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Chapter 5: Q19E (page 271)

In Problems 19–24, convert the given second-order equation into a first-order system by setting v=y’. 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).

d2xdt2-y=0

Short Answer

Expert verified

The point is an unstable saddle point (0, 0).

Step by step solution

01

Find the critical point

Here the equation is d2xdt2-y=0.

Put v=y'  and  v'=y''.

Then the system is:

y'=vy''=yv'=y

For critical points equate the system equal to zero.

v=0y=0

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

The phase plane equation is:

dvdy=yv∫vdv=∫ydyv2-y2=c

02

Sketch

This is the required result.

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Most popular questions from this chapter

In Problems 19–24, convert the given second-order equation into a first-order system by setting v=y’. 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

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

y4(t)-y3(t)+7y(t)=cost;y(0)=y'(0)=1,y''(0)=0,y3(0)=2

In Problems 3–6, find the critical point set for the given system.

dxdt=y-1dydt=x+y+5

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direction field in the phase plane and from this description the stability of the critical points (i.e., compare with Figure 5.12).

dxdt=2x+13y,dydt=-x-2y

Falling Object. The motion of an object moving vertically through the air is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - g - }}\frac{{\bf{g}}}{{{{\bf{V}}^{\bf{2}}}}}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}\left| {\frac{{{\bf{dy}}}}{{{\bf{dt}}}}} \right|\)where y is the upward vertical displacement and V is a constant called the terminal speed. Take \({\bf{g = 32ft/se}}{{\bf{c}}^{\bf{2}}}\)and V = 50 ft/sec. Sketch trajectories in the yv-phase plane for \( - 100 \le {\bf{y}} \le 100, - 100 \le {\bf{v}} \le 100\)starting from y = 0 and y = -75, -50, -25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physically; why is V called the terminal speed?
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