Let y = y(x) be the solution of the differential equation (x + 1)y' - y = e^3x(x + 1)^2, w…

Question

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 :

Accepted Answer

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

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

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

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