Differential EquationHard
Question
Let y(x) be the solution of the differential equation (x log x) dy/dx + y = 2x log x, (x ≥ 1). Then y(e) is equal to :
Options
A.2
B.2e
C.e
D.0
Solution
(x log x) 
+ y = 2x log x, (x ≥ 1)
=
= 2
I.F. =
= elog(log x) = log x
= y log x =
y log x = 2x(logx − 1) + C ...(1)
Put x = 1 y.0 = −2 + C
C = 2 ...(2)
Put x = e in (1)
y log e = 2e (log e − 1) + C
y(e) = C = 2 ........... from (2)
+ y = 2x log x, (x ≥ 1) =
= 2I.F. =
= elog(log x) = log x= y log x =
y log x = 2x(logx − 1) + C ...(1)
Put x = 1 y.0 = −2 + C
C = 2 ...(2)
Put x = e in (1)
y log e = 2e (log e − 1) + C
y(e) = C = 2 ........... from (2)
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