The following DE describes a physical problem involving electronic circuits. How can I solve it?
$$x(t)\cdot a+x'(t)\cdot b+c\cdot\ln\left(1+d\cdot x(t)\right)=0\space\Longleftrightarrow\space x(t)=\dots$$
All the variables are real and positive and $x(0)=x_0$
I've no idea where to start what so ever. Thanks for any help or ideas
On
Going by @LutzL's suggestion we can linearize the equation for small $x$
$$ bx' + (a+cd)x = 0 $$
since $\ln(1+dx) \approx dx$
Then an approximation given by
$$ x(t) = x_0\exp\left(-\frac{(a+cd)}{b}t\right) $$
which resembles a decaying exponential as you've mentioned. Note that this approximation is only accurate if $|dx_0| \ll 1$ and is only valid for $t \ge 0$
Note that this solution is only valid for $|Dx_0|<1$. The integral cannot be evaluated in closed form generally, but it can be approximated using a Taylor series.
Given $$t=-\int\frac{b\,dx}{c\ln(1+Dx)+ax}$$one can expand the logarithm as a Taylor series, and ignore all terms excluding the lowest two order terms $Dx-\frac{D^2x^2}2$. $$\begin{align}t-t_0&=-\int^x\frac{b\,du}{(a+cD)u-\frac{cD^2}2u^2}=\frac2{cD^2}\int^x\frac{b\,du}{u^2-\frac{2(a+cD)}{cD^2}u}\end{align}$$Let $A=\frac{(a+cD)}{cD^2}$. $$\begin{align}t-t_0&=\frac2{cD^2}\int^x\frac{b\,du}{(u-2A)u}\\&=-\frac{2b}{cD^2}\frac1{2A}\int^xdu\left(\frac{1}u-\frac{1}{u-2A}\right)\\&=\frac{b}{cAD^2}\ln \left(\frac{x-2A}{x}\right)\end{align}$$ Then inverting, $$x-2A=x\exp\left(\frac{cAD^2}b(t-t_0)\right)\\x=\frac{2A}{1-\exp\left(\frac{cAD^2}b(t-t_0)\right)}$$ Finally, applying the initial condition, $$x=\frac{x_0}{\exp\left(\frac{cAD^2}bt\right)+\frac{x_0}{2A}\left(1-\exp\left(\frac{cAD^2}bt\right)\right)}$$
We initially assumed $x_0$ was small, and we can see that with this assumption, the corresponding solution ensures that $x$ stays small as time passes. So for small $x_0$, this is a good approximation to the true solution.