#### Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 36 subquestion (iv)

Answer:  \begin{aligned} &\frac{1}{5} x^{4}+\frac{c}{x} \\ & \end{aligned}

Give:  $x \frac{d y}{d x}+y=x^{4} \\$

Hint: Using  $\int x^{n} d x$

Explanation:  $x \frac{d y}{d x}+y=x^{4}$

Divide by x

\begin{aligned} &\frac{d y}{d x}+\frac{y}{x}=\frac{x^{4}}{x} \\ &\frac{d y}{d x}+\left(\frac{1}{x}\right) y=x^{3} \end{aligned}

This is a first order linear differential equation of the form

\begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{1}{x} \text { and } Q=x^{3} \end{aligned}

The integrating factor $If$  of this differential equation is

$I f=e^{\int P d x}$

\begin{aligned} &=e^{\int \frac{1}{x} d x} \\ &=e^{\log x} \quad\left[\int \frac{1}{x} d c=\log |x|+C\right] \\ &=x \end{aligned}

Hence, the solution of differential equation is

\begin{aligned} &y(I f)=\int Q I f d x+C \\ &=y x=\int x^{3} x d x+C \\ &=y x=\int x^{4} d x+C \\ &=y x=\frac{x^{5}}{5}+C \end{aligned}

Divide by x, we get

$=y=\frac{x^{4}}{5}+\frac{C}{x}$