1. For the ODE (x - 1)y" - xy + y = 0, x > 1, (a) Verify that yl = ex is a solution, (b) Use the method of reduction-of-order to find a second (linearly independent) solution, (c) Verify linear...

1) (x-1)y’’ – xy’ +y = 0, x>1
a) Verify that y1 = e
x is a solution
Y’1 = e
x
Y’’1 = e
x
Substituting in the ODE:
LHS = (x-1)y’’ – xy’ +y
= (x-1) ex – x ex + ex
= ex((x-1) – x + 1)
= 0
=RHS
 y1 = e
x satisfies the ODE and hence is a solution.
b) Use the method of reduction of order to find a second solution
(x-1)y’’ – xy’ +y = 0, x>1
y1 = e
x is one solution.
Let another solution be y =u y1 = ue
x
 y’ = u’ y1 + u y’1 = u’ e
x + u ex = ex (u’ + u)
 y’’ = u’’ y1 + 2u’ y’1 + u y’’1 = u’’ e
x + 2u’ ex + u e
x = ex(u’’+2u’+ u)
Substituting we get:
(x-1)ex (u’’+2u’+ u) - x ex (u’ + u) + uex = 0
 ex((x-1)(u’’+2u’+ u) - x (u’ + u) + u) = 0
 (x-1)(u’’+2u’+ u) - x (u’ + u) + u = 0 since ex ≠ 0
 (x-1)u’’+2(x-1)u’+(x-1) u - x u’ - xu + u = 0
 (x-1)u’’+(2(x-1) – x) u’+(x-1) u – u(x – 1) = 0
 (x-1)u’’+(x-2)u’ = 0
 Let v = u’ => v’ = u’’
 (x-1)v’+(x-2)v = 0
 (x-1)v’= -(x-2)v










On integrating we get
 lnv = -(x-ln(x-1))
 lnv = -x + ln(x-1)
 v = (x-1)e-x
 v = u’ = (x-1)e-x



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