# `y' + ytanx = secx + cosx , y(0) = 1` Find the particular solution of the differential equation that satisfies the initial condition

Given` y'+ytanx=secx+cosx`

when the first order linear ordinary Differentian equation has the form of

`y'+p(x)y=q(x)`

then the general solution is ,

`y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx)`

so,

`y'+ytanx=secx+cosx--------(1)`

`y'+p(x)y=q(x)---------(2)`

on comparing both we get,

`p(x) = tanx and q(x)=secx +cosx`

so on solving with the above general solution we get:

y(x)=`((int e^(int p(x) dx) *q(x)) dx +c)/e^(int p(x) dx)`

=`((int e^(int (tanx) dx) *(secx+cosx)) dx +c)/e^(int tanx dx)`

first we shall solve

`e^(int (tanx) dx)=e^(ln(secx))= secx `

so proceeding further, we get

`y(x) =((int e^(int (tanx) dx) *(secx+cosx)) dx +c)/e^(int tanx dx)`

=`((int secx *(secx+cosx)) dx +c)/(secx )`

=`((int (sec^2 x+cosxsecx)) dx +c)/(secx )`

= `((int (sec^2 x)dx +int 1 dx) +c)/secx`
=` (tanx+x +c)/secx`
=` sinx +(x+c)/secx`
`y(x) = sinx +(x+c)/secx`

to find the particular solution of the differential equation we have `y(0)=1`

on substituting x=0 we get y=1 and so we can find the value of the c
`y(0)= sin0+(0+c)/sec0 =0+0+c/1 = c`
but `y(0)=1`
=> `1=c`
=> `c=1 `
so `y=sinx+(x+1)/secx`