# Homogeneous Linear ODE with complex roots

1. Dec 5, 2012

### sharks

1. The problem statement, all variables and given/known data
I'm trying to understand the simplification of the general solution for homogeneous linear ODE with complex roots.

2. Relevant equations
In my notes, i have the homogeneous solution given as:
$$y_h (t)= C_1 e^{(-1+i)t}+C_2e^{(-1-i)t}$$
And the simplified solution is given as:
$$y_h (t)= A e^{-t}\cos t+Be^{-t}\sin t$$

3. The attempt at a solution
First, using Euler's formula, then I expanded each part individually before summing them all up:
$$C_1 e^{(-1+i)t}=C_1(e^{-t}(\cos t +i\sin t))=C_1e^{-t}\cos t +C_1e^{-t}i\sin t \\C_2 e^{(-1-i)t}=C_1(e^{-t}(\cos t -i\sin t))=C_2e^{-t}\cos t -C_2e^{-t}i\sin t$$
Now, adding these up, i just do not understand how the imaginary terms lose the "i" along the way. Can someone please clarify this part?

For the sake of completion, adding them up, i get:
$$C_1 e^{(-1+i)t}+C_2e^{(-1-i)t} \\=C_1e^{-t}\cos t +C_1e^{-t}i\sin t+C_2e^{-t}\cos t -C_2e^{-t}i\sin t \\=(C_1+C_2)e^{-t}\cos t + (C_1-C_2)e^{-t}i\sin t$$where, $A = (C_1+C_2)$ and $B=(C_1-C_2)$. However, the "i" coefficient of the sine term should not be there, according to the answer.

Last edited: Dec 5, 2012
2. Dec 5, 2012

### pasmith

You have assumed that $C_1$ and $C_2$ are real. They can be complex.

Let $C_1 = a + ib$ and $C_2 = c + id$ for real a, b, c, and d. What conditions must you impose so that $i(C_1-C_2)$ and $C_1 + C_2$ are real?

3. Dec 5, 2012

### sharks

Hi pasmith

The conditions for $i(C_1-C_2)$ and $C_1 + C_2$ to be real would be:
In the first case: a = c
and in the second case: b = -d

The combination of those 2 conditions would give, in the first case, $i^22b$ or $-i^22d$, meaning, -2b or 2d, and in the second case, 2a or 2c.

Is that correct? I'm not sure that i grasp the meaning of it all though. I have not been given any initial conditions, otherwise i could have tested the validity of your suggestion.

Last edited: Dec 5, 2012
4. Dec 5, 2012

### pasmith

Do you require $y_h(t)$ to be real? If so, the above shows that you must take $C_2 = \bar C_1$, the complex conjugate of $C_1$.

If you don't require $y_h(t)$ to be real, then you have simply $A = C_1 + C_2$ and $B = i(C_1 - C_2)$.

5. Dec 5, 2012

### Zondrina

Usually what you do is create two new functions, lets say u(t) and v(t).

Let u(t) = y1 + y2 and v(t) = y1 - y2

You should also drop any multiplicative scalars you get and then your final real valued solution would be : y = c1u(t)+ c2v(t) for arbitrary constants c1 and c2

6. Dec 5, 2012

### sharks

From what i understood, $y_h(t)$ means the homogeneous solution of the ODE, which indicates that the auxilliary equation must be equal to 0. Therefore, the free term or forcing function must be 0. There is no mention of a real or imaginary form of $y_h(t)$ anywhere in my notes, so i would assume that $y_h(t)$ has to be real?

In that case, if $C_1 = a + ib$, then $C_2 = a - ib$, meaning that $A = 2a$ and $B = -i2b$

Continuing from the first post,
$$(C_1+C_2)e^{-t}\cos t + (C_1-C_2)e^{-t}i\sin t \\=(2a)e^{-t}\cos t + (-i2b)e^{-t}i\sin t \\=(2a)e^{-t}\cos t + (2b)e^{-t}\sin t$$
Is that correct? Is there anything missing at this point?

I have no idea what you mean by those 2 new functions. How is that related and used to solve the problem in the first post?

7. Dec 5, 2012

### pasmith

Close enough, but $C_1 - C_2 = 2ib$ so $i(C_1 - C_2) = -2b$.

8. Dec 5, 2012

### sharks

OK, then in the simplified form of the homogeneous solution: $$y_h (t)= A e^{-t}\cos t+Be^{-t}\sin t$$where, $A = 2a$ and $B = -2b$