2
$\begingroup$

Fnd a general solution to the differential equation $y'' - y' - 2y = 0$. Then, use the two solutions you found to write two other linearly independent solutions to the problem. Write a second general solution using your new linearly independent solutions.

We have the characteristic equation:

$r^2 - r - 2 = 0$

$(r-2)(r+1) = 0$

Thus, we have real, distinct roots $r_1 = 2$ and $r_2 = -1$. Our independent solutions are then $c_1e^{r_1x}$ and $c_2e^{r_2x}$ and so our general solution is $y(x) = c_1e^{r_1x} + c_2e^{r_2x}$.

Now, I'm just unsure how I would go about writing two other linearly independent solutions and the second general solution. Can I just substitute two random values for $c_1$ and $c_2$?

Thank you!

$\endgroup$
2
  • $\begingroup$ If $f_1,f_2$ are linearly independent, then so are the pair $f_1+f_2, f_1-f_2$. $\endgroup$
    – copper.hat
    Oct 3, 2012 at 21:23
  • $\begingroup$ Your independent solutions are not $c_1e^{r_1x}$ and $c_2e^{r_2x}$. Your independent solutions are $e^{r_1x}$ and $e^{r_2x}$. $\endgroup$ Oct 4, 2012 at 7:50

1 Answer 1

3
$\begingroup$

You can substitute almost random values for $c_1$ and $c_2$. You need to make sure that the two solutions you come up with are independent, which will be true unless you are very unlucky or skillful.

As best I can tell, the second general solution will be to add a new pair of multipliers times your new solutions. So if you have $f(x)=ae^{2x}+be^{-x}$ and something similar for $g(x)$, you new general solution will be $a'f(x)+b'g(x)$. Of course, this is the same as your original general solution, just expressed differently.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .