Question

For the differential equation, find the general solution:

`x log x dy/dx + y=    2/x log x`

Answer 1

The given equation

`x log x dy/dx + y = 2/x log x`

or `dy/dx + y/(x log x) = 1/x^2`            ....(i)

Comparing with `dy/dx + Py = Q`,

`P = 1/(x log x)` and `Q = 2/x^2`

∴ `I.F. = e^(intP dx) = e^(int 1/(x log x)dx)`

`= e^(log(log x)) = log x`

Hence the required solution

∴ `y × I.F. = int I.F. xx Q dx + C`

`=> y log x = 2 int 1/x^2 (log x) dx + C`

`=> y log x = 2 [log x (- 1/x) - int 1/x ((- 1)/x) dx] + C`

`=> y log x = (- 2)/x log x + 2 int 1/x^2 dx + C`

`=> y log x = (- 2)/x log x - 2/x + C`

⇒ `y = log x = (- 2)/x (1 + log |x|) + C`