The solution of the differential equation
\[{{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,\]
where 0 \[ \le \] x < \[{\pi \over 2}\], and y (0) = 1, is given by :
Solutions · 1
Official solution
Given,
\[{{dy} \over {dx}} + {y \over 2}\sec x = {{\tan x} \over {2y}}\]
\[ \Rightarrow \] \[2y{{dy} \over {dx}} + {y^2}\sec x = \tan x\]
Now, let
y2 = t
\[ \Rightarrow \] 2y\[{{dy} \over {dx}} = {{dt} \over {dx}}\]
\[ \therefore \] New equation,
\[{{dt} \over {dx}} + t\sec x = \tan x\]
\[ \therefore \] I.F = \[{e^{\int {\sec xdx} }}\]
= \[{e^{\ln \left( {\sec x + \tan x} \right)}}\]
= sec x + tan x
\[ \therefore \] Solution is,
t(sec x + tan x) = \[\int {\tan x} \] (sec x + tan x) dx
\[ \Rightarrow \] t(sec x + tan x) = sec x + tan x - x + c
\[ \Rightarrow \] t = 1 - \[{x \over {\sec x + \tan x}} + c\]
\[ \Rightarrow \] y2 = 1 - \[{x \over {\sec x + \tan x}} + c\]
Given,
y(0) = 1
\[ \therefore \] 1 = 1 - 0 + c
\[ \Rightarrow \] c = 0
\[ \therefore \] y2 = 1 - \[{x \over {\sec x + \tan x}}\]