272 Differential equations
272 Differential equations Review 1
1. A body of mass m is projected vertically upward with an initial velocity [pic] in a medium offering a resistance[pic], where k is a constant. Assume that the gravitational attraction of the earth is constant.
(a) Find the velocity [pic]of the body at any time.
(b) Use the result of part (a) to calculate the limit of [pic]as[pic], that is, as the resistance approaches zero. Does this result agree with the velocity of a mass m projected upward with an initial velocity [pic] in a vacuum?
(c) Use the result of part (a) to calculate the limit of[pic]as[pic], that is, as the mass approaches zero.
Solution; (a) the force applied to the body at any moment t is the sum of the gravitational force[pic], where [pic]( the sign minus indicates that this force is directed down), and the force of resistance[pic]. Pay attention that when the velocity v is positive, the body moves up and the force of resistance is negative and directed down; on the other hand, when v is negative the body moves down, and the force of resistance is positive and directed up.
By the second law of Newton the acceleration [pic]at moment t is equal to
[pic].
Therefore we have the following linear differential equation of the first order for the velocity v.
[pic].
To solve this equation we will follow the method on page 36. In notations on this page we have[pic]. Formula (30) on page 36 yields
[pic].
By formula (33) on the same page
[pic]
The constant C can be found from the initial condition [pic] whence [pic]and
[pic]
(b) We have to find the limit of the above expression when[pic].
[pic].
Clearly, [pic].
By L’Hospital’s rule[pic], whence
[pic].
The last expression agrees completely with equations of motion in vacuum.
(c) [pic].[pic]
2. A spring–mass system has a spring constant of[pic]. A mass of 2 kg is attached to the spring and the motion takes place in a viscous fluid that offers a resistance numerically equal to the magnitude of the instantaneous velocity. If the system is driven by an external force of[pic], solve the equation of motion and determine the steady-state response in the form[pic].
Solution; the equation of motion is (compare with formula (1) on page 207)
[pic]
where m is the mass, γ is the damping coefficient, k is the spring constant, and F(t) is the external force. In our case[pic]. The equation of motion becomes
[pic].
First we solve the homogeneous equation
[pic].
The characteristic equation [pic]has solutions [pic]and therefore the general solution of the homogeneous equation is
[pic].
Next we look for a particular solution of the non-homogeneous equation [pic] in the form[pic]. Then[pic],[pic], and [pic]. Therefore
[pic]
Solving this system we get [pic]and
[pic].
The general solution is [pic]and because [pic]is going to 0 exponentially when [pic]the steady-state response equals to[pic].
3. Solve the equation[pic].
Solution; first we will solve the homogeneous equation
[pic].
The corresponding characteristic equation [pic]has solution -2 of multiplicity 2 and therefore the general solution of the homogeneous equation is[pic].
Now we can apply Theorem 3.7.1 on page 189. We can take as two linearly independent solutions of the homogeneous equation the functions[pic]. Then the Wronskian [pic] is computed as (see page 146)
[pic]
Applying formula (28) on page 189 we see that a particular solution of our non-homogeneous equation can be found as
[pic]
The general solution can be written as
[pic].
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