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


1 Answer 1


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.

  • $\begingroup$ wow thanks a lot $\endgroup$
    – eric
    Apr 8, 2015 at 1:20
  • $\begingroup$ You're very welcome. $\endgroup$ Apr 8, 2015 at 1:25

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