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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Differential Equations

NDA 2018 · Mathematics · Differential Equations

What is the solution of (1+2 x) d y-(1-2 y) d x=0 ?

What is the solution of (1+2x)dy−(1−2y)dx=0(1+2 x) d y-(1-2 y) d x=0 ?

Options

  1. Ax−y−2xy=cx-y-2 x y=c
  2. By−x−2xy=cy-x-2 x y=c
  3. Cy+x−2xy=cy+x-2 x y=c
  4. Dx+y+2xy=cx+y+2 x y=c

Correct answer

A. x−y−2xy=cx-y-2 x y=c

Step-by-step solution

(1+2x)dy−(1−2y)dx=0⇒(1+2x)dy=(1−2y)dx⇒dy1−2y=dx1+2x\begin{aligned} & (1+2 x) d y-(1-2 y) d x=0 \Rightarrow(1+2 x) d y=(1-2 y) d x \\ & \Rightarrow \frac{d y}{1-2 y}=\frac{d x}{1+2 x} \end{aligned} Integrating both sides, ∫dy1−2y=∫dx1+2x\int \frac{d y}{1-2 y}=\int \frac{d x}{1+2 x}⇒−12log⁡(1−2y)=12log⁡(1+2x)+12log⁡c⇒log⁡(1−2y)+log⁡(1+2x)=log⁡c⇒(1+2x)(1−2y)=c⇒1+2x−2y−4xy=c⇒2x−2y−4xy=c⇒x−y−2xy=c\begin{aligned} & \Rightarrow \frac{-1}{2} \log (1-2 \mathrm{y})=\frac{1}{2} \log (1+2 \mathrm{x})+\frac{1}{2} \log \mathrm{c} \\ & \Rightarrow \log (1-2 \mathrm{y})+\log (1+2 \mathrm{x})=\log \mathrm{c} \\ & \Rightarrow(1+2 \mathrm{x})(1-2 \mathrm{y})=\mathrm{c} \\ & \Rightarrow 1+2 \mathrm{x}-2 \mathrm{y}-4 \mathrm{xy}=\mathrm{c} \\ & \Rightarrow 2 \mathrm{x}-2 \mathrm{y}-4 \mathrm{xy}=\mathrm{c} \\ & \Rightarrow \mathrm{x}-\mathrm{y}-2 \mathrm{xy}=\mathrm{c} \end{aligned}$$
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