Topic 1 DQ 2  When it comes to interacting and gaining


  Topic 1 DQ 2 

When it comes to interacting and gaining compliance with a patient that may be resistant or considered difficult such as Alma, using collaboration is beneficial to better serve this patient. Collaboration is a term used synonymously with compliance, but includes the patient as a member on their own treatment plan. This is important, because compliance has an authoritarian connotation (Falvo, 2011), and for a patient like Alma not feeling included or disengaged from her treatment plan, and expected to comply may contribute to her disposition and negative attitude toward staff. So, instead using collaboration will increase her willingness to adhere, as she will feel the mutual respect and trust that is so very important in caring for patients. Words substituted for compliance are adherence, collaboration, and concordance which have coincided with a change of approach to patient care, being a more patient-centered approach, which focuses on the needs of the patient opposed to the goals of the health professional (Falvo, 2011). This type of collaborative approach is essential for a patient like Alma, because the pan of care will be tailored to her specific needs, making the treatment plan feasible for her, which will increase adherence.

Reference

Falvo, D. (2011). Effective Patient Education: A Guide To Increased Adherence. Retrieved from https://viewer.gcu.edu/RQBKXW

 

Share This Post

Email
WhatsApp
Facebook
Twitter
LinkedIn
Pinterest
Reddit

Order a Similar Paper and get 15% Discount on your First Order

Related Questions

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′′