#### Please solve RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 68 textbook solution.

Answer : $y=x^{2}+\log\; \left | x \right |$

Hint              :  Using variable separable method

Given            :  $x \; d y=\left(2 x^{2}+1\right) d x \quad, x \neq 0$

Solution :  $x \; d y=\left(2 x^{2}+1\right) d x \quad$

\begin{aligned} &d y=\frac{\left(2 x^{2}+1\right)}{x} d x \\ &d y=\left(\frac{2 x^{2}}{x}+\frac{1}{x}\right) d x \\ &d y=\left(2 x+\frac{1}{x}\right) d x \end{aligned}

Integrating both sides

$\begin{gathered} \int d y=\int\left(2 x+\frac{1}{x}\right) d x \\ \end{gathered}$                                             ....(i)

$\begin{gathered} \int d y=\int 2 x d x+\int \frac{1}{x} d x \\ \end{gathered}$

$\begin{gathered} y=\frac{2 x^{2}}{2}+\log |x|+c \\ \end{gathered}$

$\begin{gathered} y=x^{2}+\log |x|+c \end{gathered}$                                                    .....(ii)

Since the curve passes through (1,1)

Putting $x=1, y=1$      in (ii)

\begin{aligned} &1=(1)^{2}+\log (1)+c \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\log 1=0] \\ &1=1+0+c=>c=0 \end{aligned}

Put $c=0$ in (ii)

i.e  , $\begin{gathered} y=x^{2}+\log |x|+c \\ \end{gathered}$

$y=x^{2}+\log |x|+0 \\$

$y=x^{2}+\log |x|$

Hence the equation of curve is

$y=x^{2}+\log\; \left | x \right |$