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2

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Step 1

From the given information it is needed to calculate the initial value problem using power series for the problem:

(1+ x2)y" 2xy'-2y=0, y(0) = 0, y'(0)=1
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(1+ x2)y" 2xy'-2y=0, y(0) = 0, y'(0)=1

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Step 2

The power series solution and its derivative to above initial value problem is:

y(1) -Σα
ο
y(x) Σa,r"
y'(x) = Ση(n-1) a,r"1
-n-2
n-2
8
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y(1) -Σα ο y(x) Σa,r" y'(x) = Ση(n-1) a,r"1 -n-2 n-2 8

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Step 3

Now, put these values into the given dif...

σο
3α
(1+x)Ση(n - 1) a,r" + 2xΣ mr-2Σ α" -0
n-2
n1
Σ/n-1)a,
Zn(п-1)а,x +
Σ/n-1) aχ" +Σ Σu" -Σ 2αμ.
2ах" 3D0
n=2
n-2
n1
n0
ο
σ
ο
Σ (n+2) (n+ 1) a,,x" +Ση (n -1) a, ά
σο
2па,х"
2а,х" %3D0
n-0
n-2
n-1
n-0
oο
Σ(n+2)(n+ 1) α,, + n(n-1) α, + 2na,-2α,-0
n-0
help_outline

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σο 3α (1+x)Ση(n - 1) a,r" + 2xΣ mr-2Σ α" -0 n-2 n1 Σ/n-1)a, Zn(п-1)а,x + Σ/n-1) aχ" +Σ Σu" -Σ 2αμ. 2ах" 3D0 n=2 n-2 n1 n0 ο σ ο Σ (n+2) (n+ 1) a,,x" +Ση (n -1) a, ά σο 2па,х" 2а,х" %3D0 n-0 n-2 n-1 n-0 oο Σ(n+2)(n+ 1) α,, + n(n-1) α, + 2na,-2α,-0 n-0

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