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!

  • $\begingroup$ 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 '12 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 $e^{r_2x}$. $\endgroup$ – Gerry Myerson Oct 4 '12 at 7:50

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.


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