I was solving a basic differential equation which is given below.
$$\frac{dx}{\sqrt{1-x^2}}+\frac{dy}{\sqrt{1-y^2}}=0$$
My solution is:- $-\sin^{-1}{x}=\sin^{-1}{y}+c$
I wrote it as $-\sin^{-1}{x}=\sin^{-1}{y}+\sin^{-1}{k}$
So my actual question is can I reduce it further?
If we take sine of both sides I think it will reduce to $-x=y\sqrt{1-c^2}+c\sqrt{1-y^2}$....$(1)$
Is it correct or wrong.If it is wrong then please provide me the correct one.
Further I am unable to get the actual differential equation by differentiating both sides in equation $(1)$ and this is why I am in doubt.
After separating $x$ and $y$ variables we have:
$\frac{dx}{\sqrt{1-x^2}}=-\frac{dy}{\sqrt{1-y^2}}$
We then integrate:
$\int{\frac{1}{\sqrt{1-x^2}} dx}=-\int{\frac{1}{\sqrt{1-y^2}}dy}$
by changing the parameters of $x=sint$ and $y=sint$ likewise, we calculate the integral and conclude that:
$-{sin}^{-1}x={sin}^{-1}y+c$
Let it be like:
${sin}^{-1}x+{sin}^{-1}y=c$ (Note that as $c$ is undetermined, there is no difference between $c$ or $-c$)
Let's say:
${sin}^{-1}x=\alpha , {sin}^{-1}y=\beta$
So we have:
$\alpha+\beta=c$
Taking sine from both sides will provide:
$sin(\alpha+\beta)=sin(c)$
Or simply: $sin(\alpha+\beta)=c$
We now expand the sine:
$sin(\alpha+\beta)=sin\alpha cos\beta+sin\beta cos\alpha$
We can easily find each terms below:
$sin(\alpha)=sin({sin}^{-1}x)=x$
$sin(\beta)=sin({sin}^{-1}y)=y$
$cos(\alpha)=cos({sin}^{-1}x)=\sqrt{1-x^2}$
$cos(\beta)=cos({sin}^{-1}y)=\sqrt{1-y^2}$
So this is the result of simplification:
$x \sqrt{1-y^2} + y \sqrt{1-x^2}=c$