⇒

⇒

We can see that it is a linear differential equation.

Comparing it with

P = – , Q = 1

I.F = e^∫Pdx

= e^dx

= e^dx

= e^{log|x + 1| – log|x|}

= e^log

=

Solution of the given equation is given by

y × I.F = ∫Q × I.F dx + c

⇒ y × = ∫ 1 × dx + c

⇒ y × = ∫ dx + c

⇒ y × = x + logx + c ……(1)

As the equation passing through (1,0),

0 = 1 + log1 + c

⇒ c = – 1

Putting the value of c in equation (1)

∴ y × = x + logx – 1