I found $$r_{1}$$ and $$r_{2}$$ to be -1 and 1/2 respectively. Show that every member of the family of functions $$y=ae^{r_{1}x}+be^{r_{2}x}$$ is also a solution

Question

anonymous · Answer

solution for what?

anonymous · Answer

sorry here is the first part of the question that I solved:
$$y=e^{rx}$$
$$2y"+y'-y=0$$

anonymous · Answer

first question was for what values of r does the function satisfy the differential equation

anonymous · Answer

take first and second derivatives of 
$$y=ae^{r_{1}x}+be^{r_{2}x}$$
and put them in the equation u have
what will u get?

anonymous · Answer

oh

anonymous · Answer

$$y'=.5ae^{.5x} - be^{-x}$$
$$y''=.25ae^{.5x}+be^{-x}$$
it should all cancel...i made some algebra error somewhere but I'll figure it out

anonymous · Answer

thanks!

anonymous · Answer

your welcome

anonymous · Answer

negative signs...I tell ya! got it solved finally