Solution of the following differential equation $ e^{u'(x)}(u'(x)-1)-u^2=c$ I'm trying to solve an exercises and I'm up to prove that
$$ \begin{cases} e^{u'(x)}(u'(x)-1)-u^2=c\\
u(0)=1\\
u(1)=0
\end{cases}$$
has no solution but I have no idea on how to do it.
Can you help me?
 A: First, note that the function $y\mapsto e^y(y-1)$ attains its minimum at $y=0$ (indeed the derivative $ye^y$ is negative for $y<0$ and positive for $y>0$ and zero at $y=0$), where it attains the minimum value $-1$. 
Thus, $e^y(y-1)\ge -1$ for all $y\in\mathbb{R}$. In particular, if there is such a solution $u$ to your equation, then $c+u(x)^2 = e^{u'(x)}(u'(x)-1)\ge -1$ for all $x\in[0,1]$. 
Since $u(1) = 0$, we have $c\ge -1$. 
Since $u(0) = 1$, we have $e^{u'(0)}(u'(0)-1) = c+u(0)^2\ge -1+1 = 0$. In particular, $u'(0)-1\ge 0$, i.e. $u'(0)\ge 1$, so $u$ is increasing at $0$. 
If we let $x_0$ denote the point in $[0,1]$ where $u$ attains its maximum, then $x_0$ is strictly between $0$ and $1$. Thus $u'(x_0) = 0$. But this also means that $c+u(x_0)^2 = e^{u'(x_0)}(u'(x_0)-1) = -1$, so $u(x_0)^2 = -1 - c\le 0$ since $c\ge -1$, i.e. $u(x_0) = 0$, which is impossible since $x_0$ is where $u$ should attain its maximum value, and $u(0) = 1$ is larger.
A: Assume that the given DE has solution then find the satisfying function. Then for the given initial conditions show that equation can't be satisfied for any interval of validity that contains the initial values. Can be proved directly using existence and uniqueness theorems.
Solution for the DE:
\begin{equation}
c_{1}+x=\int_{1}^{u(x)} \frac{1}{W\left(\frac{\varepsilon^{2}+c}{e}\right)+1} d \xi
\end{equation}
(W(x) is the product log function).
Inserting the boundary conditions yield no solutions. 
