Does a solution exist for $u_x u_{xx} = 1- x$?

Question

I'm trying to find a solution to the equation

$$\det(T(x))g(\nabla T(x)) = f(x)$$

where $T(x)$ has the form $T(x) = \nabla \phi$, $f(x) = 1-x$ and $g(x) = x$.

I think this leads to

$$\phi_x \phi_{xx} = 1 - x,$$

but I'm unsure if there's a solution. I tried $\phi = Ax^2 + Bx$ to no avail.

Answer 1

Integrating gives $$ 2\phi_x \phi_{xx} = 2(1-x) $$ $$ {\phi_x}^2 = C - (1-x)^2 $$

$$ \phi_x = \pm \sqrt{C-(1-x)^2} $$

You can do the rest.

Answer 2

$$\phi_x \phi_{xx} = (1 - x)$$ Substitute $p=\phi_x$ $$pp_x = 1 - x,$$ $$pdp =\int (1 - x) dx$$ $$p^2 =2\int (1 - x) dx$$ $$p^2 =2 (x - \frac {x^2}2+K)$$ $$p^2 =2 x - {x^2}+K$$ $$\phi_x =\pm\sqrt{2 x - {x^2}+K}$$

Is $\phi_x$ a partial derivative ?