On a smooth surface, a mass m_1=2 kg is attached to the first object by a spring with spring constant k_1=4 N/m . Another mass m_2=1 kg is attached to the first object by a spring with spring constant k_2=2 N/m . The objects are aligned horizontally so that the springs are their natural lengths. If both objects are displaced 3 m to the right of their equilibrium positions and then released, what are the equations of motion for the two objects?

The force on block 1 from the first spring is
F_1=-k_1x
Where x is the position to the right of the equilibrium position. Let y be the distance to the right of block 2's equilibrium position. Then the second spring exerts a force F_2 on block 1 given by
F_2=k_2(y-x)
Think about it. If x=y there should be no force from spring 2. Also if ygtx then there should be a pull to the right (positive) on block 1. If xlty then the block should pull to the left (negative).
Since block 2 is only attached to spring 2, from Newton's third law
F_3=-F2=-k_2(y-x)
Now apply Newton's second law to both the blocks.
m_1 (d^2x)/dt^2=F_1+F_2=-k_1x+k_2(y-x)
m_2 (d^2y)/dt^2=F_3=-k_2(y-x)
Plug in the values for k and m , then move the terms to the left hand side.
2(d^2x)/dt^2+6x-2y=0
(d^2y)/dt^2+2y-2x=0
I'm going to set let D:=d/dt and solve this system of equations by the elimination method.
(1):-gt (2D^2+6)x-2y=0
(2):-gt (D^2+2)y-2x=0
Now multiply eq. (1) by (D^2+2) and multiply eq. (2) by 2 .
(1):-gt(D^2+2)(2D^2+6)x-2(D^2+2)y=0
(2):-gt 2(D^2+2)y-4x=0
Add eq. (1) and eq. (2) together to eliminate y. 
(D^2+2)(2D^2+6)x-2(D^2+2)y+2(D^2+2)y-4x=0
2D^4x+6D^2x+4D^2x+12x-2D^2y-4y+2D^2y+4y-4x=0
2D^4x+10D^2x+8x=0
(3):-gt 2(d^4x)/(dt^4)+10(d^2x)/(dt^2)+8x=0
Try a solution of the form x(t)=e^(rt) .
Then eq. (3) takes the form
2(r^4+5r^2+4)e^(rt)=0
Solve for the roots of this characteristic equation.
(r^4+5r^2+4)=0
(r^2+1)(r^2+4)=0
The roots are r=i, -i, 2i, -2i .
Using Euler's formula, it follows that two linearly independent solutions are
z_1(t)=e^(it)=cos(t)+isin(t)
z_2(t)=e^(2it)=cos(2t)+isin(2t)
The general solution is a superposition of the imaginary and real parts of the linearly independent solutions.
 
x(t)=a_1cos(t)+a_2sin(t)+a_3cos(2t)+a_4sin(2t)
To find y(t) use earlier equation to put y in terms  of x ,
2(d^2x)/dt^2+6x-2y=0
y(t)=(d^2x)/dt^2+3x
Differentiate the general solution for x(t) twice and plug it in. You will find
y(t)=2a_1cos(t)+2a_2sin(t)-a_3cos(2t)-a_4sin(2t)
To determine the coefficients apply the initial conditions:
x(0)=3
(dx)/(dt)=0
y(0)=3
(dy)/(dt)=0
This will yield a system of 4 equations. You should find that,
x(t)=2cos(t)+cos(2t)
y(t)=4cos(t)-cos(2t)

x(t) is in black and y(t) in red.
http://astro.physics.ncsu.edu/urca/course_files/Lesson10/index.html

https://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf