I am doing some studying regarding Differential Equations and using the Method of Undetermined Coefficients in order to solve second order, non-linear, non-homogeneous equations. While asking this question, I realized someone had already asked the same question regarding the same exact problem on stackexchange. Their question can be found here.
The question that I have is asking us to solve:

$y'' -2y' -3y = (-3t)(1/e^t)$ , (call this $\alpha$).

The answer to $\alpha$ in the textbook is the same as what user32240 said, and then the coefficients are also listed in his/her answer.
Thus far, I have looked at the following link (if you visit this link, page 15 is the area with the information regarding this problem), and I have made little progress to see why we are using:

$(At^2)(1/e^t)+(Bt)(1/e^t)$, (call this $*$).

My Question is:
Why are we using $*$ to solve this equation? While I am attempting to solve $\alpha$, I am using $*/t$. Any help would be much appreciated!


 TITLE:     Elementary Differential Equations and Boundary Value Problems
 EDITION:   Tenth
 Authors:   William E. Boyce / Richard C. DiPrima
 Question:  pg.184, question 5
  • $\begingroup$ Then, is the rule of thumb that as the power increases/decreases, the powers of the variables on $A$ and $B$ decrease/increase along with them? $\endgroup$ – T.Woody Mar 10 '14 at 4:47

Our complementary solution has an $e^{−t}$.

We would have normally chosen $y_p=(a+bt)e^{-t}$, but since we already have $e^{-t}$ in the complementary, we need to multiply by a factor of $t$, else we would just get another complementary solution.

Hence, $y_p=t(a+bt)e^{-t}$.

I wrote my $y_p$ slightly different than you, but they are actually the same.

  • $\begingroup$ You seemed to have another surge in postings! $\endgroup$ – Namaste Mar 10 '14 at 13:30

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