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Question
Solve the following:
`("d"y)/("d"x) + y/x = x'e"^x`
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Solution
`("d"y)/("d"x) + "p"y` = Q
Here P = `1/x`
Q = xex
`int "Pd"x = int 1/x "d"x`
= log x
I.F = `"e" int "pd"x`
= elog
= x
The required solution is
y(I.F) = `int "Q" ("I.F") "d"x + "c"`
y(x) = `int x"e"^x (x) "d"x`
xy = `int x^2 "e"^x "d"x`
xy = `(x^2) ("e"^x) - 2x"e"^x + 2"e"^x + "c"`
xy = `"e"^x (x^2 - 2x + 2) + "c"`
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