Using method of annihilators to solve $y''+2y'=1+4\sin(2t)$

Question

I am trying to solve $$y''+2y'=1+4\sin(2t)$$ using the method of annihilators. I have that $D^3 + 4D$ is the annihilator for the right hand side (where $D$ is the derivative operator), and on the left side you take this annihilator and multiply it by $(D^2+2D)y$. I have the general solution to the homogeneous equation, but I'm not sure how to use the annihilator method.

Answer 1

HINT

Here is an alternative approach. First of all, I would integrate the given ODE to get: \begin{align*} y^{\prime\prime} + 2y^{\prime} = 1 + 4\sin(2t) \Longleftrightarrow y^{\prime} + 2y = t - 2\cos(2t) + c \end{align*}

Now, we can multiply such equation by the integration factor $e^{2t}$ to obtain: \begin{align*} e^{2t}y^{\prime} + 2e^{2t}y = e^{2t}[t - 2\cos(2t) + c] \Longleftrightarrow (e^{2t}y)^{\prime} = e^{2t}[t - 2\cos(2t) + c] \end{align*}

Can you proceed from here?

Answer 2

For the homogeneous solution we have $$D^2+2D=D(D+2) \implies y_h=c_1+c_2e^{-2t}$$ Then for the particular solution we have $$(D^2+2D)(D^3+4D)=D^2(D+2)(D^2+4)$$ $$\implies \color{blue}{y_h}+y_p=\color{blue}{c_1+c_3e^{-2t}}+c_2t+c_4\cos(2t)+c_5\sin(2t)$$ $$y_p=\color{green}{c_2t+c_4\cos(2t)+c_5\sin(2t)}$$ Now you plug the particular solution in the equation and find the constants $c_2,c_4,c_5$

Answer 3

$P(D)=D^2 + 2D\\ P(D)y = 1 + 4\sin 2t\\ A(D) = D^3 + 4D\\ A(D)P(D)y = A(D) (1+4\sin 2t) = 0$

$A(D)P(D) = (D + 2)(D^2)(D^2+4)$

The solutions to the homogeneous equation is a vector space $\{y_1,y_2,y_3,y_4,y_5\} = \{e^{-2t},1, t, \cos 2t, \sin 2t\}$

$y = c_1y_1 + c_2y_2 + c_3y_3 + c_4y_4 + c_5y_5%$

$P(D)y_1= 0\\ P(D)y_2 = 0\\ P(D)y_3 = 2 = 2y_2\\ P(D)y_4 = \frac{d^2}{dt^2}\cos 2t + 2\frac {d}{dt} \cos 2t = -4\cos 2t - 4\sin 2t = -4y_4 - 4y_5\\ P(D)y_5 = \frac{d^2}{dt^2}\sin 2t + 2\frac {d}{dt} \sin 2t = -4\sin 2t + 4\cos 2t = 4y_4 - 4y_5$

$P(D)y = c_3y_2 + (-4c_4+4c_5)y_4 + (-4c_4 - 4c_5)y_5 = y_2 + 4y_5\\ -4c_4+4c_5 = 0\\ -4c_4 - 4c_5 = 4\\ c_4 = c_5 = -\frac 12$

$y = c_1e^{-2t} + c_2 + \frac 12 t - \frac 12 \cos 2t - \frac 12 \sin 2t$