`y' + ysecx = secx , y(0) = 4` Find the particular solution of the differential equation that satisfies the initial condition

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Given` y'+y*secx=secx`

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


then the general solution is ,

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




on comparing both we get,

`p(x) = secx and q(x)=secx`

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

first we shall solve

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


proceeding further, we get

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

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

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

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

=`(tanx+secx +c)/(secx+tanx)`

so `y(x)=(tanx+secx +c)/(secx+tanx)=1 +c/(secx+tanx)`


Now we have to find the particular solution at y(0) =4

so ` y(x) =1 +c/(secx+tanx)`

=> `y(0) = 1+c/(sec(0)+tan(0)) =4`

=> `1+c=4`


so the particular solution is

`y(x) = 1+ 3/(secx+tanx)`

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