Name: Hoda

The equation is:

y" - 2y' + y = t et + 4

We need to use The method of Undetermined coefficients. I have tried assuming that the solution is Atet+Bet+C, but all I get is C=4 and I tried (At2+Bt+C)et+D, but again I get 0=0 when I calculate the first and second derivatives, so i get no information on the constants. Any suggestions?


Hi Hoda,

I assume hat you solved the homogeneous equation y" - 2y' + y = 0 by first solving the auxilary equation m2 -2 m + 1 = 0. The polynomial m2 -2 m + 1  has the double root m = 1. Hence the general solution of the homogeneous equation is

A et + B t et The task now is to find one solution of y" - 2y' + y = t et + 4.

Since the right side contains a term of the form t ekt, (in your case k = 1), the usual tact is to try a solution with a term of the same form, t ekt. The complication is that k = 1 and 1 is a solution of the auxilary equation. In fact it is a double root. In this case you have to increase the power of t in your trial solution by 2, the multiplicity of the root. Hence, in your trial solution you need a term of the form t3 et, thus try

C t3 et + D

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