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

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

Answers (1)

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}

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