Angeline Hayden

2022-04-23

Solution of first order integro-differential equation

Consider$\frac{dy}{dx}+p\left(x\right)y={\int}_{0}^{\mathrm{\infty}}y\left(x\right)dx=R\approx \left(say\right)$

$y\left(0\right)={\int}_{0}^{\mathrm{\infty}}f\left(x\right)y\left(x\right)dx.$

I want to solve above differential equation. Here is my try:

$y\left(x\right)=y\left(0\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)d\tau}+R{\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)d\tau}ds.$

where$R={\int}_{0}^{\mathrm{\infty}}y\left(0\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)d\tau}dx+R{\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)d\tau}dsdx$

$y\left(0\right)={\int}_{0}^{\mathrm{\infty}}f\left(x\right)y\left(0\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)d\tau}dx+R{\int}_{0}^{\mathrm{\infty}}f\left(x\right){\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)d\tau}dsdx.$

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.

Consider

I want to solve above differential equation. Here is my try:

where

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.

Tatairfzk

Beginner2022-04-24Added 12 answers

Step 1

Given: $R=y\left(0\right){\int}_{0}^{\infty}{e}^{-{\int}_{0}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}x+R{\int}_{0}^{\infty}{\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}s\u200a\mathrm{d}x$

$y\left(0\right)=y\left(0\right){\int}_{0}^{\infty}f\left(x\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}x+R{\int}_{0}^{\infty}f\left(x\right){\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}s\u200a\mathrm{d}x$

$y\left(0\right)=y\left(0\right){\int}_{0}^{\infty}f\left(x\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}x+R{\int}_{0}^{\infty}f\left(x\right){\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}s\u200a\mathrm{d}x$

you can rearrange to obtain two linear equations with two unknowns. The former equation is equivalent to

$0=y\left(0\right){\int}_{0}^{\infty}{e}^{-{\int}_{0}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}x+R\left[{\int}_{0}^{\infty}{\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}s\u200a\mathrm{d}x-1\right]$

Step 2

while the latter is equivalent to

$0=y\left(0\right)\left[{\int}_{0}^{\infty}f\left(x\right){e}^{-{\int}_{0}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}x-1\right]+R{\int}_{0}^{\infty}f\left(x\right){\int}_{0}^{x}{e}^{-{\int}_{s}^{x}p\left(\tau \right)\u200a\mathrm{d}\tau}\u200a\mathrm{d}s\u200a\mathrm{d}x$

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.

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