Second Order Differential Equation (Can you check my work?)

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 = 0*

By 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.

Re: Second Order Differential Equation (Can you check my work?)

Your answer is right but i'm not sure what you are doing with the imaginary parts.

First hint, if you have an eqn of the form $\displaystyle f(y,y'')$ then observe that $\displaystyle \frac{d^2y}{dx^2} sin(x) = -sin(x) $, so try a soln. y = sin(ax), then $\displaystyle y'' = -a^2sin(x) $ so in this casae you have $\displaystyle -a^2sin(x) + 12sin(ax) = 0 $ or solve for $\displaystyle -a^2 + 12 = 0 $ to get $\displaystyle a = \sqrt{12} $.

Reverse signs and changing sin(ax) to cos(ax) will also give the soln. so you have a gen soln of type

$\displaystyle y = asin(\sqrt{12}) - b (\sqrt{12}) $.

Now we could do this your method as well, but ur method is cumbersome, and usually should be used for eqn of the form f(y'',y',y) = 0.

To see why we do that weird thing to get rid of the imaginary parts. If we solve the auxillary eqn to get $\displaystyle e^{i\sqrt{12}} $ and $\displaystyle e^{-i\sqrt{12}} $.

$\displaystyle e^{a + ib} = e^{a}(cos(a) + isin(b)$ Since sin(-x) = -sin(x). Note that $\displaystyle e^{a - ib} = e^{a}(cos(a) - isin(b) $ so since these two are linearly independent solutions, adding them together, $\displaystyle e^{a + ib} + e^{a - ib} = e^{a}cos(a) + e^{a}cos(a) = 2e^{a}cos(a) $

Now as i said above, if cos(a) solves a linear equation, then -sin(a) also solves it, (or vice versa). To show that $\displaystyle 2e^{a}cos(a)$ and $\displaystyle -2e^{a}sin(a)$ are linearly iandependent, i can take the wronskian determinant of them. (If you haven't heard of this, go to Wronskian - Wikipedia, the free encyclopedia) $\displaystyle \begin{bmatrix} 2e^{a}cos{a} & -2e^{a}sin(a) \\ -2e^{a}sin(a) & -2e^{a}cos(a) \end{bmatrix} $ taking the determinant of this matrix, we get $\displaystyle -4e^{2a}(cos^{2}(a) + sin^{2}(a) = -4e^{2a} $ which does not equal 0 for any a. So these two are linearly independent and thus,

our general solution is of the form y = acos(a) - bsin(a).