The general solution of the differential equation $\frac{dy}{dx}=\sin ^{-1} x$ is : |
$y=x \sin ^{-1} x-\sqrt{1-x^2}+C$, where C is a constant $y=x \sin ^{-1} x+\sqrt{1+x^2}+C$, where C is a constant $y=x \sin ^{-1} x+\sqrt{1-x^2}+C$, where C is a constant $y=-x \sin ^{-1} x+\sqrt{1-x^2}+C$, where C is a constant |
$y=x \sin ^{-1} x+\sqrt{1-x^2}+C$, where C is a constant |
$\frac{dy}{dx}=\sin ^{-1} x$ so $dy = sin^{-1} dx$ → integrating both sides → $\int dy = \int 1 × sin^{-1}x dx$ using $\int uv ~dx$ $= v \int u dx - \int v' \int u dx ~dx$ $y = x sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx$ as $\frac{d}{dx}(\sqrt{1-x^2})$ $= \frac{1 × (-2x)}{2 \sqrt{1-x^2}} = \frac{-x}{\sqrt{1-x^2}}$ $y = x sin^{-1}x + \sqrt{1-x^2} + C$ |