#### Explain solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 66 Subquestion (ix) textbook solution.

Answer : $y(\sec x+\tan x)=\sec x+\tan x-x+c$

Hint : integrate by applying integration of sec x and tan x

Given : $\frac{d y}{d x}+\sec x(y)=\tan x$

Solution : differential equation is of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &\text { Where }_{L} P=\sec x, \quad Q=\tan x \end{aligned}

Finding integrating factor

\begin{aligned} &I . F=e^{\int P d x} \\ &I . F=e^{\int \sec x d x} \\ &I \cdot F=e^{\log |\sec x+\tan x|} \\ &\text { I. } F=\sec x+\tan x \end{aligned}

Solution is

$\begin{gathered} y \times I . F=\int(Q \times I . F) d x+c \\ y(\sec x+\tan x)=\int \tan x(\sec x+\tan x)dx+c \\ y(\sec x+\tan x)=\int \tan x \sec x d x+\int \tan ^{2} d x+c \end{gathered}$

\begin{aligned} &y(\sec x+\tan x)=\int \tan x \sec x d x+\int\left(\sec ^{2} x-1\right) d x+c \\ &y(\sec x+\tan x)=\sec x+\tan x-x+c \end{aligned}