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प्रश्न
Solve the following differential equation:
x cos y dy = ex(x log x + 1) dx
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उत्तर
The equation can be written as
cos y dy = `"e"^x ((xlogx + 1))/x "d"x`
cos y dy = `"e"^x [(xlogx)/x + 1/x] "d"x`
cos y dy = `"e"^x [logx + 1/x] "d"x`
Taking integration on both sides, we get
`int cos y "d"y = int "e"^x [log x + 1/x] "d"x` ........(1)
R.H.S
`int "e"^x [log x + 1/x] "d"x`
⇒ Take f(x) = log x
f'(x) = `1/x`
This of the form `int "e"^x ["f"(x) + "f'"(x)] "d"x = "e"^x "f"(x) + "C"`
∴ `int "e"^x [log x + 1/x] "d"x = "e"^x lo x + "C"`
Substituting in (1), we get
sin y = ey log x + C
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