#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 38 textbook solution.

Answer : $y=\frac{x^{2}}{4}+\frac{c}{x^{2}}$

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

Hint : Using $\int \frac{1}{x} d x \text { and } e^{\log e^{x}}=x$

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

Divide by x

\begin{aligned} &=\frac{d y}{d x}+\frac{2 y}{x}=x \\ &=\frac{d y}{d x}+\left(\frac{2}{x}\right) y=x \end{aligned}

This is a linear differential equation of the form

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

The integrating factor $If$ of this differential equation is

\begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int \frac{2}{x} d x} \\ &=e^{2 \int \frac{1}{x} d x} \\ &=e^{2 \log x} \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log x+C\right] \end{aligned}

$\begin{array}{ll} = & e^{\log x^{2}} \\ =x^{2} & {\left[e^{\log e^{x}}=x\right]} \end{array}$

Hence, the solution is

\begin{aligned} &y I f=\int Q I f d x+C \\ &=y x^{2}=\int x x^{2} d x+C \\ &=y x^{2}=\int x^{3} d x+C \\ &=y x^{2}=\frac{x^{4}}{4}+C \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}

Divide by $x^{2}$

$=y=\frac{x^{2}}{4}+\frac{c}{x^{2}}$