Homogeneous Linear ODE with complex roots

[DryRun]

Homework Statement

I'm trying to understand the simplification of the general solution for homogeneous linear ODE with complex roots.

Homework 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

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<br /> \\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}<br /> \\=C_1e^{-t}\cos t +C_1e^{-t}i\sin t+C_2e^{-t}\cos t -C_2e^{-t}i\sin t<br /> \\=(C_1+C_2)e^{-t}\cos t + (C_1-C_2)e^{-t}i\sin twhere, 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.

 

[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?

 

[DryRun]

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?

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.

 

[pasmith]

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

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).

 

[STEMucator]

Usually what you do is create two new functions, let's 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

 

[DryRun]

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).

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?

Usually what you do is create two new functions, let's 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

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?

 

[pasmith]

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?

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

 

[DryRun]

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 $B = -2b$