Question

If r_1 and r_2 are the values of r. Show that every member of the family of functions
\[y=ae^{xr_1}+be^{xr_2}\] is also a solution.

Answer 1

r_1=.5 r_2=-1

Answer 2

did u try to solve it

Answer 3

solution for what...?

Answer 4

the first part of the question reads
For what values of r does the function y=e^(rx) satisfy the differential equation
2y" + y' -y =0?

Answer 5

so I solved for y and got -1 and .5

Answer 6

solved for r ...sorry

Answer 7

its ok
for the second part just take the first and second derivative of given function and put them in the original equation

Answer 8

sorry i lost my connection

Answer 9

u want to show that
\[y=ae^{r_{1}x}+be^{r_{2}x} \ \ \ \ \ (I)\]
is also a solution
for equation
2y" + y' -y =0 (original equation)

compute y'' and y' from (I) and put them in the 2y" + y' -y =0
see what happens
and we know that e^(r1x) and e^(r2x) are solutions u should use this to show that (I) satisfies original equation