QJEEM20Ma6050: JEE Main 2020 Mathematics - Class 12 JEEM20Ma6050 Solution: The solution of the differential equation is: (where c is a…

Question

Verified Solution • 2025–26 Session moderate

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)

Accepted Answer

{{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0

Let

\ln (y + 3x) = t

{1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}

\Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}

\therefore

\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}

\Rightarrow tdt = dx

{{{t^2}} \over 2} = x + c

{1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c

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