Use the power series method to solve the given initial-value problem. (Format your final answer as an elementary function.) (x − 1)y'' − xy' + y = 0, y(0) = −7, y'(0) = 3

You're looking for a solution of the form

Differentiating twice yields

Substitute these series into the DE:

Two of these series start with a linear term, while the other two start with a constant. Remove the constant terms of the latter two series, then condense the remaining series into one:

which indicates that the coefficients in the series solution are governed by the recurrence,

Use the recurrence to get the first few coefficients:

You might recognize that each coefficient in the n-th position of the list (starting at n = 0) involving a factor of -7 has a denominator resembling a factorial. Indeed,

-7 = -7/0!

-7/2 = -7/2!

-7/6 = -7/3!

and so on, with only the coefficient in the n = 1 position being the odd one out. So we have

which looks a lot like the power series expansion for -7eˣ.

Fortunately, we can rewrite the linear term as

3x = 10x - 7x = 10x - 7/1! x

and in doing so, we can condense this solution to

Just to confirm this solution is valid: we have

y = 10x - 7eˣ   ==>   y (0) = 0 - 7 = -7

y' = 10 - 7eˣ   ==>   y' (0) = 10 - 7 = 3

y'' = -7eˣ

and substituting into the DE gives

-7eˣ (x - 1) - x (10 - 7eˣ ) + (10x - 7eˣ ) = 0

as required.

6 (3) - 4 + 32

hope this

4 packs of water and 7 packs of candy bars

step-by-step explanation:

35x4=140

140/20=7