Answer: So, here we need to verify that the given functions are solutions for the given differential equations, this is:
a) y' = -5y; y = 3e-5x
if
→ 
So it's true!
where i used the fact that if :
where a is any number.
b) y' = cos(3x); y = į sin(3x) + 7
if y = į sin(3x) + 7 → 
the equality is only true if j = 1/3.
c) y' = 2y; y = ce2x
if y =
→ 
So the function is a solution for the differential equation.
d) y" + y' – 6y = 0 ; yı = (2x), y2 = (–3x )
Here we have two functions to test; is easy to se that in both cases y'' = 0, because both are linear functions, so we need to solve: y' - 6y = 0
if the functions are the 2x and -3x, then this never will be true, because when you derive y with respect of x you only will get a constant (y1' = 2 and y2' = -3), and the difference y' - 6y = 0 (2 - 12x = 0 for the first function) will only be true for some value of x, if the functions are wrong.
e) y" + 16y = 0; yı = cos(4x), y2 = sin(4x)
if y = cos(4x) → 
so y'' + 16y = -16cos(4x) + 16cos(4x) = 0
so y = cos(4x) is a solution
if y = sin(4x) →
then: y'' + 16y = -16sin(4x) + 16sin(4x) = 0
So again; y = sin(4x) is a solution.