1. ## differential equation

a)for what values of r does the function $\displaystyle y=e^{rx}$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$

b)If $\displaystyle r_1$ and $\displaystyle r_2$ are the values of r the were found in part a, show that every member of the family of functions is $\displaystyle y=a e^{r_1x} +b e^{r_2 x}$ is also a solution.

so I took the first and second derivative getting $\displaystyle y'=re^{rx}$ and $\displaystyle y''=r^2 e^{rx}$.

Therefore:

$\displaystyle 2r^2e^{rx} + re^{rx} + e^{rx} = 0$

$\displaystyle \Rightarrow e^{rx} (2r^2+r+1) = 0$

I can't factor that any farther so I don't know how to find the roots.

2. a)for what values of r does the function $\displaystyle y=e^(rx)$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$
Try solving $\displaystyle 2m^{2}+m-1=0$

3. Originally Posted by superdude
a)for what values of r does the function $\displaystyle y=e^{rx}$ satisfy the differential equation $\displaystyle 2y''+y'-y=0$

b)If $\displaystyle r_1$ and $\displaystyle r_2$ are the values of r the were found in part a, show that every member of the family of functions is $\displaystyle y=a e^{r_1x} +b e^{r_2 x}$ is also a solution.

so I took the first and second derivative getting $\displaystyle y'=re^{rx}$ and $\displaystyle y''=r^2 e^{rx}$.

Therefore:

$\displaystyle 2r^2e^{rx} + re^{rx} {\color{red}+} e^{rx} = 0$

$\displaystyle \Rightarrow e^{rx} (2r^2+r {\color{red}+} 1) = 0$

I can't factor that any farther so I don't know how to find the roots.
Very careless. The red plus is clearly wrong.

4. Since $\displaystyle e^{rx}$ can never be zero, $\displaystyle 2r^{2}+r-1$ must be zero.