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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 13.

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 13.

Answers (1)

Answer:  y=\frac{1}{6}(x^5\tan^{-1}x^3-x^3+\tan^{-1}x^3)+C

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

Given: \frac{dy}{dx}=x^5\tan^{-1}x^3

Solution: We have \frac{dy}{dx}=x^5\tan^{-1}x^3

  dy=(x^5\tan^{-1}x^3)dx

Integrate both sides

\int dy= \int x^5\tan^{-1}(x^3)\; x

y= \int x^5\tan^{-1}(x^3)\; x

Put t=x^3

dt=3x^2dx

\begin{aligned} &y=\frac{1}{3} \int t \tan ^{-1} t d t \\ &=\frac{1}{3} \int t \times \tan ^{-1} t d x\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\right] \\ &=\frac{1}{3}\left(\tan ^{-1} t \int t d t-\int\left(\frac{d}{d t} \tan ^{-1} t \cdot \int t d t\right) d t\right) \end{aligned}

\begin{aligned} &=\frac{1}{3} \times \frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{6} \int \frac{t^{2}}{1+t^{2}} d t \\ &=\frac{t^{2} \tan ^{-1} t}{6}-\frac{1}{6} \int \frac{t^{2}+1-1}{1+t^{2}} d t \\ &=\frac{t^{2} \tan ^{-1} t}{6}-\frac{1}{6} \int d t+\frac{1}{6} \int \frac{1}{1+t^{2}} d t \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}}=\tan ^{-1} x+c\right] \end{aligned}

\begin{aligned} &=\frac{t^{2} \tan ^{-1} t}{6}-\frac{1}{6} t+\frac{\tan ^{-1} t}{6}+C \\ &=\frac{x^{6} \tan ^{-1} x^{3}}{6}-\frac{1}{6} x^{3}+\frac{\tan ^{-1} x^{3}}{6}+C \\ &=\frac{1}{6}(x^6 \tan^{-1}x^3-x^3+\tan^{-1}x^3)+C \end{aligned}     

Hence, y=\frac{1}{6}(x^5\tan^{-1}x^3-x^3+\tan^{-1}x^3)+C

Posted by

infoexpert24

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