The differential equations, find a particular solution satisfying the given condition: `(dy)/(dx)=ytanx ; y=1`when x = 0

Explanation

Given, `(dy)/(dx)=ytanx`
`implies (1)/(y)dy=tanxdx`
`implies int(1)/(y)dy=inttanxdx`
`implies log|y|=log|(secx)|+log|C|`
`implies logy=log|(Csecx)|`
`implies y=C sec x`
Now, put `y=1` and `x=0`,
`1=C sec0 implies 1=Cxx1impliesC=1`
put `C=1`,
`y=secx`
which is the required particular solution