I have an integral equation of the following form:
$y(t)=\lambda x(t) + x(t)\int_{-\infty}^{\infty}K(t,s)x(s)ds$
I haven't been able to find any discussion online of integral equations with the extra $x(t)$ term multiplying the integral (which makes going to the Fourier domain less useful). Are there any techniques to find a solution for $x(t)$ here?
I am writing here my effort to solve this problem. I couldn't finish the job, but it might give you some starting point. Let's suppose $K(t,s)$ is a normal distribution as a function of $s$ centered at $t$. We can then write:
$y(t)=\lambda x(t) + x(t) \int _{-\infty} ^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(s-t)^2}{2}} x(s)ds = \lambda x(t) + x(t) \left( (\frac{1}{\sqrt{2\pi}} e^{-\frac{t^2}{2}}) \ast x(t)\right)$
which $\ast$ is the convolution. We take the Fourier transform from both sides keeping in mind that the Fourier transform of zero-mean Gaussian is also a Gaussian.
$Y(f)=\lambda X(f) +X(f) \ast \left( e^{-\frac{f^2}{2}} X(f)\right) = \lambda X(f) + \int _{-\infty} ^{\infty} X(h-f)X(h) e^{-\frac{h^2}{2}}dh $
Considering that the Normal distribution should obey the probability condition (its sum from $-\infty$ to $\infty$ should be 1), we can write:
$Y(f)= \frac{1}{\sqrt{2\pi}}\lambda X(f) \int _{-\infty} ^{\infty} e^{-\frac{h^2}{2}}dh + \int _{-\infty} ^{\infty} X(h-f)X(h) e^{-\frac{h^2}{2}}dh = \int _{-\infty} ^{\infty} \left(\frac{1}{\sqrt{2\pi}}\lambda X(f)+X(h-f)X(h)\right) e^{-\frac{h^2}{2}}dh $
Now, maybe you can assume that the integral is again the definition of a convolution and take the Fourier transform again. This might lead to somewhere!