Question

The degree of the differential equation satisfied by the curves $\sqrt{1+x}-a\sqrt{1+y=1}$, is

Answer 1

$\sqrt{1+x}-a\sqrt{1+y}=1$

$\Rightarrow a\sqrt{1+y}=\sqrt{1+x}-1$

$\Rightarrow a=\frac{\sqrt{1+x}-1}{\sqrt{1+y}}$

$\Rightarrow \frac{1}{a}=\frac{\sqrt{1+y}}{\sqrt{1+x}-1}$

differentiating w.r.t $x$

$\Rightarrow 0=\frac{d}{dx}\left(\frac{\sqrt{1+y}}{\sqrt{1+x}-1}\right)$

$\Rightarrow \frac{\frac{1}{2\sqrt{1+y}}\frac{dy}{dx}\left(\sqrt{1+x}-1\right)-\sqrt{1+y}\times \frac{1}{2\sqrt{1+x}}}{{\left(\sqrt{1+x}-1\right)}^2}=0$

$\Rightarrow \frac{\sqrt{1+x}-1}{\sqrt{1+y}}\frac{dy}{dx}-\frac{\sqrt{1+y}}{\sqrt{1+x}}=0$

$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{1+y}}{\sqrt{1+x}}\times \frac{\sqrt{1+y}}{\sqrt{1+x}-1}$

$\Rightarrow \frac{dy}{dx}=\frac{1+y}{1+x-\sqrt{1+x}}$

Therefore the degree of the differential equation satisfying this curve is 1