# [SOLVED]ODE resonance

#### dwsmith

##### Well-known member
$y''+y=\alpha\cos x + \cos^3x$

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

##### Well-known member
$y''+y=\alpha\cos x + \cos^3x$

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

Last edited:

#### dwsmith

##### Well-known member
[HR][/HR]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

##### Well-known member
[HR][/HR]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

#### dwsmith

##### Well-known member
So $\alpha = -\dfrac{3}{4}$.