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AP EAMCET · Maths · Differential Equations

The solution of \(x d y-y d x=\sqrt{x^2+y^2} d x\) when \(y(\sqrt{3})\) \(=1\) is

  1. A \(y^2+\sqrt{x^2+y^2}=x^2\)
  2. B \(5 y-\sqrt{x^2+y^2}=x^2\)
  3. C \(y+\sqrt{x^2+y^2}=x\)
  4. D \(5 y^2-\sqrt{x^2+y^2}=x\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(y+\sqrt{x^2+y^2}=x\)

Step-by-step Solution

Detailed explanation

Since, \(x d y-y d x=\sqrt{x^2+y^2} d x\) \(\begin{aligned} & \Rightarrow x d y=\left(y+\sqrt{x^2+y^2}\right) d x \\ & \Rightarrow \frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x} \end{aligned}\) Let \(y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\) So,…