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99 Percentile Qs Bank for JEE MainMathematicsDifferential Equations

The solution of d y d x = x - 1 2 + y - 2 2 t a n - 1 y - 2 x - 1 x y - 2 x - y + 2 t a n - 1 y - 2 x - 1 is equal to

Options

  1. A( x - 1 ) 2 +   y - 2 2 t a n - 1 y - 2 x - 1 - x - 1 y - 2 = 2 x - 1 2 log x - 1 + C
  2. Bx - 1 ) 2 + y - 1 2 - 2 x - 1 y - 2 t a n - 1 y - 2 x - 1 = 2 x - 1 2 log C
  3. Cx - 1 2 + y - 1 2 tan - 1 y - 2 x - 1 - 2 x - 1 y - 2 = log C x - 1
  4. DNone of these

Correct answer

A. ( x - 1 ) 2 +   y - 2 2 t a n - 1 y - 2 x - 1 - x - 1 y - 2 = 2 x - 1 2 log x - 1 + C

Step-by-step solution

Given, d y d x = x - 1 2 +   y - 2 2 t a n - 1 y - 2 x - 1 x y - 2 x - y + 2 t a n - 1 y - 2 x - 1 ⇒ d y d x = x - 1 2 +   y - 2 2 t a n - 1 y - 2 x - 1 x y - 2 - y - 2 t a n - 1 y - 2 x - 1 ⇒ d y d x = x - 1 2 + y - 2 2 t a n - 1 y - 2 x - 1 x - 1 y - 2 t a n - 1 y - 2 x - 1 Let, m = x - 1   &   n = y - 2 ⇒ d m = d x   &   d n = d y ⇒ d n d m = m 2 + n 2 t a n - 1 n m m n t a n - 1 n m ⇒ d n d m = 1 + n m 2 t a n - 1 n m n m t a n - 1 n m . Let, n

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