Differential Equations [Show every member is a solution]

1. The problem statement, all variables and given/known data
PART A: For what values of r does the function y= e^ (rx) satisfy the differential equation 2y" + y' - y = 0?

PART B: If r1 and r2 are the values of r that you found in part A, show that every member of the family of functions y=ae^(r1*x) + be^(r2*x) is also a solution


2. Relevant equations
derivatives, factoring.


3. The attempt at a solution

I got part A, by taking the 1st and 2nd derivative of y and plugging into the equation and factoring to obtain r= -1 and r =1/2.

but what I'm really stuck on is how to show the answer for PART B?

Do you just plug in your r1 & r2 values from part a?

Please Help.

You should have gotten two specific values for r₁ and r₂, one of which happens to be a fraction. Show that the function y = ae^{r_1 x} + be^{r_2 x} is a solution to your differential equation. That's it.

I did get 2 answers, r= -1 and r=1/2. So now just plug those into the equation in part b ?

Yes. Then for that function, show that 2y'' + y' - y = 0.

Yes. Then for that function, show that 2y'' + y' - y = 0.

When I Plugged back into 2y'' + y' -y = 0 I get 2ae^-x+1/2be^1/2x -ae^-x +1/2be^1/2x -ae^-x -be^1/2x=0

and theres nothing to cancel out ?

Help?

2ae^-x+1/2be^1/2x -ae^-x +1/2be^1/2x -ae^-x -be^1/2x = e^-x(2a - a - a) + e^.5x(.5b + .5b - b) = ?