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Unformatted text preview: age 215 —————————————————————————— CHAPTER 5. —— Hence the solution of the initial value problem is The solution is bounded, as , if . 2
results
380. Substitution of are given by in the quadratic equation
Formally, the roots The roots
will be complex, if
as
, we need
The roots will be equal, if
zero as long as
.
The roots will be real and distinct, if . . For solutions to approach zero,
In this case, all solutions approach
. It follows that
. For solutions to approach zero, we need
.
Hence all solutions approach zero, as
23 . Given that , That is, , as long as
. By the chain rule,
. Similarly, . Direct substitution results in
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page 216 —————————————————————————— CHAPTER 5. —— ,
that is,
.
The associated characteristic equation is
it follows that
. If the roots , are real and distinct, then . If the roots . Since are real and equal, then . If the roots are complex conjugates, then 24. Based on Prob. , the change of variable , and transforms the ODE into
. The associated characteristic equation is
Hence
, and
26. The change of variable , with roots ,. transforms the ODE into
. The associated characteristic equation is
, with roots
,
.
Hence
Since the right hand side is not a solution of the
homogeneous equation, we can use the method of undetermined coefficients to show
that a particular solution is
. Therefore the general solution is given by
, that is,
27. The change of variable transforms the given ODE into ________________________________________________________________________
page 217 —————————————————————————— CHAPTER 5. —— .
The associated characteristic equation is
, with roots
,.
Hence
Using the method of undetermined coefficients, let
. It follows that the general solution is given by
, that is, 28. The change of variable transforms the given ODE into
. The solution of the homogeneous equation is
The right
hand side is not a solution of the homogeneous equation. We can use the method of
undetermined coefficients to show that a particular solution is
. Hence
the general solution is given by
, that is, 29. After dividing the equation by , the change of variable
into transforms the ODE .
The associated characteristic equation is
. Hence the general solution is , with complex roots ,
and therefore
. 30. Let
and Since . Setting , successive differentiation gives
. It follows that , we find that ________________________________________________________________________
page 218 —————————————————————————— CHAPTER 5. —— Given that and
are roots of
Therefore
and
differential equation,
, for , we have
.
are linearly independent solutions of the
, as long as
. ________________________________________________________________________
page 219 —————————————————————————— CHAPTER 5. ——
Section 5.6
1.
when
. Since the three coeff...
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This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue.
 Spring '08
 Staff

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