Find the particular solution of the differential equation ` cos x (dy )/(dx) + y = sin x `, given that ` y = 2 `when ` x = 0 `.

Find the particular solution of the differential equation ` cos x (dy

1.	Find the particular solution of the differential equation ` cos x (dy )/(dx) + y = sin x `, given that ` y = 2 `when ` x = 0 `.
Answer» The given differential equation may be written as ` (dy )/(dx) + (sec x )y = tan x" "`... (i) This is of the form ` (dy)/(dx) + Py = Q`, where `P = secx and Q = tanx ` Thus, the given differential equation is linear. ` IF = e ^(int Pdx) = e ^(int sec x dx ) = e ^(log (sec x + tan x )) = (sec x + tanx )` So, its solution is given by `y xx IF = int (Q xx IF) dx + C ` i.e. `y (sec x + tan x ) = int (tanx (secx + tanx )dx + C ` ` " " = int sec x tanx dx + int tan^(2) x dx + C ` `" " = sec x + int (sec ^(2) x - 1 ) dx + C ` ` " " = sec x + int sec ^(2) x dx - int dx +C ` ` " " = sec x + tanx - x + C`. Thus, `y (sec x + tanx ) = sec x + tan x - x + C " " ` ... (ii) It is given that what ` x = 0 `, then `y = 2 ` ` therefore ` putting ` x = 0 and y = 2 ` in (ii), we get ` C = 1 ` Hence, `y (sec x + tanx ) = secx + tanx - x + 1` is the required solution.