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 [tex]y = 3*e^{-5x}[/tex] → [tex]y' = \frac{dy}{dx} = -5*3*e^{-5x} = -5y[/tex]
So it's true!
where i used the fact that if : [tex]f(x) = e^{a*x} ------> df/dx = a* e^{a*x}[/tex] where a is any number.
b) y' = cos(3x); y = į sin(3x) + 7
if y = į sin(3x) + 7 → [tex]\frac{dy}{dx} = 3*j*cos(3x) + 0 = j*3cos(3x)[/tex]
the equality is only true if j = 1/3.
c) y' = 2y; y = ce2x
if y = [tex]c*e^{2x}[/tex] → [tex]\frac{dy}{dx} = 2*c*e^{2x} = 2*(c*e^{2x}) = 2*y[/tex]
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) → [tex]\frac{dy}{dx} = -4*sin(4x) --> \frac{d^{2}y }{dx^{2} } = -16*cos(4x)[/tex]
so y'' + 16y = -16cos(4x) + 16cos(4x) = 0
so y = cos(4x) is a solution
if y = sin(4x) →[tex]\frac{dy}{dx} = 4*cos(4x) --> \frac{d^{2}y }{dx^{2} } = -16*sin(4x)[/tex]
then: y'' + 16y = -16sin(4x) + 16sin(4x) = 0
So again; y = sin(4x) is a solution.
