# Is the following integro-differential equation solvable? $\frac{dx}{dt} = a_0 + (a_1+a_2x)e^{-a_3 \int_0^t x(s)} + a_4x(t)$

I have an integro-differential equation of the following form: $$\frac{dx}{dt} = a_0 + (a_1+a_2x(t))e^{-a_3 \int_0^t x(s)} + a_4x(t)$$ In an attempt to solve this equation, I transformed it first to an equation of the form $$\frac{d^2g}{dt^2} = a_0 + (a_1+a_2 \frac{dg}{dt})e^{-a_3 g(t)} + a_4\frac{dg}{dt},$$ by defining $$g(t) = \int_0^t x(s)ds$$ (I know that $$x(s)>0$$ for all $$s$$) and then to an equation of the form $$u(g)\frac{du}{dg} = a_0 + (a_1+a_2 u(g))e^{-a_3 g} + a_4u,$$ by defining $$u(g) = \frac{dg}{dt}(t(g))$$ (since g(t) is monotonically increasing, it is invertible). Then further substituting $$h=u^2$$, I get an equation of the form $$\frac{dh}{dg} = b_0 + b_1 h^{\frac{1}{2}} + (b_2 h^{\frac{1}{2}} + b_3)e^{b_4g}.$$

I am stuck here however. The "homogenous" part of this equation is solvable with solution of the form $$h_{hom} = b_1g + \frac{b_2}{b_4}e^{b_4g} + c_0,$$ but I am not sure how to proceed from here since the equation is not-linear. Also in general $$b_0=b_2$$ and $$b_1 \neq b_2$$ so the equation is not seperable as well.