Get Answers to all your Questions

header-bg qa

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}


If 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] \\ & \end{aligned}

=e^{-\int \frac{d t}{t}} \\

=e^{-\ln t} \\

=e^{\ln \left(t^{-1}\right)} \\

=t^{-1} \\


\begin{aligned} &=\frac{1}{1+x^{2}} \\ & \end{aligned}

y \text { If }=\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 \\

=\int d x+\tan ^{-1} x+C

\begin{aligned} &=x+\tan ^{-1} x+C \\ & \end{aligned}

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

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support