#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 37 subquestion (vi)

Answer:  \begin{aligned} & y=x^{2}+\cos x \\ & \end{aligned}

Give:   $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x \\$

Hint: Using  $\int \frac{1}{1+x^{2}} d x \\$

Explanation:  $\frac{d y}{d x}+y \tan x=2 x+x^{2} \tan x$

$=\frac{d x}{d y}+(\tan x) y=2 x+x^{2} \tan x$

This is a linear differential equation of the form

\begin{aligned} &\frac{d x}{d y}+P x=Q \\ &P=\tan x \text { and } Q=2 x+x^{2} \tan x \end{aligned}

The integrating factor  $If$  of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \tan x d x} \\ &=e^{\log |\sec x|} \end{aligned}                      $\quad\left[\int \tan x d x=\sec x+C\right] \\$

$=\sec x \quad\left[e^{\log e^{x}}=x\right]$

Hence, the solution is

\begin{aligned} &y I f=\int \text { QIf } d x+C \\ &=y \sec x=\int\left(2 x+x^{2} \tan x\right) \sec x d x+C \\ &=y \sec x=I+C \ldots(i) \\ &=I=\int\left(2 x+x^{2} \tan x\right) \sec x d x \\ &=\int 2 x \sec x d x+\int x^{2} \tan x \sec x d x \end{aligned}

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

Substituting in (i)

$=y \sec x=x^{2} \sec x+C$

Dividing by $\sec x$

\begin{aligned} &=y=x^{2}+\frac{C}{\sec x} \\ &=y=x^{2}+C \cos x \ldots(i i) \end{aligned}

Now   \begin{aligned} &y(0)=1 \text { when } x=0, y=1 \\ & \end{aligned}

$\quad=1=0^{2}+C \cos (0) \\$

$\quad=1=+C(1) \quad[\cos 0=1] \\$

$\quad=C=1$

Substituting in (ii)

\begin{aligned} &=y=x^{2}+(1) \cos x \\ &=y=x^{2}+\cos x \end{aligned}