Solve the differential equation in the case where there is no resonance.

Question

Problem:

$$y'' + 20y = 100sin(wt)$$

Initial Condition: $$y=0$$ when $$t=0$$

Attempt:

I know $w = 2\sqrt{5}$ when resonance occurs.

But what do you mean by "no resonance occurs"

Thank you

Answer 1

The general solution of homogeneous equation is $y_h=A\sin(w_0t+\phi)$ with $w_0=2\sqrt{5}$ and $A,\phi$ some parameters.

Now let's search a particular solution in the case $w\neq w_0$.

We can hope to find it under the form $y_p=B\sin(wt)$.

$y_p''+20y_p=-Bw^2\sin(wt)+20B\sin(wt)=100\sin(wt)\iff-Bw^2+20B=100$

$y_p=\frac{100}{20-w^2}\sin(wt)$

Now you have your final solution $y(t)=y_h+y_p=A\sin(w_0t+\phi)+\frac{100}{20-w^2}\sin(wt)$

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You can now apply initial conditions : $y(0)=0=A\sin(\phi)$

$A=0$ is possible, but this would be very reductive. So I guess $\sin(\phi)=0\iff \phi=k\pi$ is more appropriate.

$A\sin(w_0t+k\pi)=\pm A\sin(w_0t)$ but since $A$ is a dummy parameter, $A$ or $\pm A$ means the same.

Finally $y(t)=A\sin(w_0t)+\frac{100}{20-w^2}\sin(wt)$

We do not loose in generality setting $\phi$ to $0$, because it is still possible to set $A=0$, but an additional initial condition would be required to be absolutely certain of the value of $A$ anyway.

Answer 2

Observe that a solution of your ODE is the imaginary part of the solution of the (complex) ODE $$y''(t)+20y(t)=100e^{i\omega t}.$$ If $\omega$ is such that $(i\omega)^2+20\neq 0$, which means $\omega\neq 2\sqrt{5}$ then $$y_p(t)=\frac{100}{P(i\omega)}e^{i\omega t}=\frac{100}{-\omega^2+20}e^{i\omega t},$$ where $P(z)=z^2+20$ is the to the ODE associated polynomial, is a particular solution. Now just take the imaginary part of this and adjust it to the initial value problem.

The general solution is given by the general one (which can be found in the same way) addided to the particular solution.