x log x

dy

dx

+ y =

2

x

log x
Dividing all the terms of the equation by xlogx, we get
⇒

dy

dx

+

y

xlogx

=

2

x2

This equation is in the form of a linear differential equation

dy

dx

+ Py = Q , where { =

1

xlogx

and Q =

2

x2

Now, I.F. = e∫Pdx = e∫

1

xlogx

dx = elog(logx) = log x
The general solution of the given differential equation is given by
y × I.F. = ∫ (Q × I.F.) dx + C
⇒ y log x = ∫ (

2

x2

logx) dx
⇒ y log x = 2 ∫ (logx×

1

x2

) dx
= 2 [log x × ∫

1

x2

dx - ∫ {

d

dx

(log‌x)×∫

1

x2

dx} dx]
= 2
= 2 [−

logx

x

+∫

1

x2

dx]
= 2 [−

logx

x

−

1

x

] + C
So the required general solution is y log x = −

2

x

(1 + log x) + C
OR

dy

dx

= y tan x
⇒

dy

y

= tan x dx
On integration, we get
∫

dy

y

= ∫ tan x dx
⇒ log y = log (sec x) + log C ... (1)
⇒ log y = log (C sec x)
⇒ y = C sec x
Now,it is given that y = 1 when x = 0
⇒ 1 = C × sec 0
⇒ 1 = C × 1
∴ C = 1
Substituting C = 1 in equation (1), we get
y = sec x as the required particular solution.