#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 48 maths

Answer: $2 y \sin y=2 x^{2} \log x+x^{2}-1$

Hint: Separate the terms of x and y and then integrate them.

Given: $\frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y}, y=0 \text { when } x=1$

Solution:

\begin{aligned} &\frac{d y}{d x}=\frac{2 x(\log x+1)}{\sin y+y \cos y} \\\\ &\Rightarrow(\sin y+y \cos y) d y=2 x(\log x+1) d x \end{aligned}

Integrating both sides

\begin{aligned} &\Rightarrow \int(\sin y+y \cos y) d y=\int 2 x(\log x+1) d x \\\\ &\Rightarrow \int \sin y d y+\int y \cos y d y=\int 2 x \log x d x+\int 2 x d x \end{aligned}                    [?Integration by parts]

\begin{aligned} &\Rightarrow-\cos y+\left[y \sin y-\int \sin y d y\right]=2\left[\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \frac{x^{2}}{2} d x\right] \\\\ &\Rightarrow-\cos y+y \sin y+\cos y=x^{2} \log x-\frac{x^{2}}{2}+c \\\\ &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}+c \end{aligned}   ...........(1)

Given that $y=0 \text { when } x=1$

\begin{aligned} &\Rightarrow 0 \sin 0=1 \cdot \log 1+\frac{1}{2}+c \\\\ &\Rightarrow 0=0+\frac{1}{2}+c \\\\ &\Rightarrow c=-\frac{1}{2} \end{aligned}

Put in (1)

\begin{aligned} &\Rightarrow y \sin y=x^{2} \log x+\frac{x^{2}}{2}-\frac{1}{2} \\\\ &\Rightarrow 2 y \sin y=2 x^{2} \log x+x^{2}-1 \end{aligned}