What is the meaning of having imaginary solutions to a differential equation I am trying to solve this equation
$$x+e^{xt}=0$$
for different values of x and see when there is no real solution. Plotting this on a graph and assuming x is real gives me the different values of t. I want to find the value of t when the above equation has no real solution. To do this, I first differentiate the equation and set that value equal to $0$. This gives me the relation for the minimum value of $$x+e^{-xt}$$. When the value of t is greater than this the curve doesnt cut the x axis and there is no real solution. But now assuming x is imaginary I can still solve to get a solution which consists of complex terms. But I dont understand what this means in the real domain. If there is no real value of $x$ for which there is a solution, then what does allowing imaginary values of $x$ do? How does a system like this work in real life?

I don't know about "real life", but one place your equation can come up is in solving a PDE such as
$$ \dfrac{\partial u}{\partial t} = \dfrac{\partial^2 u}{\partial x^2} $$
with boundary condition
$$ \dfrac{\partial u}{\partial x}(0,t) + u(L,t) = 0$$
If you assume a solution of the form $u(x,t) = \exp(r x + r^2 t)$ you get
from the boundary condition 
$$ r + e^{rL} = 0 \tag{1}$$
You may only be interested in real solutions of your PDE, but note that the real and imaginary parts of a solution are solutions.  Thus if $r = \alpha + i \beta$ is a complex solution of (1), you get real solutions of your PDE of the forms
$$ \eqalign{\text{Re}(\exp(r x + r^2 t)) &= \exp(\alpha x + (\alpha^2-\beta^2)t) \cos(\beta x + 2 \alpha \beta t)\cr
\text{Im}(\exp(r x + r^2 t)) &= \exp(\alpha x + (\alpha^2-\beta^2)t) \sin(\beta x + 2 \alpha \beta t)\cr}$$
