How to solve an intro-differential equation with integrals.

Question

$y'(t)-y(t)-3\cdot\int[e^{x-t}\cdot y(x),x,0,t]=16\cdot t, y(0)=16$

I am having a difficult time figuring out how to evaluate the integral to solve with the rest of the problem.

Answer 1

If your problem is to solve $$y'(t)-y(t)-3\mathrm{e}^{-t}\int_{0}^{t}\mathrm{e}^{x}y(x)\mathrm{d}x=16t$$ with $y(0)=16$, then you can differentiate wrt $t$ and get $$y''(t)-y'(t)-3y(t)+3\mathrm{e}^{-t}\int_{0}^{t}\mathrm{e}^{x}y(x)\mathrm{d}x=16.$$ Now you can use the first equation to replace the integral term and obtain the new ODE $$y''(t)-4y(t)-16t-16=0.$$ The homogenous solution is given in general by $$y(t)=A_+\mathrm{e}^{2t}+A_-\mathrm{e}^{-2t}\quad\quad A_+,A_-\in\mathbb{R}.$$ To find a particular solution, you may check it as a polynomial in $t$.