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Chapter 5: Q23E (page 271)

In Problems 19鈥�24, convert the given second-order equation into a first-order system by setting v=y鈥�. Then find all the critical points in the yv-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12).

y''(t)+y(t)-y(t)4=0

Short Answer

Expert verified

The critical points are (0,0),(1,0).

Step by step solution

01

Find the critical point

Here the equation is y''t+yt-yt4=0.

Put v=y'鈥�鈥�and鈥�鈥�v'=y''.

Then the system is;

y'=vy''=-y+y4v'=-y+y4

For critical points equate the system equal to zero.

v=0-y+y4=0y-1+y3=0y=0鈥�or鈥�-1+y3=0

If y鈮�0鈥�鈥�then

-1+y3=0y=1

So, the critical point is (0, 0) and (1, 0).

02

Sketch the directional field. 

Therefore, the critical points are (0, 0), (1, 0).

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