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

Let y = y(x) be solution of the differential equation

${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$, with y(0) = 0.

If $y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$, then the value of $\alpha$ is equal to :

  1. A $- {1 \over 4}$ Correct answer
  2. B ${1 \over 4}$
  3. C $2$
  4. D $- {1 \over 2}$

Solution

$${{dy} \over {dx}} = {e^{3x}}.{e^{4y}} \Rightarrow \int {{e^{ - 4y}}dy = \int {{e^{3x}}dx} } $$<br><br>$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} + C \Rightarrow - {1 \over 4} - {1 \over 3} = C \Rightarrow C = - {7 \over {12}}$$<br><br>$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} - {7 \over {12}} \Rightarrow {e^{ - 4y}} = {{4{e^{3x}} - 7} \over { - 3}}$$<br><br>$${e^{4y}} = {3 \over {7 - 4{e^{3x}}}} \Rightarrow 4y = \ln \left( {{3 \over {7 - 4{e^{3x}}}}} \right)$$<br><br>$4y = \ln \left( {{3 \over 6}} \right)$ when $x = - {2 \over 3}\ln 2$<br><br>$y = {1 \over 4}\ln \left( {{1 \over 2}} \right) = - {1 \over 4}\ln 2$

About this question

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

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