Verify that the function(s) solve the following differential equations (des): a) y' = -5y; y = 3e-5x b) y' = cos(3x); y = į sin(3x) + 7 c) y' = 2y; y = ce2x , where c is any real number. d) y" + y' – 6y = 0 ; yı = (2x, y2 = (–3x e) y" + 16y = 0; yı = cos(4x), y2 = sin(4x)

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.

3x3 = 9 and 2x2 = 4

step-by-step explanation:

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following