Question

The degree of the differential equation satisfying $\sqrt{1-x^4}+\sqrt{1-y^4}=a(x^2-y^2)$, is

• A. 1
• B. 2
• C. 3
• D. None of these

Answer 1

The correct option is A

1

Put $x^2=sin\alpha ,y^2=sin\beta$
$\therefore$ Given equation reduces to $cos\alpha +cos\beta =a(sin\alpha -sin\beta )$
$\Rightarrow 2cos\left(\frac{\alpha +\beta }{2}\right)cos\left(\frac{\alpha -\beta }{2}\right)=2acos\left(\frac{\alpha +\beta }{2}\right)sin\left(\frac{\alpha -\beta }{2}\right)$
$\Rightarrow cot\left(\frac{\alpha -\beta }{2}\right)=a$
$\Rightarrow \alpha -\beta =2co{t}^{-1}a$
$\Rightarrow si{n}^{-1}{x}^2-si{n}^{-1}{y}^2=2co{t}^{-1}a$
On differentiating w.r.t.x, we get
$\frac{1}{\sqrt{1-x^4}}.2x-\frac{1}{\sqrt{1-y^4}}.2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{x}{y}\sqrt{\frac{1-y^4}{1-x^4}}$
which is a differential equation of first order and first degree
Hence, (A) is the correct answer.