Medium MCQ +4 / -1 PYQ · JEE Mains 2021

Let y = y(x) be the solution of the differential equation ${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$, y(1) = $-$1. Then the value of (y(3))2 is equal to :

  1. A 1 $-$ 4e<sup>3</sup>
  2. B 1 $-$ 4e<sup>6</sup> Correct answer
  3. C 1 + 4e<sup>3</sup>
  4. D 1 + 4e<sup>6</sup>

Solution

${e^x}\sqrt {1 - {y^2}} dx + {y \over x}dy = 0$<br><br>$\Rightarrow {e^x}\sqrt {1 - {y^2}} dx + {{ - y} \over x}dy$<br><br>$$ \Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx} $$<br><br>$\Rightarrow \sqrt {1 - {y^2}} = {e^x}(x - 1) + c$<br><br>Given : At x = 1, y = $-$1<br><br>$\Rightarrow$ 0 = 0 + c $\Rightarrow$ c = 0<br><br>$\therefore$ $\sqrt {1 - {y^2}} = {e^x}(x - 1)$<br><br>At x = 3 <br><br>$1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}$

About this question

Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree

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