Hi,

I just started working second order DE's, and I want to make sure I'm doing it correctly. Do you see any problem with the process below? Sorry that it looks so messy, I don't know how to code.

Question:

y'' + 12y = 0By definition,

y'' + 12y= y'' + 0y' + 12y = (r^2) + 0r + 7 = 0

First I found the discriminate, given by the formula D = b^2 - 4(a)(c) = 0^2 - 4(1)(12). Since D is negative, there are no real roots.

Now I can find r_1 and r_2 by the equation r_1,2 = ( -b (plus/minus) sqrt(D) )/2(a) = (-0 (plus/minus) sqrt(-48) )/2

r_1 = (-sqrt(48))/2 = -i(sqrt(48)/2)

r_2 = (sqrt(48))/2 = i(sqrt(48)/2)

The general formula can be expressed as y(x) = C_1( e^(r_1*x) ) + C_2( e^(r_2*x) ) = C_1(e^(-i(sqrt(48)/2)x) + C_2(e^(i(sqrt(48)/2)x)

Now to remove the imaginary units I can use the equation e^ix = cos(x) + i*sin(x) (from here on I am not sure if I did correctly)

y(x) = C_1(cos(sqrt(48/2)x) + -i*sin(sqrt(48/2)x) + C_2(cos((sqrt(48/2)x) + i*sin(sqrt(48/2)x)

And multiplying out...

y(x) = C_1(cos(sqrt(48/2))x) -i*C_1(sin(sqrt(48/2))x) + C_2(cos(sqrt(48/2))x) + i*C_2(sin(sqrt(48/2))x)

Now we can denote A = C_1 + C_2 and B = i*C_2 - i*C_1 to get...

y(x) = Acos(sqrt(48/2)x) + Bsin(sqrt(48/2))x)

That's my final answer.

Again, I apologize that it is hard to read, but ANY help is MUCH appreciated!!!

If anyone could explain to me how using denotations like A and B make the imaginary part of the complex numbers go away (basically, why the last step is done), that would be helpful.