Consider $$ \frac{dy}{dx} + p(x)y = \int_{0}^{\infty} y(x)dx = R~~(say) $$ $$ y(0) = \int_{0}^{\infty} f(x)y(x)dx.$$
I want to solve above differential equation. Here is my try:
$$ y(x) = y(0) e^{-\int_{0}^{x}p(\tau)d \tau} + R \int_{0}^{x} e^{-\int_{s}^{x} p(\tau) d \tau} ds. $$ where $$ R = \int_{0}^{\infty} y(0) e^{-\int_{0}^{x}p(\tau)d \tau} dx + R \int_{0}^{\infty} \int_{0}^{x} e^{-\int_{s}^{x} p(\tau) d \tau} ds dx $$ $$ y(0) = \int_{0}^{\infty} f(x)y(0) e^{-\int_{0}^{x}p(\tau)d \tau} dx + R \int_{0}^{\infty}f(x)\int_{0}^{x} e^{-\int_{s}^{x} p(\tau) d \tau} ds dx. $$ Now, I am not able to find $R$ which is independent of $y(0)$ and $y(0)$ which is independent of $R$. Please let me know if it is possible to find solution of this integro-differential equation.
Given $$R=y(0)\int_0^{\infty}e^{-\int_0^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}x+R\int_0^{\infty}\int_0^xe^{-\int_s^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}s\,\mathrm{d}x$$ $$y(0)=y(0)\int_0^{\infty}f(x)e^{-\int_0^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}x+R\int_0^{\infty}f(x)\int_0^xe^{-\int_s^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}s\,\mathrm{d}x$$ you can rearrange to obtain two linear equations with two unknowns. The former equation is equivalent to $$0=y(0)\int_0^{\infty}e^{-\int_0^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}x+R\left[\int_0^{\infty}\int_0^xe^{-\int_s^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}s\,\mathrm{d}x-1\right]$$ while the latter is equivalent to $$0=y(0)\left[\int_0^{\infty}f(x)e^{-\int_0^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}x-1\right]+R\int_0^{\infty}f(x)\int_0^xe^{-\int_s^xp(\tau)\,\mathrm{d}\tau}\,\mathrm{d}s\,\mathrm{d}x$$ This can be solved with just elementary algebra, or using matrix algebra. You will get $y(0)$ and $R$ exclusively in terms of $f$ and $p,$ provided suitable integrability conditions for these functions. It depends on the determinant of the corresponding matrix. If the matrix is invertible, then $y(0)$ and $R$ are unique.