Question
Asked Jul 19, 2019

Differential Equations

Solve the initial value problem using the power series method:

(x-1)y"-xy'+y=0,    y(0)=-2, y'(0)= 6

check_circleExpert Solution
Step 1

The differential equation is given as:

(x-1)y"
0, y (0)=-2 y'(0) 6
Consider the power series as ycx
Obtain the first and second derivatives as follows.
ykxy=2k(k-1)c,x*-2
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(x-1)y" 0, y (0)=-2 y'(0) 6 Consider the power series as ycx Obtain the first and second derivatives as follows. ykxy=2k(k-1)c,x*-2

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Step 2

Substitute y and its derivatives in the differential equation and obtain the recurrence relation as follows.

co
Τ
Τ
( x-)Σ (k- 1)ς,x*3 -Σkr,x*
-1
Σc,x-0
k-2
-0
k-1
στ
ΣΣk(k-1)c, πς -Σk(t-1)c,xt-3 xΣ k,x* +Σοχ =0
-2
-2
t-l
k-0
Ε
m
Σ (k-1)ς,xα -Σ (k -1)c, x-Σk,x' +Σq -0
t-2
-0
t-0
Σ(k+ 1)c,
Σ(k+2) (k + 1)c, ,-Σ&,χ' + Σς,χ' -0
t-0
-0
k-0
-0
help_outline

Image Transcriptionclose

co Τ Τ ( x-)Σ (k- 1)ς,x*3 -Σkr,x* -1 Σc,x-0 k-2 -0 k-1 στ ΣΣk(k-1)c, πς -Σk(t-1)c,xt-3 xΣ k,x* +Σοχ =0 -2 -2 t-l k-0 Ε m Σ (k-1)ς,xα -Σ (k -1)c, x-Σk,x' +Σq -0 t-2 -0 t-0 Σ(k+ 1)c, Σ(k+2) (k + 1)c, ,-Σ&,χ' + Σς,χ' -0 t-0 -0 k-0 -0

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Step 3

Substitute various values of k and o...

:с, 3у (0)—-2
At k %3D0, -2с, + с, %3D 0%3D с, %3-1
Atk%31, (1)(2)c, —(3)(2)с, -(1)с, +, %3D0%3D с, %3
:с, %3D-1
Atk 2
1
:с, %3-1 and c,
3
(2)(3)с, - (4)(3)с, - (2)с, +с, 3D0—с,
12
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:с, 3у (0)—-2 At k %3D0, -2с, + с, %3D 0%3D с, %3-1 Atk%31, (1)(2)c, —(3)(2)с, -(1)с, +, %3D0%3D с, %3 :с, %3D-1 Atk 2 1 :с, %3-1 and c, 3 (2)(3)с, - (4)(3)с, - (2)с, +с, 3D0—с, 12

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