# For the Following Differential Equation, Find the General Solution:- D Y D X = Sin − 1 X - Mathematics

Sum

For the following differential equation, find the general solution:- $\frac{dy}{dx} = \sin^{- 1} x$

#### Solution

We have,

$\frac{dy}{dx} = \sin^{- 1} x$

$\Rightarrow dy = \left( \sin^{- 1} x \right)dx$

Integrating both sides, we get

$\int dy = \int\left( \sin^{- 1} x \right)dx$

$\Rightarrow \int dy = \sin^{- 1} x\int1 dx - \int\left[ \frac{d}{dx}\left( \sin^{- 1} x \right)\int1 dx \right]dx$

$\Rightarrow y = x \sin^{- 1} x - \int\frac{x}{\sqrt{1 - x^2}}dx$

$\text{Putting }t^2 = 1 - x^2,\text{ we get}$

$2t\ dt = - 2x\ dx$

$\Rightarrow - t\ dt = x\ dx$

$\therefore y = x \sin^{- 1} x + \int dt$

$\Rightarrow y = x \sin^{- 1} x + t + C$

$\Rightarrow y = x \sin^{- 1} x + \sqrt{1 - x^2} + C$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 64.5 | Page 146