For each of the following differential equations, find a particular solution satisfying the given condition :
it being given that y = 1 when x = 0.
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