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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.

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