`y' + ytanx = secx` Solve the first-order differential equation

Given` y'+y tanx = secx`

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'+y tanx = secx--------(1)`

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

on comparing both we get,

`p(x) = tanx and q(x)=sec x`

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))...

Given` y'+y tanx = secx`

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'+y tanx = secx--------(1)`

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

on comparing both we get,

`p(x) = tanx and q(x)=sec x`

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)) dx +c)/e^(int tanx dx) `

first we shall solve

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

as we know`int tanx dx = ln(secx)`  

So, proceeding further, we get

y(x) =`((int secx *secx) dx +c)/secx `

=`(int sec^2 x  dx +c)/secx`

=`(tanx+c)/secx`

=`tanx/secx + c/secx`

`y(x)=sinx+ c*cosx`