Question

Find the particular series solution with terms to order 6 of the equation that satisfies the given initial conditions.
(1-x)y'''-2xy'+3y=0; y(0)=1,y'(0)=-1,y''(0)=2

Answer 1

Assume \(y\) is analytic about \(x=0\) so$$y=\sum_{n=0}^\infty a_nx^n$$Differentiating yields$$y'=\sum_{n=0}^\infty(n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty (n+1)(n+2)a_{n+2}x^n\\y'''=\sum_{n=0}^\infty(n+1)(n+2)(n+3)a_{n+3}x^n$$ergo we find from \((1-x)y'''-2xy'+3y=0\)$$(1-x)\sum_{n=0}^\infty(n+1)(n+2)(n+3)a_{n+3}x^n-2x\sum_{n=0}^\infty(n+1)a_{n+1}x^n+3\sum_{n=0}^\infty x^n=0$$

Answer 2

let's focus on our first term for a second:$$(1-x)\sum(\dots)=\sum(\dots) -x\sum(\dots)$$so:$$(1-x)\sum(\dots)=\sum_{n=0}^\infty (n+1)(n+2)(n+3)a_{n+3}x^n-\sum_{n=0}^\infty (n+1)(n+2)(n+3)a_{n+3}x^{n+1}$$

Answer 3

the second series here can be rewritten for convenience:$$\sum_{n=0}^\infty(n+1)(n+2)(n+3)a_{n+3}x^{n+1}=\sum_{n=0}^\infty n(n+1)(n+2)a_{n+2}x^n$$where \(n+1\mapsto n\)

Answer 4

We can similarly transform our second term in the original ODE:$$2x\sum_{n=0}^\infty(n+1)a_{n+1}x^n=\sum_{n=0}^\infty 2(n+1)a_{n+1}x^{n+1}=\sum_{n=0}^\infty2na_nx^n$$under the map \(n+1\mapsto n\)

Answer 5

Since every summation has identical indices and each term is of \(x^n\) we can condense these all by 'collecting' coefficients:$$\dots=\sum_{n=0}^\infty((n+1)(n+2)(n+3)a_{n+2}-n(n+1)(n+2)a_{n+2}-2na_n+3)x^n$$For this summation to be identically \(0\) for all \(x\) requires that the coefficient is \(0\):$$(n+1)(n+2)(n+3)a_{n+2}-n(n+1)(n+2)a_{n+2}-2na_n+3=0$$

Answer 6

oops, that first term should be \(a_{n+3}\). anyways, we get:$$a_{n+3}=\frac{n(n+1)(n+2)a_{n+2}+2na_n-3}{(n+1)(n+2)(n+3)}=\frac{n}{n+3}\left(a_{n+2}+\frac{2na_n-3}{(n+1)(n+2)}\right)$$

Answer 7

we're told \(y(0)=1\implies a_0=1\\y'(0)=-1\implies a_1=-1\\y''(0)=2\implies a_2=1\)

use these to compute \(a_3,a_4,a_5,\dots\) using the recurrence

Answer 8

wow, thanks so much