Show that $c$ is a constant solution of $(E)$, then that $(E)$ admits exactly two constant solutions denoted $c _ { 1 }$ and $c _ { 2 }$ such that $c _ { 1 } < 0 < c _ { 2 }$. Deduce the value of $c$ as a function of $c _ { 1 }$ and $c _ { 2 }$. We admit that $y$ is decreasing on $\mathbf { R } _ { + }$ and $\lim _ { x \rightarrow + \infty } y ( x ) = c$, where $c \in \mathbf { R }$. The equation $(E)$ is: $$( E ) : \quad y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }.$$
Show that $c$ is a constant solution of $(E)$, then that $(E)$ admits exactly two constant solutions denoted $c _ { 1 }$ and $c _ { 2 }$ such that $c _ { 1 } < 0 < c _ { 2 }$. Deduce the value of $c$ as a function of $c _ { 1 }$ and $c _ { 2 }$.
We admit that $y$ is decreasing on $\mathbf { R } _ { + }$ and $\lim _ { x \rightarrow + \infty } y ( x ) = c$, where $c \in \mathbf { R }$. The equation $(E)$ is:
$$( E ) : \quad y ^ { \prime } ( x ) + y ( x ) + 1 = \frac { 1 } { 2 } \mathrm { e } ^ { y ( x ) }.$$