# Find the pa

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}

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