Get Answers to all your Questions

Name: Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 29
Uploaded: Fri, 2021-08-27 17:18
Description: Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 29

Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 29

Answers (1)

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

Hint: To solve this equation we use $\frac{d y}{d x}+P y=Q$ 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: $\frac{d y}{d x}+\left(\frac{-2 x y}{1+x^{2}}\right)=\frac{\left(x^{2}+2\right)\left(x^{2}+1\right)}{1+x^{2}}$

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