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

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

Answers (1)

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 |

Posted by

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