Unformatted text preview: — CHAPTER 5. —— 15. If and are solutions, then substituting into the ODE results in
. Setting
, we find that
. Similarly, substituting into the ODE results
in
. Therefore
and
may not be analytic. If they were,
Theorem
would guarantee that and were the only two solutions. But note
that an arbitrary value of
cannot be a linear combination of
and
Hence
must be a singular point.
16. Let . Substituting into the ODE, That is, Setting the coefficients equal to zero, we obtain for
is The coefficient . It is easy to see that . Therefore the general solution , which can be arbitrary. 17. Let . Substituting into the ODE, That is, Combining the series, we have ________________________________________________________________________
page 198 ————————————————————————— CHAPTER 5. —— Setting the coefficient equal to zero,
Note that the indices differ by two, so for and for . and Hence the general solution is .
The coefficient , which can be arbitrary. 19.
21 Let . Substituting into the ODE, That is, Combining the series, we have Setting the coefficients equal to zero,
Hence the general solution is The coefficient
2
191. Let and for . , which can be arbitrary.
. Substituting into the ODE, ________________________________________________________________________
page 199 —————————————————————————— CHAPTER 5. —— That is, Combining the series, and the nonhomogeneous terms, we have Setting the coefficients equal to zero, we obtain
, , , and
. The indices differ by two, so for
,
and for Hence the general solution is Collecting the terms containing , Upon inspection, we find that Note that the given ODE is first order linear, with integrating factor
general solution is given by . The ________________________________________________________________________
page 200 —————————————————————————— CHAPTER 5. —— 23. If If 24 , then . If , then . Based on Prob. for , then for . As a result, . As a result, , Normalizing the polynomials, we obtain Similarly, ________________________________________________________________________
page 201 —————————————————————————— CHAPTER 5. —— . . has no roots.
The zeros of by has one root at
are
, . The zeros of
. The roots of ,
and . The roots of
, are at
are given
are given by . 25. Observe that .
But for all nonnegative integers . 27. We have
,
which is a polynomial of degree . Differentiating n times,
, in which the lower index is . Note that if , then . ________________________________________________________________________
page 202 —————————————————————————— CHAPTER 5. ——
Now shift the index, by setting
.
Hence Based on Prob. , 29. Since the
polynomials , ,
of
is , any polynomial, , of degree ,
are linearly independent, and the degree
can be expressed as a linear combin...
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 Spring '08
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 Taylor Series, lim, Complex number

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