MHB Are There Complex Solutions for a 2nd Order ODE?

[ognik]

Hi - I have y''+9y=0 which is a constant coefficients ODE, the CE is then $r^2+9=0$ and I get a general solution $ y=C_1e^{3ix}+C_2e^{-3ix} $

But I have seem these solutions written as ACos3x+BSin3x. If I use Euler on my solution, I get $ C_1(Cos3x + iSin3x) +C_2(Cos3x-iSin3x) $ ... Are they just using $ A=C_1+C_2 $ and $ B=i(C_1-C_2) $ or am I missing something?

 

[MarkFL]

Suppose we are given the ODE:

$$ay''+by'+cy=0\tag{1}$$

Let $z(x)=u(x)+iv(x)$ be a solution to (1), where $a$, $b$, and $c$ are real numbers. Then the real part $u(x)$ and the imaginary part $v(x)$ are real valued solutions of (1).

Proof:

By assumption $az''+bz'+cz=0$, and hence:

$$a(u''+iv'')+b(u'+uv')+c(u+iv)=0$$

Which we can arrange as:

$$(au''+bu'+cu)+i(av''+bv'+cv)=0$$

A complex number is zero iff both its real and imaginary parts are zero. Thus, we must have:

$$au''+bu'+cu=0$$

$$av''+bv'+cv=0$$

And this means that both $u(x)$ and $v(x)$ are real-valued solutions of (1).

 

[ognik]

Thanks, very clear. Question - A 2nd order ODE has at most 2 solutions, so are u an v those 2 solutions, or are there other complex solutions (besides u+iv)?

 

[Prove It]

Hi - I have y''+9y=0 which is a constant coefficients ODE, the CE is then $r^2+9=0$ and I get a general solution $ y=C_1e^{3ix}+C_2e^{-3ix} $

But I have seem these solutions written as ACos3x+BSin3x. If I use Euler on my solution, I get $ C_1(Cos3x + iSin3x) +C_2(Cos3x-iSin3x) $ ... Are they just using $ A=C_1+C_2 $ and $ B=i(C_1-C_2) $ or am I missing something?

Yes that is exactly what they are doing. Even complex constants are still constants.

 

[MarkFL]

Thanks, very clear. Question - A 2nd order ODE has at most 2 solutions, so are u an v those 2 solutions, or are there other complex solutions (besides u=iv)?

A 2nd order ODE will have a general 2 parameter family of solutions. For example, consider the homogeneous ODE:

$$ay''+by'+cy=0$$

Let $r_1$ and $r_2$ be the roots of the associated characteristic equation. Then, we know, by the principle of superposition that the general solution is:

$$y(x)=c_1e^{r_1x}+c_2e^{r_2x}$$

Since there are an infinite number of ways we can choose the two parameters, there are an infinite number of solutions. What we do, is identify 2 linearly independent solutions using the roots of the characteristic equation, and then affix a parameter to each to get the general solution.