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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 16.

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 16.

Answers (1)

Answer:  y=\frac{1}{2}\sin^{-1}x^2+C

Hint: You have to integrate by applying integration of xn.

Given: \sqrt{1-x^4}dy=x\; dx

Solution:

\sqrt{1-x^4}dy=x\; dx

dy=\frac{x}{\sqrt{1-x^4}}\; dx

Integrating both sides,

\int dy=\int \frac{x}{\sqrt{1-x^4}}\; dx

Let x^2=\sin \theta

 \begin{aligned} &2 x d x=\cos \theta d \theta \\ &y=\frac{2 x d x}{\sqrt{1-\left(x^{2}\right)^{2}}} \\ &=\frac{1}{2} \int \frac{\cos \theta}{\sqrt{1-\sin ^{2} \theta}} d \theta\left[\sqrt{1-\sin ^{2} \theta}=\cos \theta\right] \\ &=\frac{1}{2} \int d \theta \end{aligned}

=\frac{1}{2} \theta+C

=\frac{1}{2} \sin^{-1}x^2+C   

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