Medium MCQ +4 / -1 PYQ · JEE Mains 2020

The solution of the differential equation

$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:

(where c is a constant of integration)

  1. A $x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$ Correct answer
  2. B $y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$
  3. C x – log<sub>e</sub>(y+3x) = C
  4. D x – 2log<sub>e</sub>(y+3x) = C

Solution

${{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0$<br><br>Let $\ln (y + 3x) = t$<br><br>${1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}$<br><br>$$ \Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}$$<br><br>$\therefore$ $\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}$<br><br>$\Rightarrow tdt = dx$<br><br>${{{t^2}} \over 2} = x + c$<br><br>${1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c$

About this question

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

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