# Solve the following intial value probelm, specifying the specific values of A0 and A1.(x-1)y"-xy'+y=0y(0)=-2y'(0)=6

mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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Let `y(x)=a_0 + a_1 x + a_2 x^2 + ...`

Then:

`y(0) = a_0 = -2`

`y'(x) = a_1 + 2a_2 x + 3 a_3 x^2 + ...`

`y'(0) = a_1 = 6`

`y''(x)= 2a_2 + 6 a_3 x + 12 a_4 x^2 + ...`

Putting all of this into the equation `(x-1)y'' - xy' + y = 0` we have:

`(x-1)(2a_2 + 6a_3 x + ... + n(n-1)a_n x^(n-2) + ...) -
x(6 + 2a_2 x + 3 a_3 x^2 + ... + n a_n x^(n-1) ) +
(-2 + 6x + a_2 x^2 + a_3 x^3 + ... + a_n x^n) =0`

We look for the only terms not containing an x:

`-2a_2 - 2 = 0` Thus `a_2 = -1`

Our equation becomes:

`(x-1)(-2 + 6a_3 x + ... + n(n-1)a_n x^(n-2) + ...) -
x(6 -2 x + 3 a_3 x^2 + ... + n a_n x^(n-1) ) +
(-2 + 6x -x^2 + a_3 x^3 + ... + a_n x^n) =0`

Now we look for the `x^1` terms (that is, not `x^2` or `x^3`, etc)

`-6a_3 x -2x - 6x + 6x = 0`

`a_3 = -(1)/(3)`

We may solve for any term given the previous terms:

For example, we look for the coefficient of the `x^n` term:

(x-1)(... (n+1)na_(n+1)x^(n-1) + (n+2)(n+1)a_(n+2)x^n + ...) - x(... na_nx^(n-1) + ...) + (... + a_n x^n + ...) = 0

So the coefficient of the `x^n` term is:

`(n+1)na_(n+1) - (n+2)(n+1)a_(n+2) - na_n + a_n = 0`

`(a_n (1-n)+a_(n+1)(n+1)n)/((n+1)(n+2))=a_(n+2)`

So, if we know the n and n+1 terms, we can get the n+2 term:

`a_0 = -2`

`a_1 = 6`

`a_2 = (-2*1+6*1*0)/(2) = (-2)/(2) = -1`

`a_3 = (0*a_1+a_2*2*1)/(2*3) = -(1)/(3)`

In this way, we may build up to any term we want, and we have:

`y=-2+6x-x^2-(1)/(3)x^3+...`

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