Name: Hoda

Question:

The equation is:

y" - 2y' + y = t e^{t} + 4

We need to use The method of Undetermined coefficients. I have tried assuming that the
solution is Ate^{t}+Be^{t}+C, but all I get is C=4 and I tried (At^{2}+Bt+C)e^{t}+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?

Thanks,

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

A e^{t} + B t e^{t}
The task now is to find one solution of y" - 2y' + y = t e^{t} + 4.
Since the right side contains a term of the form t e^{kt}, (in your case k = 1), the usual tact is to try a solution with a term of the same form, t e^{kt}. 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 t^{3} e^{t}, thus try

C t^{3} e^{t} + D
Harley