View the step-by-step solution to:

Question

NEED HELP!!!

Screen Shot 2019-10-14 at 1.46.20 PM.png

Screen Shot 2019-10-14 at 1.46.20 PM.png

(a) What can you say about a solution of the equation y' = —(1/2)y2 just by looking at the differential equation? A The function y must be strictly decreasing on any interval on which it is defined. A The function y must be strictly increasing on any interval on which it is defined.
0 The function y must be decreasing (or equal to 0) on any interval on which it is defined.
A The function y must be increasing (or equal to 0) on any interval on which it is defined. A The function y must be equal to O on any interval on which it is defined.
J (b) Verify that all members of the family y = 2/(x + C) are solutions of the equation in part (a). 2 . 2 J
y= = y=-—-
x+C (x+C)2
. 2 J 1 2 q 2 1 2
LHS=y=——=-—( )=——y =RHS
(x+C)2 2 x+C 2 (c) Can you think of a solution of the differential equation y' = —(1/2)y2 that is not a member of the family in part (b)? A y = e2X is a solution of y' = _(1/2)y2 that is not a member of the family in part (b).
A Every solution of y' = _(1/2)y2 is a member of the family in part (b).
A y = 2 is a solution of y' = _(1/2)y2 that is not a member of the family in part (b).
= 0 is a solution of r: _ 1 2 that is not a member of the family in part (b).
y y ( / )
A y = x is a solution of y' = _(1/2)y2 that is not a member of the family in part (b).
a! (d) Find a solution of the initial-value problem. Y' = —(1/2)y2 Y(0) = 0-2 x
y— 2+5 3:

Top Answer

Hello the answer is:... View the full answer

Sign up to view the full answer

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question