the equations I need to show they are identical are..
$$\frac{dy}{dt}+r\int_{0}^{t}dt'\alpha e^{-\alpha (t-t')}y(t')=0$$ $$\frac{d^2y}{dt^2}+\alpha\frac{dy}{dt}+r \alpha y=0$$
I think differentiate the first term and used the condition $$\frac{dy}{dt}=-r\int_{0}^{t}dt'\alpha e^{-\alpha (t-t')}y(t')$$ can get the second equation.
my first attempt was changing the integral term of first equation with integration by parts, but it seems not working.
please help me
Since you have a mixture of derivatives and integrals (and you want a second order derivative in your ODE), you should think about differentiating your integro-differential equation. First we rewrite it slightly to make it easier for us:
$$\frac{dy}{dt} + r\alpha e^{-\alpha t}\int_0^t e^{\alpha t'}y(t')\,dt'.$$
Differentiating, we have (by using product rule and the fundamental theorem of calculus)
$$ 0 = \frac{d^2y}{dt^2} - r\alpha^2 e^{-\alpha t}\int_0^t e^{\alpha t'}y(t')\,dt' + r\alpha e^{-\alpha t} e^{\alpha t}y(t).$$
This isn't quite what you have but if you use your integro-differential equation, you can get the ODE.