To continue the questions:
Approximation of the convolution operator
Convolution using Integral Transforms
I have a differential equation, for example:
$\frac{dx}{dt}= \mu \cdot e^{-x^2} \cdot a_1 \cdot \sin(\omega_1 \cdot t) + a_1 \cdot \cos(\omega_1 \cdot t)$
{\[Alpha] = 0.1, \[Omega] = 2 Pi 0.1, \[Mu] = 5
His numerical solution is as follows:
The figure shows that one of the components of the solution can be:
$x(t) ≈ e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t}$
If we represent the solution of the equation in the form:
$x(t) = e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t} + \Delta(t)$
And substitute in the original equation, we get:
$\frac{e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t} + \Delta(t)}{dt}= \mu \cdot e^{-(e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t} + \Delta(t))^2} \cdot a_1 \cdot \sin(\omega_1 \cdot t) + a_1 \cdot \cos(\omega_1 \cdot t)$
Is it possible to somehow analytically and very approximately get a solution for $\Delta(t)$
EDIT:
I thought about this for a long time and came to the following assumption.
If we take the integral from both sides and transfer the $- e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t}$ to the right side, then we can get the following equation:
$\Delta(t) - $$\int_{}^{} \mu \cdot e^{-(e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t} + \Delta(t))^2} \cdot a_1 \cdot \sin(\omega_1 \cdot t) + a_1 \cdot \cos(\omega_1 \cdot t) dt = - e^{-\mu \cdot \frac{\alpha^2}{2} \cdot t}$
But is it possible to solve this equation, like the Fredholm equation, and find $\Delta(t)$?
I will be glad to be advised and helped.
