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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Differential Equations

BITSAT 2015 · Mathematics · Differential Equations

The degree of the differential equation satisfying 1-x^ 2 + 1+y^ 2...

The degree of the differential equation satisfying 1−x2+1+y2=a(x−y)\sqrt{1-x^{2}}+\sqrt{1+y^{2}}=a(x-y) is

Options

  1. A1
  2. B2
  3. C3
  4. D4

Correct answer

A. 1

Step-by-step solution

\quad Put x=sin⁡θx=\sin \theta and y=sin⁡ϕ⇒cos⁡θ+cos⁡ϕ=a(sin⁡θ−sin⁡ϕ)⇒2cos⁡y=\sin \phi\Rightarrow \cos \theta+\cos \phi=a(\sin \theta-\sin \phi)\Rightarrow 2 \cosθ+ϕ2cos⁡θ−ϕ2=2acos⁡θ+ϕ2sin⁡θ−ϕ2⇒cot⁡θ−ϕ2=a⇒θ−ϕ=2cot⁡−1a⇒sin⁡−1x−sin⁡−1y=2cot⁡−1a Differentiate 11−x2−11−y2dydx=0\begin{aligned} & \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=2 a \cos \frac{\theta+\phi}{2} \sin \frac{\theta-\phi}{2} \\ \Rightarrow \cot \frac{\theta-\phi}{2}=a \Rightarrow \theta-\phi=2 \cot ^{-1} \mathrm{a} \\ \Rightarrow \sin ^{-1} \mathrm{x}-\sin ^{-1} \mathrm{y}=2 \cot ^{-1} \mathrm{a} \\ \text{ Differentiate } \frac{1}{\sqrt{1-\mathrm{x}^{2}}}-\frac{1}{\sqrt{1-\mathrm{y}^{2}}} \frac{\mathrm{dy}}{\mathrm{dx}}=0 \end{aligned}$$ so the degree is one
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