Find the characteristic curves of: $P(D) = x_1^2 D_1^2 + x^2 D_2^2 -2 x_1 x_2 D_1 D_2 +e^{x_1} D_1 + e^{x_2} D_2$
To begin with, the principal part of $P$ is: $$ P_0 (D) = x_1^2 D_1^2 + x_2^2 D_2 ^2 $$
Let $n = \left( \frac{dx_2}{dt}, -\frac{dx_1}{dt} \right)$ be the vector perpendicular to the tangent of the characteristic curve of $P$ with $x_1 = f_1 (t)$ and $x_2 = f_2(t)$. Then we must solve the equation:
$$
x_1 ^2 (dx_2)^2 = x_2^2 (dx_1)^2 \iff x_1 dx_2 = \pm x_2 dx_1
$$
Thus the characteristic curves of $P$ are described by:
$$
\ln(x_1) = \pm \ln(x_2) + c \iff x_1 = c_1 x_2 \quad \text{and} \quad x_1 = c_2 \frac{1}{x_2}
$$
Is the principle part of $P$ and the rest of the approach correct?
Hint
$$ x_1^2 D_1^2 + x_2^2 D_2^2 -2 x_1 x_2 D_1 D_2 = \left(x_1 D_1-x_2D_2\right)^2 $$