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Plugging the equation $y''+y'=0$ into Wolfram Alpha yields the following solution:

$$y(x) = c_2-c_1 e^{-x}.$$

This has me stumped because my textbook states that in the case of $b^2-4ac > 0$.

"If root 1 $r_1$ and root 2 $r_2$ are two real and unequal roots to the auxiliary equation $ar^2 + br + c =0$, then

$$y = c_1e^{r_1x} + c_2e^{r_2x}$$

is the general solution."

Based of my book, the solution to the above problem would be

$$y=c_2+c_1 e^{-x}$$

so which one's right?

  • 7
    Both are right. Arbitrary constant means arbitrary constant.2011-07-11
  • 0
    Be careful with signs-- the equation you typed into Wolfram Alpha is y''+y'=0, and your question is about y''-y'=0.2011-07-11

4 Answers 4