I NEED RESPONSE TO THIS ASSIGNMENT2 REFERENCESpresentation on an adolescent who is anxious, depressed and withdrawn and also has problems with coping with online learning. I agree with your approach and I would like to add that cognitive therapy can be added. Treatment approaches for adolescent depression include cognitive behavioral therapy, interpersonal therapy, and family system approaches (Sander & McCarty, 2005). There are two specific types of family systems approaches that have specifically been evaluated with depressed adolescents. One is systemic-behavioral family therapy (SBFT), and the other is the attachment-based family therapy (Sander & McCarty, 2005). SBFT includes skill-building in communication, incorporating reinforcement, and distinct aspects of cognitive restructuring (Sander & McCarty, 2005). Attachment-based family therapy incorporates family systems and attachment theory in a manualized treatment for depression in youth, based on the premise that a “negative family environment” inhibits children from developing the internal and interpersonal coping skills needed to buffer against the family, social, and community stressors that can cause or exacerbate depression (Sander & McCarty, 2005).ReferencesSander, J. B., & McCarty, C. A. (2005). Youth depression in the family context: familialrisk factors and models of treatment. Clinical child and family psychology review, 8(3), 203–219.https://doi.org/10.1007/s10567-005-6666-3Thapar, A., Collishaw, S., Pine, D. S., & Thapar, A. K. (2012). Depression in adolescence.Lancet (London, England), 379(9820), 1056–1067.https://doi.org/10.1016/S0140-6736(11)60871-4
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′′