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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Differential Equations

NDA 2016 · Mathematics · Differential Equations

What are the degree and order respectively of the differential...

What are the degree and order respectively of the differential equation satisfying \mathrmey1−x2+x1−y2=c\mathrmex\mathrme^{\mathrm{y} \sqrt{1-\mathrm{x}^2}+\mathrm{x}} \sqrt{1-\mathrm{y}^2}=c \mathrme^{\mathrm{x}}, (where c>0,∣x∣<1,∣y∣<1)\mathrm{c} > 0,|\mathrm{x}| < 1,|\mathrm{y}| < 1) ?

Options

  1. A1,1
  2. B1,2
  3. C2,1
  4. D2,2

Correct answer

A. 1,1

Step-by-step solution

\begin{aligned} & e^{y\left(\sqrt{1-x^2}+x \sqrt{1-y^2}\right)-x=c} \\ & \begin{aligned} \Rightarrow & y \sqrt{1-x^2}+x \sqrt{1-y^2}-x=\log c \\ \Rightarrow & \frac{d y}{d x} \sqrt{1-x^2}+y \cdot \frac{1}{\not 2 \sqrt{1-x^2}}(-\not 2 x)+\sqrt{1-y^2} \\ & +x \cdot \frac{1}{\not 2 \sqrt{1-y^2}}(-\not 2 y) \cdot \frac{d y}{d x}-1=0\end{aligned}\end{aligned}⇒dydx1−x2−xy1−x2+1−y2−xy1−y2⋅dydx=1Degree =1, order =1\begin{aligned} & \Rightarrow \frac{d y}{d x} \sqrt{1-x^2}-\frac{x y}{\sqrt{1-x^2}}+\sqrt{1-y^2}-\frac{x y}{\sqrt{1-y^2}} \cdot \frac{d y}{d x}=1 \\ & \text{Degree }=1, \text{ order }=1\end{aligned}$
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