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

The differential equation satisfied by the system of parabolas

y2 = 4a(x + a) is :

  1. A $$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
  2. B $$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$$
  3. C $$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$ Correct answer
  4. D $$y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

Solution

${y^2} = 4ax + 4{a^2}$<br><br>differentiate with respect to x<br><br>$\Rightarrow 2y{{dy} \over {dx}} = 4a$<br><br>$\Rightarrow a = \left( {{y \over 2}{{dy} \over {dx}}} \right)$<br><br>So, required differential equation is <br><br>$${y^2} = \left( {4 \times {y \over 2}{{dy} \over {dx}}} \right)x + 4{\left( {{y \over 2}{{dy} \over {dx}}} \right)^2}$$<br><br>$$ \Rightarrow {y^2}{\left( {{{dy} \over {dx}}} \right)^2} + 2xy\left( {{{dy} \over {dx}}} \right) - {y^2} = 0$$<br><br>$$ \Rightarrow y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

About this question

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

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