Question

An integrating factor of the differential equation `x/y ("d"y)/("d"x) + log x = ("e"^x x^(-tanx))/y, (x > 0)`, is ______.

Options

• `x^logx`

• `(sqrt(x))^logx`

• `(sqrt("e"))^logx`

• `"e"^(x^2)`

Answer 1

An integrating factor of the differential equation `x/y ("d"y)/("d"x) + log x = ("e"^x x^(-tanx))/y, (x > 0)`, is `(sqrt(x))^logx`.

Explanation:

`x/y ("d"y)/("d"x) + log x = ("e"^x x^(-tanx))/y`

⇒ `("d"y)/("d"x) + logx/x*y = "e"^x x^-tanx`

∴ I.F. = `"e"^(int logx/x  "d"x)`

= `"e"^(1/2(logx)^2`

= `("e"^(1/2 logx))^logx`  ......[∵ (a^m)ⁿ = a^mn]

= `(sqrt(x))^logx`