Verify that implicitly defined function satisfies differential equation Problem 1. on page 9 in George Simmons' textbook "Differential equations with applications and historical notes" reads "verify that the following functions (explicit or implicit) are solutions of the corresponding differential equations" and further (h):
$$y=\sin^{-1}xy\quad\quad xy'+y=y'\sqrt{1-x^2y^2}$$
So $y$ is implicitly defined. Could you provide me with an approach/Ansatz to this?
 A: If you use implicit differentiation i.e assume $y = y(x)$ then,
\begin{align*} \frac{d}{dx} (y) &= \frac{d}{dx} (\textrm{arcsin}(xy)) \end{align*}
A: We can differentiate $y(x)=\arcsin xy(x)$
using the chain rule:
$$
y'=\frac{(xy)'}{\sqrt{1-x^2y^2}}=\frac{xy'+y}{\sqrt{1-x^2y^2}}
$$
it follows that
$$
y'\sqrt{1-x^2y^2}=xy'+y
$$
A: A complete solution:
$$ y=\sin^{-1}(xy)$$
$$\frac{dy}{dx}=\frac{d}{dx}\sin^{-1}(xy)$$
with chain rule and $y'=\frac{1}{\sqrt{1-x^2}}$:
$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2y^2}}\frac{d(xy)}{dx}$$
$$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2y^2}}(y+x\frac{dy}{dx})$$
Thus,
$$y'=\frac{1}{\sqrt{1-x^2y^2}}(y+xy')$$
$$y'\sqrt{1-x^2y^2}=y+xy'$$
To compute the derivative of:
$$y=\sin^{-1}{x}$$
$$\sin{y}=x$$
$$\frac{d}{dx}\sin{y}=\frac{d}{dx}x$$
$$\cos{y}\frac{dy}{dx}=1$$
$$\frac{dy}{dx}=\frac{1}{\cos{y}}$$
with the identiy: $\sin^2{y}+\cos^2{y}=1$, and as $\sin{y}=x$, we know: $\cos{y}=\pm\sqrt{1-x^2}$. We just need the positive part to verify the solution, thus:
$$y'=\frac{1}{\sqrt{1-x^2}}$$
