Chapter 6: Q3E (page 251)
In Problems, 3–6 find two power series solutions of the given differential equation about the ordinary point x 5 0. Compare the series solutions with the solutions of the differential equations obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solutions.
Short Answer
Answer:
The required power series solutions of the given differential equation are given below:
Step by step solution
Step 1:Given Information.
The given differential equation is:
Determining the two power series solution
We can always discover two linearly independent solutions in the form of a power series centered at, that is, y is following, according to theorem 6.2.1 Existence of Power Series Solutions.
The first and second derivatives of this power series solution are calculated first.
and
When y and y" are substituted into the given differential equation, the result is
The summation's indices must now be shifted. We combine the summations when the indices of summation have the same beginning point and the powers ofagree.
We shall havefor all
in some interval since the given linear first order differential equation is
for all
in some interval.
We must haveor
,
as an identity.
A formal power series system is solved
We discover this by considering successive integer values of k starting with k=0.
and so on, with arbitraryand
.
We derive a formal power series solution by utilizing the original solution form and the aforementioned results.
The solution is rewritten in the form;
,
Where,
Use the Maclaurin series
The summationis the expansion of
, and the summation
is the expansion of
, according to the Maclaurin Series. As a result, we can simplify the solution to
Let's try to solve the problem using the method described in Section 4.3.which is the auxiliary equation for the equation
. The roots of this equation are
and
after factoring and solving. As a result, the overall solution is;
which is the same as the general answer obtained using the procedure described in section 4.3 with and
.
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