Finding the solution to differential equation dilemma.

Question

Suppose $\ddot{x_1} =\frac{k(x_2-2x_1)}{3m}$ and $\ddot{x_2}=\frac{k(x_1-x_2)}{2m}$ now how do I solve for $x_1 $ and $x_2$ where $x_2 $& $x_1$ aren't independent of each other ie can't be held constant.

Answer 1

Hint.

Assuming you know the Laplace transform method we have

$$ s^2 \left(\begin{array}{c} X_1(s)\\ X_2(s) \end{array} \right) -\left(\begin{array}{c} s \dot x_1(0)+x_1(0)\\ s \dot x_2(0)+x_2(0) \end{array} \right)= K \left(\begin{array}{c} X_1(s)\\ X_2(s) \end{array} \right) $$

or

$$ \left(I_2 s^2-K\right)\left(\begin{array}{c} X_1(s)\\ X_2(s) \end{array} \right) = \left(\begin{array}{c} s \dot x_1(0)+x_1(0)\\ s \dot x_2(0)+x_2(0) \end{array} \right) $$

then

$$ \left(\begin{array}{c} X_1(s)\\ X_2(s) \end{array} \right) = \left(I_2 s^2-K\right)^{-1}\left(\begin{array}{c} s \dot x_1(0)+x_1(0)\\ s \dot x_2(0)+x_2(0) \end{array} \right) $$

Here

$$ I_2 s^2-K = \left(\begin{array}{cc}s^2-k_{11}& -k_{12}\\ -k_{21}& s^2-k_{22}\end{array}\right) $$

and

$$ \left(I_2 s^2-K\right)^{-1} = \frac{1}{(s^2-k_{11})(s^2-k_{22})+k_{12}k_{21}}\left(\begin{array}{cc}s^2-k_{22}& k_{12}\\ k_{21}& s^2-k_{11}\end{array}\right) $$

and

$$ K = \left(\begin{array}{cc}-\frac{2k}{3m}& \frac{k}{3m}\\ \frac{k}{2m}& -\frac{k}{2m}\end{array}\right) $$

Answer 2

You can convert to a 1st order ODE through $x_3 := \dot{x_1}$ and $x_4 := \dot{x_2}$. Then let $y = (x_1,x_2,x_3,x_4)^T$ so that $$\dot{y} = \begin{bmatrix} 0_{2\times 2} & I_{2\times 2}\\ \frac{k}{6m} \begin{bmatrix} -4 & 2 \\ 3 & -3\end{bmatrix} & 0_{2\times 2} \end{bmatrix}y := Ay$$ Then use Jordan Normal from to write $$A = P^{-1} D P$$ Where $D$ is diagonal. Then change variables again so that $$\bar{y} := P \bar{y}$$ solves $$\dot{\bar{y}} = D \bar{y}$$ Then $$\bar{y}_i = e^{ d_i t}$$ with $d_i$ the possible complex entries of $\bar{D}$ and convert back to $x_i$'s.