A) 1
B) 3
C) 4
D) 0
Correct Answer: A
Solution :
Idea To express given differential equation as polynomial in derivatives. Use trigonometric formulae. Hence, we know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] Given that\[\sqrt{1-{{x}^{2}}}+\sqrt{1-{{y}^{2}}}=b(x-y)\] Put\[x=\sin \alpha ,y=\sin \beta \]in Eq. (i) \[\Rightarrow \]\[\alpha ={{\sin }^{-1}}x,\beta ={{\sin }^{-1}}y\] \[\cos \alpha +\cos \beta =b(\sin \alpha -\sin \beta )\] Use identities\[2\cos \left( \frac{\alpha +\beta }{2} \right).\cos \left( \frac{\alpha -\beta }{2} \right)\] \[=b\cdot 2\cos \frac{\alpha +\beta }{2}\cdot \sin \frac{\alpha -\beta }{2}\] \[\alpha -\beta =2{{\cot }^{-1}}b\] \[{{\sin }^{-1}}x-{{\sin }^{-1}}y=2\,{{\cot }^{-1}}b\] Differentiating w.r.t x, we get \[\frac{1}{\sqrt{1-{{x}^{2}}}}-\frac{1}{\sqrt{1-{{y}^{2}}}}\frac{dy}{dx}=0\] Degree of above differential equation is one. TEST Edge Order and degree related to question are asked. To solve such types of questions students are advised to understand basic concept of differentiation and express the differential equation as a polynomial in derivatives.
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