Question
For each of the differential equations in find the general solution:
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\text{sin}^{-1}\text{x}$
$\frac{\text{dy}}{\text{dx}}=\sin^{-1}\text{x}$
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.Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.