How do we find $u(x)$? I want to know how to find $u(x)$ in the below question:
$$u''(x)+{e^u}^{(x)} = 0\\ x \in[0,1]\\u(0) = u(1) = 0$$
Please explain briefly how this was done??
Thanks!!
 A: According to Maple,
$$ u(x) =  - \ln \left(2 c^2\;\cosh^2\left(\dfrac{2 c \; \text{arctanh}\left(\sqrt{1-2c^2}\right)+x}{2c}\right)\right)$$
where $c \approx -.6591242642$.
A: This problem looks extremely difficult (at least to me). Without any other solutions, I sued a CAS for solving $$u''(x)+{e^u}^{(x)} = 0$$ This leads to the terrible expression $$u(x)=\log \left(\frac{1}{2} \left(c_1-c_1 \tanh ^2\left(\frac{1}{2} \sqrt{c_1
   \left(c_2+x\right){}^2}\right)\right)\right)$$ Applying the condition $u(0)=0$ leads to $$c_1-c_1 \tanh ^2\left(\frac{1}{2} \sqrt{c_1 c_2^2}\right)=2$$ and $u(1)=0$ leads to $$c_1-c_1 \tanh ^2\left(\frac{1}{2} \sqrt{c_1 \left(c_2+1\right){}^2}\right)=2$$ which seems to lead to some contradiction as  Kainui pointed out in his/her comment.
A: $$ \frac{d^2}{dx^2}u(x)+e^{u(x)}=0 $$
$$ \frac{d^2}{dx^2}u(x)=-e^{u(x)}$$
$$ \left[\frac{d^2}{dx^2}u(x)\right] \left[\frac{d}{dx}u(x)\right] =-e^{u(x)}\left[\frac{d}{dx}u(x)\right] $$
$$ \left[\frac{d^2}{dx^2}u(x)\right] \left[\frac{d}{dx}u(x)\right] dx =-e^{u(x)}\left[\frac{d}{dx}u(x)\right]dx $$
$$ \left[\frac{d}{dx}u(x)\right] d\left[\frac{d}{dx}u(x)\right]=-e^{u(x)}du(x)$$
$$ \int \left[\frac{d}{dx}u(x)\right] d\left[\frac{d}{dx}u(x)\right]=-\int e^{u(x)}du(x)$$
$$ \frac{1}{2}\left[\frac{d}{dx}u(x)\right]^2+C_1=-e^{u(x)}+C_2 $$
$$ \left[\frac{d}{dx}u(x)\right]^2=-2e^{u(x)}+2C_2-2C_1 $$
$$ \frac{d}{dx}u(x)=\pm\sqrt{-2e^{u(x)}+2C} $$
$$ \frac{\frac{d}{dx}u(x)}{\sqrt{-2e^{u(x)}+2C}}=\pm 1 $$
$$ \frac{\frac{d}{dx}u(x)}{\sqrt{-2e^{u(x)}+2C}}dx=\pm dx $$
$$ \int \frac{\frac{d}{dx}u(x)}{\sqrt{-2e^{u(x)}+2C}}dx=\pm \int dx $$
$$ \int \frac{1}{\sqrt{-2e^{u(x)}+2C}}du(x)=\pm \int dx $$
Can you take it from here?
