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Question
Solve the following Linear differential equation:
`("d"y)/("d"x) + y/(xlogx) = (sin2x)/logx`
Solution
The given differential equation may be written as
`("d"y)/("d"x) + y/(xlogx) = (sin2x)/logx`
This is of the form `("d"y)/("d"x) + "P"y` = Q
Where P = `1/(xlogx)`
Q = `(sin2x)/logx`
Thus, the differential equation is linear.
I.F = `"e"^(int "Pd"x)`
= `"e"^(int 1/(xlogx) "d"x)`
= `"e"^(int 1/"t" "dt")`
= `"e"^(log "t")`
= log x
So, the solution of the given differential equation is
y × I.F = `int ("Q" xx "I.F") "d"x + "c"`
`y xx log x = int (sin2x)/logx xx log x "d"x + "c"`
= `int sin 2x "d"x + "c"`
`y log x = (- cos 2x)/2 + "c"`
`y log x + (cos2x)/2` = c is a required solution.
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