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  • Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 29 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 29 maths textbook solution.

Answers (1)

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

Hint : To solve this equation we use \frac{dy}{dx}+Py=Q where P,Q are constants.

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

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

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

I\: f of differential equation is

           \begin{aligned} &\text { If }=e^{-\int \frac{2 x}{1+x^{2}} d x} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[1+x^{2}=t, 2 x d x=d t\right] \\ &=e^{-\int \frac{d t}{t}} \\ &=e^{-\ln t} \end{aligned}

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

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

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

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