Could you help me with the following system of two equations? I have browsed through the forum, but I haven't found a similar problem. $$m_1x_1''(t) + k(x_1 - x_2)+ cx_1'(t) = 0$$ $$m_1x_2''(t) + k(x_2 - x_1) = D_0 \sin(\omega \cdot t)$$
I just need a method for solving it. An idea for a substitution. I have tried $x_1(t) = r(t)\cos(at),x_2(t) = r(t)\sin(at)$, but I still can't solve it.
Put $v_1 = x_1' $ and $v_2 = x_2'$. Then (after rearranging) your system becomes a first order system of ODEs $$ \begin{pmatrix}x_1'\\x_2'\\v_1'\\v_2'\end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -\frac{k}{m_1} & \frac{k}{m_1} & -\frac{c}{m_1} & 0 \\ \frac{k}{m_2} & -\frac{k}{m_2} & 0 & 0 \end{pmatrix} \begin{pmatrix}x_1\\x_2\\v_1\\v_2\end{pmatrix} + \begin{pmatrix}0\\0\\0\\\frac{D_0}{m_2}\sin \omega t\end{pmatrix}, $$ which can be solved by standard methods (see e.g. Solving nonhomogeneous systems of ODEs).