General solution of $y''+y'=0$

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?

• Both are right. Arbitrary constant means arbitrary constant. – André Nicolas Jul 11 '11 at 13:42
• Be careful with signs-- the equation you typed into Wolfram Alpha is y''+y'=0, and your question is about y''-y'=0. – Jonas Kibelbek Jul 11 '11 at 13:44

Both are correct.The only difference is the sign of the constant $c_1$, which does not matter once you plug in your initial conditions.

You can of course verify that both are correct solutions. Let's call the solutions $y_1$ and $y_2$: \begin{align} y_1 = c_2 - c_1 e^{-x} \\ y_2 = d_2 + d_1 e^{-x} \end{align}

Now you get: $$y_1'' + y_1' = -c_1e^{-x} + c_1e^{-x} = 0$$ and $$y_2'' + y_2' = d_1e^{-x} - d_1e^{-x} = 0$$

• oh, that makes an incredible amount of sense. Thanks a bunch! – Fred W. Jul 11 '11 at 14:00

HINT $\rm\quad \langle 1,\:e^{-x}\rangle\ \cong\: \langle 1, -e^{-x}\rangle\$ as vector spaces.

$y''+y'=0 \\ y'=t(x) \ \to y''=t'(x) \\ t'+t=0 \\ \frac{dt}{t}+dx=0 \\ \ln |t|+x=C \\ \ln|t| + \ln e^x = \ln e^{C} \\ t \cdot e^x=C_1 \\ y' \cdot e^x =C_1 \\ dy=C_1\frac{dx}{e^x} \\ \int dy=C_1 \int \frac{dx}{e^x} \\ ...$

so see from your definition that if $b^2-4ac$ is more than zero, let's check it,

$y^{''}-y'=0$ is in form $k^2-k=0$ yes? so $k(k-1)=0$ from here there is two chance $k=0$ or $k=1$ or use $r_1$ and $r_2$ as you like so $r_1=0$ and $r_2=1$ and put them in general solution