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Suppose I have the following third-order difference equation:

Y[t]+aY[t-1]+bY[t-2]+cY[t-3]=G d^t

I need to convert it into a differential equation and I followed this method:

y' = y(t+1)-y(t);
y'' = y'(t+1)-y'(t) = y(t+2)-2y(t+1)+y(t)
y'''=y''(t+1)-y''(t) = y(t+3)-3y(t+2)+3y(t+1)-y(t)

Rearranging the orginal expression:

y(t+3)-3y(t+2)+3y(t+1)-y(t)+(a+3)(y(t+2)-2y(t+1)+y(t))+(3-2a+b)(y(t+1)-y(t))+(1+c+b-3a)y(t) = G d^t

Which can be converted to: y'''+(a+3)y''+(3-2a+b)y'+(1+c+b-3a)y = G d^t

Is this procedure correct? How do I treat G d^t ? Thanks a lot for your help.

Lorenzo Di Domenico
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 29 '22 at 08:42
  • Assuming by d^t you mean $d^t$ you got an ODE already. – Kurt G. Sep 29 '22 at 10:32
  • Thanks, the problem is that the result of the difference equation is different from the differential one. For instance, if we take a simpler version of the original function: Y[t]+aY[t-1]+bY[t-2]+cY[t-3]=G. Using mathematica, the result of the differential equation is: G/(1 - 3 a + b + c) + E^(t Root[1 - 3 a + b + c + (3 - 2 a + b) #1 + (3 + a) #1^2 + #1^3 &, 1]) C[1] + E^(t Root[ 1 - 3 a + b + c + (3 - 2 a + b) #1 + (3 + a) #1^2 + #1^3 &, 2]) C[2] + E^(t Root[ 1 - 3 a + b + c + (3 - 2 a + b) #1 + (3 + a) #1^2 + #1^3 &, 3]) C[3]]}} – Lorenzo Di Domenico Sep 29 '22 at 11:38
  • While the result of the difference equation is: G/(1 + a + b + c) + C[1] Root[c + b #1 + a #1^2 + #1^3 &, 1]^t + C[2] Root[c + b #1 + a #1^2 + #1^3 &, 2]^t + C[3] Root[c + b #1 + a #1^2 + #1^3 &, 3]^t. And this solution is correct, while the solution produced by solving the differential equation isn't – Lorenzo Di Domenico Sep 29 '22 at 11:40
  • Is that surprising ? With a huge time step of $\Delta t=1$ is it true that $y'(t)=y(t+1)-y(t)$ ? – Kurt G. Sep 29 '22 at 12:49

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