Q. 16

For each of the following differential equations, find a particular solution satisfying the given condition :

it being given that y = 1 when x = 0.

Answer :


Rearranging the terms we get:




log|y| = log|secx| + logc


log|y| - log|secx| = logc


log|y| + log|cosx| = logc


ycosx = c


y = 1 when x = 0


1×cos0 = c


c = 1


ycosx = 1


y = 1/cosx


y = secx


Ans: y = secx


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