Question

(Reduction of order to find second particular solution to a differential equation)

(x - 1)y'' - xy' + y = sin, x > 1

Find y_2(x) given that y_1(x) = e^x.

Correct answer: y_2(x) = x

Here is my work: http://f.imgtmp.com/DLOo8.jpg

What am I doing wrong? Or, if what I did is correct so far, what do I do next?

Answer 1

if i recall, reduction order would simply say like: z = y' and sub in the appropriate parts

Answer 2

http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx

this might help me remember some of the forgotten things

Answer 3

ok, assume there is a function u such that uy1 is a soluton

Answer 4

y2 = u e^x

y2' = u' e^x + u e^x

y2'' = u'' e^x + u' e^x + u'(x) e^x + u e^x
= u'' e^x + 2u' e^x +u e^x

and sub a dub dub

Answer 5

(x - 1) (u'' e^x + 2u' e^x +u e^x)

- x (u' e^x + u e^x)

+ u e^x
-------------------------------
= sin


u'' xe^x + 2xu' e^x + u xe^x
-u'' e^x - 2 u' e^x - u e^x
- u' xe^x -u xe^x
+ u e^x
--------------------------
= sin

u'' xe^x + xu' e^x
-u'' e^x - 2u' e^x
--------------------------
(x-1)u''e^x + (x-2)u' e^x = sin

Answer 6

looks fine to me so far

Answer 7

u'' + (x-2)u' e^x = sin(x)
---------- ---------
(x-1)e^x (x-1)e^x


u'' + (x-2) u' = sin(x)
---- ---------
(x-1) (x-1)e^x

Answer 8

u' = w ; u'' = w' to match your notes; and just rewrite u to v to match as well

Answer 9

Thanks a lot :)