DES加密算法中的一个S盒为: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 0 1 2 3 14 4 13 1 2 15 11 8 3 10 6 12 5 9 0 7 0 15 7 4 14 2 13 1 10 6 12 11 9 5 3 8 4 1 14 8 13 6 2 11 15 12 9 7 3 10 5 0 15 12 8 2 4 9 1 7 5 11 A: 1010 B: 0001 C: 1011 D: 0111