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TS EAMCET · Maths · Differential Equations

The solution of the differential equation \(x d y-y d x=\sqrt{x^2+y^2} d x\), given that \(y=1\) when \(x=\sqrt{3}\), is

  1. A \(\left(x^2-y^2\right)^2=x^2+y^2\)
  2. B \(\left(x^2-y^2\right)^2=x^2+y^2\)
  3. C \(\left(x^2+y\right)^2=x^2-y^2\)
  4. D \(x^2-y=\left(x+y^2\right)^2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left(x^2-y^2\right)^2=x^2+y^2\)

Step-by-step Solution

Detailed explanation

Given, differential equation \(x d y-y d x=\sqrt{x^2+y^2} d x\) \(\ldots\) (i) Putting, \(y=v x\) and differentiating w.r.t. \(x\), we get \(d y=v d x+x d v\) \(\ldots\) (ii) From Eq. (ii), \(x(v d x+x d v)-v x d x=\sqrt{x^2+(v x)^2} d x\)…