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)
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
This question is part of PrepWiser's free JEE Main question bank. 172 more solved questions on Differential Equations are available — start with the harder ones if your accuracy is >70%.