QJEEM22Ma1170: JEE Main 2022 Mathematics - Class 12 JEEM22Ma1170 Solution: Let be the solution of the differential equation , with . Then, …

$(x + 1){{dy} \over {dx}} - y = {e^{3x}}{(x + 1)^2}$

${{dy} \over {dx}} - {y \over {x + 1}} = {e^{3x}}(x + 1)$

If ${e^{ - \int {{1 \over {x + 1}}x} }} = {e^{ - \log (x + 1)}} = {1 \over {x + 1}}$

$\therefore$ $y\left( {{1 \over {x + 1}}} \right) = \int {{{{e^{3x}}(x + 1)} \over {x + 1}}dx}$

${y \over {x + 1}} = \int {{e^{3x}}dx}$

${y \over {x + 1}} = {{{e^{3x}}} \over 3} + c$

$\because$ $y(0) = {1 \over 3}$

${1 \over 3} = {1 \over 3} + c$

$\therefore$ $c = 0$

So : $y = {{{e^{3x}}} \over 3}(x + 1)$

$y' = {e^{3x}}(x + 1) + {{{e^{3x}}} \over 3} = {e^{3x}}\left( {x + {4 \over 3}} \right)$

$y'' = 3{e^{3x}}\left( {x + {4 \over 3}} \right) + {e^{3x}} = {e^{3x}}(3x + 5)$

$y' = 0$ at $x = {{ - 4} \over 3}$ & y'' = {e^{ - 4}}(1) > 0 at $x = {{ - 4} \over 3}$

$\Rightarrow x = {{ - 4} \over 3}$ is point of local minima