Solve the differential equation: $\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} } $

Question

$$\frac{dy}{dx} = \sqrt{ \frac{1-y^2}{1-x^2} }$$

Simplified it to: $$ \int \frac{1}{\sqrt{1-y^2}} dy = \int \frac{1}{\sqrt{1-x^2}} dx$$

$$\implies\sin^{-1} y = ( \sin^{-1} x + C)$$

$$y = (\sin (\sin^{-1} x) \cos C_1) + \cos(\sin^{-1} x) (\sin C_1)$$

How to I simplify $\cos(\sin^{-1} x)$ to get the final answer ?

Answer 1

$\cos^2(\arcsin(x))=1-\sin^2(\arcsin(x))=1-x^2$

Answer 2

Put,

$y$ = $\sin \alpha$ and $x$ = $\sin \beta$

calculate $dy$ and $dx$,

$dy$ = $\cos \alpha$$.d \alpha$ and $dx$ = $\cos \beta$$.d\beta$.

The equation will be something like,

=> $\int \cot\alpha $ $d\alpha$= $\int \cot\beta$ $d\beta$
=> $\log$ |$sin$ $\alpha$| + $c_1$ = $\log$ |$sin$ $\beta$| + $c_2$
=> $\left(\frac{sin \alpha} {sin\beta}\right)$ = $e^{c_2 - c_1}$ = $c$
=> $sin$ $\alpha$ = $sin$ $\beta$ $.c$

Now, replace $sin$ $\alpha$ and $sin$ $\beta$, with $y$ and $x$.

$y$ = $x$ $.c$

Answer 3

I think what you have come up to looks like an identity, $$sinAcosB+cosAsinB=sin(A+B)$$

Answer 4

You can simplify it like this

$$ y = \sin(\sin^{-1} x + C) = \cos (C) \sin(\sin^{-1}x) + \sin (C)\cos(\sin^{-1}x) \\ = \cos (C)x + \sin(C)\sqrt{1-x^2} $$

The explanation is

$$ \cos(\sin^{-1}x) = \sqrt{1-\sin^2(\sin^{-1}x)} = \sqrt{1-x^2} $$