THE CLIENT IVE BEEN USING IS TIFFANI BRADLEY.By Day 7Upload your Kaltura video of you presenting your PowerPoint presentation that addresses the following:Identify the theoretical orientation you have selected to use.Describe how you would assess the problem orientation of the client (Tiffani Bradley)in your selected case study (i.e., how the client perceives the problem). Remember to keep the theoretical orientation in mind in this assessment stage.Discuss the problem definition and formulation based on the theoretical orientation you have selected.Identify and describe two solutions from all the solutions possible. Remember, some of these solutions should stem from the theoretical orientation you are utilizing.Describe how you would implement the solution. Remember to keep the theoretical orientation in mind.Describe the extent to which the client is able to mobilize the solutions for change.Discuss how you would evaluate whether the outcome is achieved or not. Remember to keep the theoretical orientation in mind.Evaluate how well the problem-solving model can be used for short-term treatment of this client.Evaluate one merit and one limitation of using the problem-solving model for this case.Your 11- to 12-slide PowerPoint presentation should follow these guidelines:Each slide should be written using bullet points, meaning no long paragraphs of written text should be in the slides.Include a brief narration for each slide (i.e., the narration takes the place of any written paragraphs, while the bullet points provide context and cues for the audience to follow along).Record both audio and video for presentation.Be sure to:Identify and correctly reference the case study you have chosen.Use literature to support your claims.Use APA formatting and style.Include the reference list on the last slide.Speak clearly.Dress professionally.
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′′