![](https://www.physics.smu.edu/fattarus/coupled_oscillator_fig2.png)
The equations of motion that describe the above situation is given by:

$$ m \ddot{x\_1} = -2kx\_1 + kx\_2 $$

$$ m \ddot{x\_2} = -2kx\_2 + kx\_1 $$

Now I want to work out the potential energy of this system. How would I use the equation:

$$ V(x) = - \int F(x) \hspace{.1cm} dx$$

to calculate the potential energy?

My attempt is to first calculate the integral for each coupled equation wrt each position:

$$ - \int (- 2kx\_1 + + kx\_2) \hspace{.1cm} dx\_1= kx^{2}\_1 - k x\_1 x\_2 + c$$
$$ - \int (- 2kx\_2 + + kx\_1) \hspace{.1cm} dx\_2= kx^{2}\_2 - k x\_2 x\_1 + d.$$

Now adding them together to get the total potential energy gives:

$$ V(x\_1, x\_2) = kx^{2}\_1 + kx^{2}\_2 + - 2 k x\_1 x\_2 + f.$$

The actual solution is given by:

$$ V(x\_1, x\_2) = kx^{2}\_1 + kx^{2}\_2 + - k x\_1 x\_2$$

Now I have inkling as to why my answer is wrong. When adding the potentials together to calculate the total potential I have double added the potential associated with the middle spring. However, mathematically how do I fix this issue to get the correct solution?