For decades, relational databases remained essentially unchanged; data was segmented into specific chunks for columns, slots, and repositories, also called structured data. However, in this Internet of Things (IoT) era, databases need to be reengineered because the very nature of data has changed. Today’s databases need to be developed with the needs of IoT in mind and have the ability to perform real-time processing to manage workloads that are dynamic. For example, relational databases should be able to work with real-time data streaming and big data (an example was presented in the Unit III Lesson).Scenario: Falcon Security wants their customers to be able to view security video footage in real-time and provide customers with the ability to query video footage for viewing. Choosing a database solution such as MongoDB would allow Falcon Security to store customer video footage in the same database as the metadata.To do this, Falcon Security needs a way to manage the demands of real-time data streaming for real-time analytics. Conduct some research for a NoSQL database application, such as MongoDB or Cassandra, that could meet this need. How would switching to a real-time database solution help Falcon Security remain competitive? Create a PowerPoint presentation that includes the components listed below.Provide a brief introduction to IoT.Present the argument to the Falcon Security CEO that switching to a more dynamic database structure (NoSQL real-time database) will meet the demands of IoT.Introduce some features of the database you chose, whether it is MongoDB, Cassandra, or another database.Describe how switching to a more dynamic database will give Falcon Security a competitive advantage.
Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′ − 4y′ + 3y = x ; use y1 = e3x vi. y′′ + 5y′ + 6y = e2xcos(x) vii. y′′ + y = sec(x) tan(x)
Obtain the general solution of the following DEs: i. y′′′ + y′′ − 4y′ + 2y = 0 ii. y(4) + 4y(2) = 0 iii. x(x − 2)y′′ + 2(x − 1)y′ − 2y = 0; use y1 = (1 − x) iv. y′′ − 4y = sin2(x) v. y′′