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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 18.

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 18.

Answers (1)

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

Hint: You have to integrate by applying integration of \tan^nx.

Given: (1+x^2)\frac{dy}{dx}-x=2\tan^{-1}x

Solution: (1+x^2)\frac{dy}{dx}-x=2\tan^{-1}x

\frac{ dy}{dx}=\frac{2\tan^{-1}x+x}{1+x^2}

Integrate on both sides

  \int dy=\int \frac{2\tan^{-1}x+x}{1+x^2}\; dx

Let  \tan^{-1}x=t               \left [1+x^2=y, 2x=dy, x=\frac{1}{2}dy \right ]

\frac{1}{1+x^2}dx=dt

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

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