Solution: Solving Harmonic Oscillation Differential Equation

[Runei]

Hello,

When I have the differential equation

$$\frac{dY(x)}{dx} = -k^2 Y(x)$$

The solution is of course harmonic oscillation, however, looking at various places I see the solution given as:
$$Y(x) = A cos(kx) + B sin(kx)$$
instead of
$$Y(x) = A cos(kx + \phi_1) + B sin(kx + \phi_2)$$

Isnt Equation 2 a more general solution than Equation 1? Or is there some reasoning (probably is) to make the phase angles go away?

Thank you.

 

[AlephZero]

The general solution to a second order differential equation has 2 arbitrary coefficients, so you can't have a "more general solution" than $A \cos kx + B \sin kx$.

For your second solution, you have $\cos(kx + \phi_1) = \cos\phi_1\cos kx - \sin\phi_1\sin kx$, and a similar expression for $\sin(kx + \phi_2)$, so you can rewrite the whole expression as $P\cos kx + Q\sin kx$, where $P$ and $Q$ are constants containing $A$, $B$, and $\cos$ and $\sin$ of $\phi_1$ and $\phi_2$. That is the same as your first solution.

Note, $A\cos(kx + \phi)$ or $A\sin(kx + \phi)$ are both general solutions (with two arbitrary constants), and those forms are sometimes nicer to use than $A \cos kx + B \sin kx$.

 

[Runei]

Thank you AlephZero!

As a side-note then, if I wanted to rewrite the solution in terms of complex exponentials, that the solution would be

$Y(x) = A \cdot e^{ikx} + A^*\cdot e^{-ikx}$

Where the constant A this time is complex.

 

[JJacquelin]

Hello,

When I have the differential equation

$$\frac{dY(x)}{dx} = -k^2 Y(x)$$

The solution is of course harmonic oscillation, however, looking at various places I see the solution given as:

Isnt Equation 2 a more general solution than Equation 1? Or is there some reasoning (probably is) to make the phase angles go away?

Thank you.

I suppose that the equation is $$\frac{d^2Y(x)}{dx^2} = -k^2 Y(x)$$
The two equations :
$$Y(x) = A cos(kx) + B sin(kx)\\<br /> Y(x) = A cos(kx + \phi_1) + B sin(kx + \phi_2)$$
are equivalent :
$$A cos(kx + \phi_1) + B sin(kx + \phi_2) = A' cos(kx) + B' sin(kx)\\<br /> A' =A cos(\phi_1)+B sin(\phi_2)\\<br /> B' =-A sin(\phi_1)+B cos(\phi_2)$$

 

[AlephZero]

Thank you AlephZero!

As a side-note then, if I wanted to rewrite the solution in terms of complex exponentials, that the solution would be

$Y(x) = A \cdot e^{ikx} + A^*\cdot e^{-ikx}$

Where the constant A this time is complex.

There is no mathematical reason why the solution has to be real, so you could just write
$Y(x) = A \cdot e^{ikx} + B\cdot e^{-ikx}$
where $A$ and $B$ are complex constants.

In this case, both the real and imaginary parts of $Y(x)$ satisfy the differential equation.
You can interpret the real part of $Y(x)$ as a physical displacement, and the real part of $dY(x)/dx$ as a physical velocity, etc.