# How to change the system in 4x4 system of first-order equations? [duplicate]

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:

$$y_1''=-\frac{(k_1+k_2)}{m_1}y_1+\frac{k_2}{m_1}y_2$$

$$y_2''=\frac{k_2}{m_2}y_1-\frac{(k_1+k_2)}{m_2}y_2$$

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

a)use 4x4 system of first order equations to model this system of two second order equations. (hint: $$x_1=y_1,x_2=y_2,x_3=y_1',x_4=y_2'$$)

## marked as duplicate by LutzL differential-equations StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 5 '18 at 12:35

Introducing a whole new set of $$x_i$$ variables is somewhat confusing. Simpler to just introduce two new variables $$y_3$$ and $$y_4$$ where $$y_3=y_1'$$ and $$y_4=y_2'$$. So you now have:
$$y_1' = y_3$$
$$y_2' = y_4$$
$$y_3' = y_1'' = \dots$$
$$y_4'=y_2''= \dots$$