Question

Find the first four nonzero terms in a power series expansion about x = 0 for the solution to the given initial value problem. y'' +(x-3)y'-y=0; y(O) = -5, y'(0) = 0 y(x) = (Type an expression that includes all terms up to order 4.)

Answer 1

SOLUTION:-

Given IVP y''+(x-3)y'-y =0 ........(1) with conditions y(0) = -5, y'(0)=0

Let

y Σαηα"....... (2) n=0

be power series expansion of the solution y of (1) about x = 0.

From (2)

ηαηα2-1, 9" = Ση(η – 1)α"-2- m=1 n=2

Substituting y, y' and y'' in (1) we get,

X ΟΧΟ Ση(η – 1)αα-2 + (a – 3)Σ παη"-1 Σα αηλ = 0 n=2 n=1 n=0

i.e,
X ΧΟ Ση(η – 1)αα2-2 + 2 +Σπα,α" - 3Σ ηαηα2-1 IM8 and" = 0 n=2 n=1 n=1 n=0

X ΟΧΟ ΟΧΟ i.e, Σ(n + 2) (n + 1)αη+2α2 +Σπαα” – 3Σ(n + 1)an+1α” –Σ and" = 0 n=0 n=1 n=0 n=0
Equating coeffiecients of $\boldsymbol{x^{n}}$

(n + 2)(n + 1)an+2 + nan - 3(n + 1)an+1 - an = = 0

(n + 2)(n+1)an+2 – 3 n +1)an+1 + (n - 1)an = 0..........(3)

Now applying initial condition y(0)=-5 in (2) we get
—5 — 0р

Using initial condition y'(0)=0 in the series of y' we get
Ο = αι

Now putting n=0,1,2, ... in succession in (3)

$\boldsymbol{n=0\Rightarrow (2)(1)a_{2}-3(1)a_{1}+(-1)a_{0}=0}$

$\boldsymbol{\therefore 2a_{2}=3a_{1}+a_{0}=3(0)+-5=-5}$   $\boldsymbol{\therefore a_{2}=\frac{-5}{2}}$

n=1 (3)(2)az 3(2)a2 + (0)a1 = 0

$\boldsymbol{\therefore 6a_{3}=6a_{2}-(0)a_{1}=6(-5/2)-0=-15}$   $\boldsymbol{\therefore a_{3}=-15/6=-5/2}$

$\boldsymbol{n=2\Rightarrow (4)(3)a_{4}-3(3)a_{3}+(1)a_{2}=0}$

.. 12a4 = 9uვ – (1)22 (1)42 = 9(-5/2) – (-5/2) = – 20

1.04 = -20/12 = -5/3

$\boldsymbol{n=3\Rightarrow (5)(4)a_{5}-3(4)a_{4}+(2)a_{3}=0}$

$\boldsymbol{\therefore 20a_{5}=12a_{4}-2a_{3}=12(-5/3)-2(-5/2)=-15}$

. Q5 = -15/20 =-3/4

and so on

Thus the Fourier series solution of (1) upto the term containing
T
is,

5 3 5 3 4 y(x) = -5 5 - - 2 2 2 3 4