Let $y = y(x)$ be the solution of the differential equation $(x + 1)y' - y = {e^{3x}}{(x + 1)^2}$, with $y(0) = {1 \over 3}$. Then, the point $x = - {4 \over 3}$ for the curve $y = y(x)$ is :

<p>$(x + 1){{dy} \over {dx}} - y = {e^{3x}}{(x + 1)^2}$</p> <p>${{dy} \over {dx}} - {y \over {x + 1}} = {e^{3x}}(x + 1)$</p> <p>If ${e^{ - \int {{1 \over {x + 1}}x} }} = {e^{ - \log (x + 1)}} = {1 \over {x + 1}}$</p> <p>$\therefore$ $y\left( {{1 \over {x + 1}}} \right) = \int {{{{e^{3x}}(x + 1)} \over {x + 1}}dx}$</p> <p>${y \over {x + 1}} = \int {{e^{3x}}dx}$</p> <p>${y \over {x + 1}} = {{{e^{3x}}} \over 3} + c$</p> <p>$\because$ $y(0) = {1 \over 3}$</p> <p>${1 \over 3} = {1 \over 3} + c$</p> <p>$\therefore$ $c = 0$</p> <p>So : $y = {{{e^{3x}}} \over 3}(x + 1)$</p> <p>$$y' = {e^{3x}}(x + 1) + {{{e^{3x}}} \over 3} = {e^{3x}}\left( {x + {4 \over 3}} \right)$$</p> <p>$y'' = 3{e^{3x}}\left( {x + {4 \over 3}} \right) + {e^{3x}} = {e^{3x}}(3x + 5)$</p> <p>$y' = 0$ at $x = {{ - 4} \over 3}$ & $y'' = {e^{ - 4}}(1) > 0$ at $x = {{ - 4} \over 3}$</p> <p>$\Rightarrow x = {{ - 4} \over 3}$ is point of local minima</p>