Cfrgtkky

Consider the coupled spring-mass system with a frictionless table, two masses $m_1$ and $m_2$, and three springs with spring constants $k_1, k_2$, and $k_3$ respectively. The equation of motion for the system are given by:

$m_1frac{d^2x_1}{dt^2}=-(k_1+k_2)x_1+k_2x_2$

$m_2frac{d^2x_2}{dt^2}=k_2x_1-(k_2+k_3)x_2$

Assume that the masses are $m_1 = 2$, $m_2 = frac{9}{4}$, and the spring constants are $k_1=1,k_2=3,k_3=frac{15}{4}$.

I substitute all value in the system above, and able to get

$frac{d^2x_1}{dt^2}=-2x_1+frac{3}{2}x_2$

$frac{d^2x_2}{dt^2}=frac{4}{3}x_1-3x_2$

Then how to convert the system about into four first order equations?

To convert it into a system of four first-order ODEs, you simply define the new variables $$x_3 = frac{dx_1}{dt} qquad x_4 = frac{dx_2}{dt}$$

The system then becomes

$$frac{dx_1}{dt} = x_3 qquad frac{dx_2}{dt} = x_4 qquad frac{dx_3}{dt}=-2x_1+frac{3}{2}x_2 qquad frac{dx_4}{dt}=frac{4}{3}x_1-3x_2$$

However, I am not sure this actually helps you to solve the system. Instead, you might consider using

begin{align}
frac{d^2x_1}{dt^2} & =-2x_1+frac{3}{2}x_2 tag{1}\
frac{d^2x_2}{dt^2} & =frac{4}{3}x_1-3x_2 tag{2}
end{align}

Do $frac{d^2}{dt^2}$ of equation $(1)$:

$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 frac{d^2x_2}{dt^2}$$

Plug in the expression for $frac{d^2x_2}{dt^2} $ from equation $(2)$:

$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2}+frac 32 bigg(frac{4}{3}x_1-3x_2bigg) = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 x_2$$

Then use equation $(1)$ to plug in expression for $x_2$:

$$frac{d^4x_1}{dt^4} = -2frac{d^2x_1}{dt^2} + 2x_1-frac 92 bigg(frac 23 frac{d^2x_1}{dt^2} + frac 43 x_1 bigg) = -5frac{d^2x_1}{dt^2} -4x_1$$

$$implies frac{d^4x_1}{dt^4}+5frac{d^2x_1}{dt^2} + 4x_1=0$$

This is now a homogeneous ODE with constant coefficients in $x_1$, which can easily be solved, giving

$$x_1(t) = Acos (2t)+Bsin (2t) + Ccos (t) + Dsin (t)$$

Plugging this back into equation $(1)$, we find

$$x_2(t) = -frac 43 Acos (2t) -frac 43 Bsin (2t)+ frac 23 Ccos (t) + frac 23 Dsin (t)$$

Finally, use the initial conditions (you should have four of them) to find the constants $A,B,C,D$.

The matrix $A=pmatrix{-2&frac32\frac43&-3}$ on the right side of the system has the characteristic polynomial
$$
0=(q+2)(q+3)-2=q^2+5q+4=(q+4)(q+1).
$$
For the eigenvalue $q=-1$ a left eigenvector is $(4,3)$ so that
$$
(4x_1+3x_2)''=-(4x_1+3x_2)implies 4x_1+3x_2=a_1cos(t)+b_1sin(t)
$$
For the eigenvalue $q=-4$ a left eigenvector is $(2,-3)$ so that
$$
(2x_1-3x_2)''=-4(2x_1-3x_2)implies 2x_1-3x_2=a_2cos(2t)+b_2sin(2)
$$
Now the solutions $x_1$, $x_2$ can be recombined and used to compare against the numerical solution of the first order system as per the other answer.