3
$\begingroup$

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!


BOOK INFO

 TITLE:     Elementary Differential Equations and Boundary Value Problems
 EDITION:   Tenth
 Authors:   William E. Boyce / Richard C. DiPrima
 Question:  pg.184, question 5
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.