Professional Platform for Ethics and LeadershipThe role of the health care professional includes being a moral agent or a person whose actions affect themselves and others at a moral level. It is important to have a personal ethic or moral framework in which you ground your practice and professional relationships. The purpose of this assignment is to explore and create a foundation for leadership and ethics in your professional practice.Write a reflection on the nature, sources, and implications of your values, beliefs, and ethical perspectives that guide your personal life and nursing practice. Please note, grading is based on the clarity and depth of your writing and the apparent effort given the assignment, not on the rightness or wrongness of your position. You are encouraged to be honest in your self-assessments and conclusions.Each of the following points must be addressed in your essay:Primary influences (childhood and adult)Ethical principles that influence you personally and professionallyEthical practice of professional nursingEthical leadership and professional development plan. Include both positive and negative aspects of your character that emerge when you are in a position of authorityAssignment Expectations:Length: 1500 to 2000 words total.Structure: Include a title page and reference page in APA format. These do not count towards the minimum word count for this assignment. Your essay must include an introduction and a conclusion.References: Use appropriate APA style in-text citations and references for all resources utilized to answer the questions. Your assignment should include at least two (2) scholarly sources PLUS the textbook
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′′