MHB Find $\alpha$ to Make ODE Resonance Free

[Dustinsfl]

$y''+y=\alpha\cos x + \cos^3x$

What value of $\alpha$ makes this resonance free?

$\cos^3 x = \frac{1}{4}\cos 3x+\frac{3}{4}\cos x$

So $y''+y=(\alpha+\frac{3}{4})\cos x + \frac{1}{4}\cos 3x$

What am I supposed to do to find alpha?

 

[CaptainBlack]

$y''+y=\alpha\cos x + \cos^3x$

What value of $\alpha$ makes this resonance free?

$\cos^3 x = \frac{1}{4}\cos 3x+\frac{3}{4}\cos x$

So $y''+y=(\alpha+\frac{3}{4})\cos x + \frac{1}{4}\cos 3x$

What am I supposed to do to find alpha?

To make this resonance free you need that the forcing term on the right does not contain a component at a natural frequency of the homogeneous equation. The natural angular frequencies of the homogeneous equation are the (imaginary part of the) roots of the characteristic equation, which here are \(\pm 1\), so you need the coefficient of \(\cos(x)\) to be zero.

You can see the resonance in the contribution of the \(e^it\) component of the forcing in your other ODE thread where the resonance term grows with time.

CB

 

[Dustinsfl]

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The roots should be $\pm i$ but $|\pm i | = 1$ still.

We want to remove the $\cos x$. Is that, because if not, we would have terms $\cos x$ and $x\cos x$ in the solution? Therefore, the variable coefficient we cause the resonance?

 

[CaptainBlack]

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The roots should be $\pm i$ but $|\pm i | = 1$ still.

We want to remove the $\cos x$. Is that, because if not, we would have terms $\cos x$ and $x\cos x$ in the solution? Therefore, the variable coefficient we cause the resonance?

Only the \(x \cos(x)\) term is a problem as this is a oscillatory term with amplitude that grows without bound.

CB

 

[Dustinsfl]

So $\alpha = -\dfrac{3}{4}$.

 

[CaptainBlack]

So $\alpha = -\dfrac{3}{4}$.

Yes.

CB