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MA2051--C97 --- Solutions for TEST II

**1.**
**(a)**
The roots for the characteristic equation are -1 and 3/2, so
two solutions are *e*^{-t} and *e*^{(3/2)
t}. To verify independence,
compute the Wronskian and show that it is not zero.

**(b)**
Use the method of undetermined coefficients to find
*y*_{p}(*t*) = 1/2
*e*^{-3t}.

**(c)**
The general solution is
The initial data give *c*_{1} = -6/5 and
*c*_{2} = +6/5,
so *y*(*t*) = (6/5) *e*^{ (3/2)t } -
(6/5) *e*^{- t} + (1/2)
*e*^{-3t}.

**(d)**
For large values of *t*, the solution approaches the exponential term
(6/5)*e*^{(3/2)t} -- everything else is transient. There is no
limiting periodic solution.
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**2.**
**(a)**
The data give you *m* = 6 and *k* = (6)(9.8)/(0.02) = 2940,
so the model reduces to

**(b)**
The characteristic equation has roots (approximately)
and so the general solution is
. The initial data
determine *c*_{1} = 0 and *c*_{2} = -0.03.

**(c)**
The amplitude is 0.03.
The period is and it takes one half of
a period to travel from highest position to the lowest position.

**(d)**
The velocity is and
the mass passes through equilibrium at . The
velocity is therefore (meters per
second).

**3.**
**(a)**
The characteristic equation reduces to (*r*+4)(*r*-2) = 0 so the
roots are *r*_{1} = -4 and *r*_{2} = +2.
The first solution pair is (*y*_{1},*z*_{1}) = (*e*^{-4t},3*e*^{-4t}).
The second solution pair is (*y*_{2},*z*_{2}) = (*e*^{2t},*e*^{2t}).
(You can test linear independence with the Wronskian.)
The general solution is
(*y*,*z*) =
*c*_{1}(*y*_{1},*z*_{1}) +
*c*_{2}(*y*_{2},*z*_{2}).

**(b)**
The initial data gives and so *y*= 50*e*^{2t}, *z*= 50*e*^{2t}.
(This is a straight line and the solution moves away from the origin
as *t* increases.)

**(c)**
If you choose initial data so that the constants in the general
solution become and *c*_{2} = 0, then you get convergence
to the origin; the solution becomes (*y*,*z*) = *c*_{1}(*e*^{-4t}, 3 *e*^{-4t}).
(You must ``remove'' that nasty *e*^{+2t}.)
Any initial data of the form *y*(0) = (1/3) *z*(0) works.
The solution curve in this case follows a straight line to the origin.

**4.**
The critical damping coefficient is *p* = 4 and the solution in this
case is
*x*(*t*) = *x*_{i} *e*^{-2t} + ( *v*_{i} + 2*x*_{i}) *t e*^{-2t}

Set this equal to zero and solve for exactly one value:
If *x*_{i} > 0, this value of *t* is positive if and only if *v*_{i} <
-2*x*_{i}. (You have to push the mass
*toward* the origin to be sure it gets there.)