Answer :

Given: 

Separating the variables, we get,

ln |ex - 2| = ln |tan y| + ln C

ln |ex - 2| = ln (C tan y )

ex - 2 = C tan y

Given x = 0, y =

= C tan

1 – 2 = C × 1

C = – 1

ex - 2 = – tan y {since C = – 1}

ex - 2 + tan y = 0.


OR



Let P = and Q =


So, this is a linear equation in one-variable of the form,  

We know that the solution of this equation is given by,

y(I.F) =


I.F =
= = =


y


y


y


Given y = 0 when x =





y { since C = – 2}


y

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