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# differential equation P6 watch

1. given that

d^2y/dx^2 + y(dy/dx) = x

y = 0, dy/dx = 2 at x = 1

how do I find a series solution of the differential equation in ascending powers of (x-1) up to and including the term in (x-1)^3

??
2. (Original post by lesser weevil)
given that

d^2y/dx^2 + y(dy/dx) = x

y = 0, dy/dx = 2 at x = 1

how do I find a series solution of the differential equation in ascending powers of (x-1) up to and including the term in (x-1)^3

??
stick in dy/dx, x=1 and y=0 into the second order one and find y''

use y, y' and y'' to find (y''', after differentiating the expression)
Basically use teh info you have - sub back into the eqtns, and plug into teh below standard eqtn

now the general solution is y= 0+ (x-1)y' +(1/2!)y''(x-1)^2 +(1/3!)y'''(x-1)^3

see pg12/13 in P6 edexcel book if you have it

phil
3. and that standard equation is in the formula booklet as well.

this topic should be easy marks. hopefully.

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