For the equation 3x(x+1)y\'\' +2y\'+3y = 0 find the first four nonzero terms in
ID: 2985681 • Letter: F
Question
For the equation 3x(x+1)y'' +2y'+3y = 0 find the first four nonzero terms in BOTH Frobenius series solutions y1(x) and y2(x).
Please upload a picture of your written work if you can and show it step by step. Please don't skip any steps! And also explain any important things along the way (if something is arbitrary, for example)
BTW, I use a shifting method to change/shift the indices. I do something like m = n+2, n= m-2. If you can do it this way that would help me understand better.
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Explanation / Answer
Find the rst four nonzero terms in a Frobenius series solution of x2y00+xy0+(x2+1)y = 0.
Then use the reduction of order technique (as in Example 4) to nd the logarithmic term and the
rst three nonzero terms in a second linearly independent solution.
We rst note that the roots of the indical equation, which is (r) = r(r%uDBC0%uDC001)%uDBC0%uDC00r+1,
are r1 = 1 and r2 = 1. From here, choose:
y1 =
1X
n=0
cnxn+1
y01
=
1X
n=0
(n + 1)cnxn
y00
1 =
1X
n=0
(n + 1)ncnxn%uDBC0%uDC001
As always, we substitute these into the original dierential equation, and obtain the
coecient relation cn = %uDBC0%uDC00cn%uDBC0%uDC002
n2 . From here, we compute:
c0 = c0
c1 = 0
c2 = %uDBC0%uDC00c0
22
c3 = 0
c4 = c0
2242
c5 = 0
c6 = %uDBC0%uDC00c0
224262
Letting c0 = 1, we have that the rst four nonzero terms are 1; %uDBC0%uDC001
22 = %uDBC0%uDC001
4 ; 1
2242 =
1
64 ; %uDBC0%uDC001
224262 = %uDBC0%uDC001
2304 , and
y1(x) = x
1 %uDBC0%uDC00
x2
4
+ x4
64 %uDBC0%uDC00
x6
2304
Now, we use the reduction of order technique to nd the rst terms of a second
linearly independent solution. We rst notice that P(x) = %uDBC0%uDC001
x , and e%uDBC0%uDC00
R
P(x)dx = x.
Then, by formula (23):
1
y2 = y1
Z
x
x2
1 %uDBC0%uDC00 x2
4 + x4
64 %uDBC0%uDC00 x6
2304 + : : :
2
y2 = y1
Z
1
x
1 %uDBC0%uDC00 x2
2 + 3x4
32 %uDBC0%uDC00 5x6
576 + : : :
y2 = y1
Z
1
x
+ x
2
+
5x3
32
+
23x5
576
+ : : : =)
y2(x) = y1(x)
ln x + x2
4
+
5x4
128
+
23x6
3456
+ : : :
8.4 # 13 Find the rst four nonzero terms in a Frobenius series solution of x2y00+(2x2%uDBC0%uDC003x)y0+3y =
0. Then use the reduction of order technique (as in Example 4) to nd the logarithmic term and
the rst three nonzero terms in a second linearly independent solution.
We proceed exactly as in the previous problem. The indical equation is (r) =
r(r%uDBC0%uDC001)%uDBC0%uDC003r+3, and the roots are r1 = 1 and r2 = 3. We use as our ansatz the function
y =
P1
n=0 cnxn+r. From here, we see that the coecients are:
cn = %uDBC0%uDC002r(n + r %uDBC0%uDC00 1)cn%uDBC0%uDC001
(n + r %uDBC0%uDC00 3)(n + r %uDBC0%uDC00 1)
.
We now look at the linearly independent solution corresponding to r2 = 3. We
obtain the recurrence cn = %uDBC0%uDC002cn%uDBC0%uDC001
n . Computing the rst coecients, we see that:
c0 = c0
c1 = %uDBC0%uDC002c0
c2 = 2c0
c3 = %uDBC0%uDC004
3 c0
Letting c0 = 1, we see that y1(x) = x3
%uDBC0%uDC00
1 %uDBC0%uDC00 2x + 2x2 %uDBC0%uDC00 4
3x3 + : : :
.
Now, we obtain the second linearly independent solution using reduction. P(x) =
2 %uDBC0%uDC00 3
x =) e
R
P9xdx = x3e%uDBC0%uDC002x. From here, using the Taylor series for e%uDBC0%uDC002x, we see:
y2 = y1
Z
x3e%uDBC0%uDC002x
x6
%uDBC0%uDC00
1 %uDBC0%uDC00 2x + 2x2 %uDBC0%uDC00 4
3x3 + : : :
2 + : : :
y2 = y1
Z
1
x3 +
2
x3 +
2
x
+
4
3
+
2x
3
+ : : :
y2(x) = y1(x)
2ln x %uDBC0%uDC00
1
2x2
%uDBC0%uDC00
2
x
%uDBC0%uDC00
4x
3 %uDBC0%uDC00
x2
3
+ : : :
8.4 # 16 Find two linearly independent Frobenius series solutions of Bessel's equation of order
3
2 ; x2y00 %uDBC0%uDC00 x(1 + x)y0 + y = 0.
2
Again, we rst note that the roots of the indical equation, (r) = r(r %uDBC0%uDC00 1) + r %uDBC0%uDC00 9
4
are r1 = 3
2 and r2 = %uDBC0%uDC003
2 . As always we let y =
P1
n=0 cnxn+r. We substitute y, y0, and
y00 in the original dierential equation, and obtain the following formula for coecients:
cn = %uDBC0%uDC00cn%uDBC0%uDC002
(n + r)2 %uDBC0%uDC00 9
4
Now, looking at the root r1 = 3
2 , we see that the recursive coecient formula be-
comes:
cn = %uDBC0%uDC00cn%uDBC0%uDC002 %uDBC0%uDC00
n + 3
2
2
%uDBC0%uDC00 9
4
.
Computing the coecients gives us:
c0 = c0
c1 = 0
c2 = %uDBC0%uDC00c0
10
c3 = 0
c4 = c0
280
c5 = 0
c6 = %uDBC0%uDC00c0
15120
c7 = 0
c8 = c0
1330560
From here, we deduce that the general formula is:
c2n =
(%uDBC0%uDC001)n
2nn!(5)(7) : : : (2n + 3)
Letting c0 = 1, we obtain:
y1(x) = x
3
2
1 +
1X
n=1
(%uDBC0%uDC001)n
2nn!(5)(7) : : : (2n + 3)x2n
!
Looking at the root r2 = %uDBC0%uDC0032, the equation becomes cn = %uDBC0%uDC00cn%uDBC0%uDC002
(n%uDBC0%uDC003
2 )2
%uDBC0%uDC009
4
. We obtain the
following coecents:
3
c0 = c0
c1 = 0
c2 = c0
2
c3 = 0
c4 = %uDBC0%uDC00c0
8
c5 = 0
c6 = c0
144c7 = 0
c8 = %uDBC0%uDC00c0
5760
From here, we infer that the general formula for the coecients is:
c2n =
(%uDBC0%uDC001)n
2nn!(%uDBC0%uDC001)(1)(3)(5) : : : (2n %uDBC0%uDC00 3); c2n+1 = 0
Letting c0 = 1, we obtain:
y2(x) = x
%uDBC0%uDC003
2
1 +
1X
n=1
(%uDBC0%uDC001)n
2nn!(1)(3) : : : (2n + 3)x2n
!