For each of the differential equations in find the general soluti… - Vidyadip

Question

For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$

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Answer

The given differential equation is
$\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
Separating the variables and integrating,

$\int \text{dy}=\int\text{sin}^{-1}\text{x dx}$
$\therefore\ \int \text{1 dy}= \int \sin^{-1}\text{x}\cdot 1\ \text{dx}$
$\therefore \ \text{y}=(\text{sin}^{-1}\text{x}).\text{x}-\int\frac{1}{\sqrt{1-\text{x}^2}}.\text{x dx}$
$\text{or}\ \text{y = x sin}^{-1}\text{x}+\frac{1}{2}\int(1-\text{x}^2)^{-\frac{1}{2}}(-2\ \text{x}) \ \text{dx}$
$\therefore\ \text{y}=\text{x sin}^{-1}\text{x}+\frac{1}{2}\frac{(1-\text{x}^2)\frac{1}{2}}{\frac{1}{2}}+\text{c}$
$\therefore\ \text{y}= \text{x sin}^{-1}\text{x}+\sqrt{1-\text{x}^2}+\text{c}$
which is the required solution.

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