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

Answer : $y=\frac{x^{2}}{16}(4 \log |x|-1)+c x^{-2}$

Hint : integrate by applying integration of $x^{n}$ and $log\; x$

Given : $x \frac{d y}{d x}+2 y=x^{2} \log x$

Solution : convert the given differential equation is of the form  $\frac{d y}{d x}+P y=Q$

$x \frac{d y}{d x}+2 y=x^{2} \log x$

Divide both sides by x

\begin{aligned} &\frac{d y}{d x}+2 \frac{y}{x}=x \log x \\ &\text { Now } P=\frac{2}{x} \quad, Q=x \log x \end{aligned}

Differentiate

\begin{aligned} &\text { I. } F=e^{\int P d x} \\ &\text { I. } F=e^{\int \frac{2}{x} d x} \\ &\text { I.F }=e^{2 \int \frac{1}{x} d x} \\ &\text { I. } F=e^{2 \log x} \\ &\text { I. } F=e^{\log x^{2}}=x^{2} \end{aligned}

Solution is

$\begin{gathered} y \times \text { I.F }=\int(Q \times I . F) d x+c \\ y x^{2}=\int x \log x \times x^{2} d x+c \\ y x^{2}=\int \log x \times x^{3}+c \end{gathered}$

\begin{aligned} &\text { Use } \int f(x) \times g(x) d x=f(x) \int g(x) d x-\int\left\{f^{\prime}(x) \int g(x) d x\right\} d x \\ &y x^{2}=\log x \int x^{3} d x-\int\left\{\frac{d}{d x} \log x \int x^{3} d x\right\} d x \\ &y x^{2}=\log x\left(\frac{x^{4}}{4}\right)-\int \frac{1}{x}\left(\frac{x^{4}}{4}\right) d x+c \end{aligned}

\begin{aligned} &y x^{2}=\frac{\log x \times x^{4}}{4}-\int \frac{x^{3}}{4} d x+c \\ &y x^{2}=\frac{x^{4} \log x}{4}-\frac{x^{4}}{16}+c \end{aligned}

\begin{aligned} &y=\frac{x^{4} \log x}{4 x^{2}}-\frac{x^{4}}{16 x^{2}}+\frac{c}{x^{2}} \\ &y=\frac{x^{2} \log |x|}{4}-\frac{x^{2}}{16}+c x^{-2} \\ &y=\frac{x^{2}}{16}(4 \log |x|-1)+c x^{-2} \end{aligned}