`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`. 

`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.

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.