My answer to this problem does not agree with the text's answer: $y=sin(t)-3cos(t)$
Problem to solve is to find the steady state oscillations of the vibrating system governed by the following equation:
$$y''+3y'+2y=10sin(t)$$
$$(my''+cy'+ky=r(x))$$
The theory and steps for solving this problem are given in this link on Chapter 2 Section 8 entitled 2.8 Modeling: Forced Oscillations. Resonance. The problem that I need to solve is from the same text but from a much earlier edition hence you will not find the same problem in the link given. There is also an instructor's solution manual for this very same text but again the problem I need solved will not be found here either. I thought this section in the text was poorly written in that it was very confusing and gave no concrete examples.
I shall solve this as a regular non-homogeneous linear differential equation by method of Undetermined Coefficients just to demonstrate how I arrived at my answer.
From the characteristic equation I get the following homogeneous solution:
$$y_h=c_1e^{-t}+c_2e^{-2t}$$
The following steps document my process for solving for the particular solution:
$$y_p=Acos(t)+Bsin(t)$$
$$y_p'=-Asin(t)+Bcos(t)$$
$$y_p''=-Acos(t)-Bsin(t)$$
$$(-Acos(t)-Bsin(t))+3(-Asin(t)+Bcos(t))+2(Acos(t)+Bsin(t))=10sin(t)$$
$$-3A+B=10$$
$$A+3B=0$$
$$A=0, B=10$$
My final answer is:
$$y=c_1e^{-t}+c_2e^{-2t}+10cos(t)$$
I am aware that this is not the solution i am trying to find. Perhaps someone can look up the text and guide me by pointing out which direction I need to go.
Best Answer
The solution of the homogeneous equation doesn't change. Basically you have to set $r(x)$ to zero and solve the equation $y''+3y'+2y=0$. The solution can be found by replacing $y_h$ and its derivative into the homogeneous differential equation. When doing that you get the homogeneous solution you wrote above. Now the last thing to do consists in finding the particular integral. First of all we write the solution of the homogeneous equation as $y_h=c_1 y_1(t)+c_2 y_2(t)$ where $y_1(t)=e^{-t}$ and $y_2(t)=e^{-2t}$. We now look for a particular integral in the form: \begin{equation} y_p(t)= f_1(t) y_1(t)+f_2(t) y_2(t) \end{equation} subject to the condition (choosen for convenience) \begin{equation} f^{'}_{1}(t) y_1(t)+ f^{'}_{2}(t) y_2(t)=0 \end{equation} Now differentiatin $y_p$ we have: \begin{equation} y^{'}_p=f^{'}_{1}(t) y_1(t)+ f_1(t) y^{'}_{1}(t)+f^{'}_{2}(t) y_2(t)+ f_2(t) y^{'}_{2}(t) \end{equation} which reduces to \begin{equation} y^{'}_p= f_1(t) y^{'}_{1}(t)+ f_2(t) y^{'}_{2}(t) \end{equation} you can now compute $y^{"}_p$ and replace the $y^{'}_p$ and $y^{"}_p$ in the original differential equation. Remember to use the equation $f^{'}_{1}(t) y_1(t)+ f^{'}_{2}(t) y_2(t)=0$ and differentiate it again. When doing that you end up with the following system which allows you to determine $f_1$ and $f_2$: \begin{equation} f^{'}_{1}(t) y_1(t)+ f^{'}_{2}(t) y_2(t)=0 \qquad f^{'}_{1}(t) y^{'}_1(t)+ f^{'}_{2}(t) y^{'}_2(t)=r(t) \end{equation}