- $\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
Solution:
Given differential equation is
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$
This is a linear differential equation
Here, $\text{P}=\tan\text{x},\text{Q}=\sec\text{x},$
$\therefore\text{I.F.}=\text{e}^{\int\tan\text{xdx}}$
$=\text{e}^{\log|\sec\text{x}|}=\sec\text{x}$
Thus, the general solution is
$\text{y}.\sec\text{x}=\int\sec\text{x}.\sec\text{x}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\int\sec^2\text{x}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\sec\text{x}=\tan\text{x}+\text{C}$