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  • Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (vi) maths textbook solution.

Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 36 subquestion (vi) maths textbook solution.

Answers (1)

Answer : \left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right)

Give : \frac{d y}{d x}-\frac{2 x y}{1+x^{2}}=x^{2}+2

Hint: Using \int \frac{1}{1+x^{2}}dx

Explanation : \frac{d y}{d x}-\frac{2 x y}{1+x^{2}}=x^{2}+2

                     =\frac{d y}{d x}-\left(\frac{2 x}{1+x^{2}}\right) y=x^{2}+2

This is a first order linear differential equation of the form

                 \begin{aligned} &\frac{d y}{d x}+P y=Q \\ &P=\frac{-2 x}{1+x^{2}} \text { and } Q=x^{2}+2 \end{aligned}

The integrating factor If  of this differential equation is

            \begin{aligned} &I f=e^{\int P d x} \\ &=e^{\int-\frac{2 x}{1+x^{2}} d x} \\ &=e^{-2 \int \frac{x}{1+x^{2}} d x} \\ &=e^{-\log \left|1+x^{2}\right|} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{x} d x=\log |x|+C\right] \end{aligned}

             \begin{aligned} &=e^{\log \left|1+x^{2}\right|^{-1}} \\ &=\left(1+x^{2}\right)^{-1} \\ &=\frac{1}{1+x^{2}} \end{aligned}

Hence, the solution of different equation is

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

           \begin{aligned} &=\frac{y}{1+x^{2}}=\int 1+\frac{1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=\int 1 d x+\int \frac{1}{x^{2}+1} d x+C \\ &=\frac{y}{1+x^{2}}=x+\tan ^{-1} x+C \quad\left[\int \frac{1}{x^{2}+1} d x=\tan ^{-1} x\right] \\ &=y=\left(1+x^{2}\right)\left(x+\tan ^{-1} x+C\right) \end{aligned}

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