3rd Order Differential Equation Help
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3rd Order Differential Equation Help

[From: ] [author: ] [Date: 11-11-06] [Hit: ]
Therefore y = Ae^2t + Be^t - te^t + 1/2 t + 3/4, and, by integrating both sides,y = 1/2 Ae^2t + Be^t - (te^t - e^t) + 1/4 t^2 + 3/4 t + C(remember you need to integrate -te^t using integration by parts)And you can rearrange this a bit to make it a bit nicer:y = 1/2 Ae^2t + Be^t - te^t + e^t + 1/4 t^2 + 3/4 t + Cy = 1/2 Ae^2t + (B - t + 1)e^t + 1/4 t(t+3) + Cand, if you want, you can let D = B + 1 to make the solution even neater:y = 1/2 Ae^2t + (D - t)e^t + 1/4 t(t+3) + CHope this helps!......
2a = 3/2
a = 3/4.
So the particular integral is y = a + bt + cte^t = 3/4 + 1/2 t - te^t.

The general solution = the complementary function + the particular integral, so the general solution is y = Ae^2t + Be^t - te^t + 1/2 t + 3/4.

Recall that we replaced y' with y to transform the initial 3rd order differential equation into a 2nd order. We now need to replace y with y' again to get the solution to the initial problem.
Therefore y' = Ae^2t + Be^t - te^t + 1/2 t + 3/4, and, by integrating both sides,
y = 1/2 Ae^2t + Be^t - (te^t - e^t) + 1/4 t^2 + 3/4 t + C
(remember you need to integrate -te^t using integration by parts)
And you can rearrange this a bit to make it a bit nicer:
y = 1/2 Ae^2t + Be^t - te^t + e^t + 1/4 t^2 + 3/4 t + C
y = 1/2 Ae^2t + (B - t + 1)e^t + 1/4 t(t+3) + C
and, if you want, you can let D = B + 1 to make the solution even neater:
y = 1/2 Ae^2t + (D - t)e^t + 1/4 t(t+3) + C

Hope this helps!
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