Quick linear equation question

[rygza]

So i have

(r^4)-16 = 0
r^4 = 16
r = 2
So i only have one root

As the general solution i have

y = C(sub1) * (e^(2x)) + C(sub2) * (e^(2x))

Is this correct? If not please lead me in the right direction

 

[hunt_mat]

Can you write down the original equation that you were trying to solve.

Mat

 

[rygza]

Can you write down the original equation that you were trying to solve.

Mat

((D^4)-16)*(y) = 0

This is in operator notation, from which i got the characteristic equation ((r^4)-16) = 0. The "r" is actually a lambda but i don't have that on my keyboard.

 

[hunt_mat]

Have you tried a route of unity approach? write $$1=\exp (2\pi i)$$ and see if you get any other solutions? The other suggestion is look at solutions of the form $$xe^{2x},x^{2}e^{2x},x^{3}e^{2x}$$ and see if they solve your equation at all.

 

[HallsofIvy]

So i have

(r^4)-16 = 0
r^4 = 16
r = 2
So i only have one root

As the general solution i have

y = C(sub1) * (e^(2x)) + C(sub2) * (e^(2x))

Is this correct? If not please lead me in the right direction

If you are going to do differential equations, you will really need to remember your basic algebra!
r^4- 16= 0 has four roots.

r^4- 16= (r^2- 4)(r^2+ 4)= (r- 2)(r+ 2)(r- 2i)(r+ 2i)

 

[rygza]

If you are going to do differential equations, you will really need to remember your basic algebra!
r^4- 16= 0 has four roots.

r^4- 16= (r^2- 4)(r^2+ 4)= (r- 2)(r+ 2)(r- 2i)(r+ 2i)

Oh yeeaa. I should have known it had four roots because the r is to the fourth power.

so here's what i have:

y = [C(sub1)cos(2x) + C(sub2)sin(2x)] + C(sub3)e^(-2x) + C(sub4)e^(2x)