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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 288
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

Referring to the coupled mass-spring system discussed in Example , suppose an external force E(t)=37cos3t is applied to the second object of mass 1 kg. The displacement functions xt and yt now satisfy the system


(a) Show that xt satisfies the equation (18)x(4)(t)+5x''(t)+4x(t)=37cos3t

(b) Find a general solution xt to the equation (18). [Hint: Use undetermined coefficients with xp=Acos3t+Bsin3t.]

(c) Substitute xt back into (16) to obtain a formula for yt.

(d) If both masses are displaced 2m to the right of their equilibrium positions and then released, find the displacement functions xt and yt.

(a) It has been shown that xt satisfies the equation (18)x(4)(t)+5x''(t)+4x(t)=37cos3t.

(b) x(t)=c1cost+c2sint+c3cos2t+c4sin2t+3740cos3t

(c) y(t)=2c1cost+2c2sint-c3cos2t-c4sin2t-3×3720cos3t

(d) The displacement functions x and y are


See the step by step solution

Step by Step Solution

Solving the given equation 

Let's rewrite the given system in operator form:


One wants to obtain the equation for x so we will eliminate y from the system. To do so one will multiply the first equation by D2+2 and the second by 2 and then add them together:



Rewriting this back in standard form we have that xt satisfies


Finding roots

To find a general solution, one will first find a homogeneous solution for xt.

The auxiliary equation is:

r4+5r2+4=0 , and its roots are:

r4+5r2+4=r2+1r2+4=0r2=-1, r2=-4r1,2=±i,r3,4=±2i

Therefore, the homogeneous solution for x is:


Finding derivatives

One will find a particular solution by using the method of undetermined coefficients and assume that a particular solution has a form of xp(t)=Acos3t+Bsin3t.

One needs the second and the fourth derivative of xpt:

x'=-3Asin3t+3Bcos3t, x''=-9Acos3t-9Bsin3tx'''=27Asin3t-27Bcos3t, x(4)=81Acos3t+81Bsin3t

Substituting The values

Now, one has that,

xp(4)(t)+5xp''(t)+4xp(t)=81Acos3t+81Bsin3t+5(-9Acos3t-9Bsin3t)+4(Acos3t+Bsin3t)=40Acos3t+40Bsin3t=37cos3t40A=37, 40B=0A=3740,B=0

So, the particular solution for x is xp(t)=37/40cos3t and the general solution for x is x(t)=xh+xp=c1cost+c2sint+c3cos2t+c4sin2t+3740cos3t.

Finding derivatives 

The first equation of the given system gives us that y(t)=x''(t)+3x(t)

So, one need to find the second derivative of x and substitute it into the previous equation.


Substituting this into y(t)=x''(t)+3x(t)

One has,

y(t)=-c1cost-c2sint-4c3cos2t-4c4sin2t-9×3740cos3t +3c1cost+c2sint+c3cos2t+c4sin2t+3740cos3ty(t)=2c1cost+2c2sint-c3cos2t-c4sin2t-3×3720cos3t

Finding derivatives

Both masses are displaced 2m to the right of their equilibrium position, so x0=y0=2, and since they are released, one has that x'(0)=y'(0)=0.

First, one will find the first derivative of y:


So, from this initial condition,one has that


Substituting the values

Adding the previous two equations together one will get


Substituting this into the second equation one will have c2=2c1-6×3740-2c3=-95

The third and the fourth initial condition give us


Finding c2,c4

Adding those equations together one gets that 3c2=0c2=0

Substituting this into the first equation one has that 2c4=0c4=0

Finally, one has that the displacement functions x and y is:


Most popular questions for Math Textbooks

Sticky Friction. An alternative for the damping friction model F = -by′ discussed in Section 4.1 is the “sticky friction” model. For a mass sliding on a surface as depicted in Figure 5.18, the contact friction is more complicated than simply -by′. We observe, for example, that even if the mass is displaced slightly off the equilibrium location y = 0, it may nonetheless remain stationary due to the fact that the spring force -ky is insufficient to break the static friction’s grip. If the maximum force that the friction can exert is denoted by m, then a feasible model is given by

\({{\bf{F}}_{{\bf{friction}}}}{\bf{ = }}\left\{ \begin{array}{l}{\bf{ky,if}}\left| {{\bf{ky}}} \right|{\bf{ < }}\mu {\bf{andy' = 0}}\\\mu {\bf{sign(y),if}}\left| {{\bf{ky}}} \right| \ge {\bf{0andy' = 0}}\\ - \mu {\bf{sign(y'),ify'}} \ne 0.\end{array} \right.\)

(The function sign (s) is +1 when s 7 0, -1 when s 6 0, and 0 when s = 0.) The motion is governed by the equation (16) \({\bf{m}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - ky + }}{{\bf{F}}_{{\bf{friction}}}}\)Thus, if the mass is at rest, friction balances the spring force if \(\left| {\bf{y}} \right|{\bf{ < }}\frac{\mu }{{\bf{k}}}\)but simply opposes it with intensity\(\mu \)if\(\left| {\bf{y}} \right| \ge \frac{\mu }{{\bf{k}}}\). If the mass is moving, friction opposes the velocity with the same intensity\(\mu \).

  1. Taking m =\(\mu \) = k = 1, convert (16) into the firstorder system y′ = v (17)\({\bf{v' = }}\left\{ \begin{array}{l}{\bf{0,if}}\left| {\bf{y}} \right|{\bf{ < 1andv = 0}}{\bf{.}}\\{\bf{ - y + sign(y),if}}\left| {\bf{y}} \right| \ge {\bf{1andv = 0}}\\{\bf{ - y - sign(v),ifv}} \ne 0\end{array} \right.\) ,
  2. Form the phase plane equation for (17) when v ≠ 0 and solve it to derive the solutions\({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\).where the plus sign prevails for v>0 and the minus sign for v<0.
  3. Identify the trajectories in the phase plane as two families of concentric semicircles. What is the centre of the semicircles in the upper half-plane? The lower half-plane?
  4. What are the critical points for (17)?
  5. Sketch the trajectory in the phase plane of the mass released from rest at y = 7.5. At what value for y does the mass come to rest?

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