Find a general solution\({\bf{x}}\left( {\bf{t}} \right){\bf{, y}}\left( {\bf{t}} \right)\)for the given system.

\(\begin{array}{c}{\bf{x' + y'' + y = 0}}\\{\bf{x'' + y' = 0}}\end{array}\)

From the first equation of the given system, one has that\({\bf{x' = - y'' - y'}}\), so when one differentiates the previous equation, one will get that\({\bf{x'' = - y''' - y'}}.\)

Substituting it into the second equation of the given system one will get:

\(\begin{array}{c}{\bf{x'' + y' = 0}}\\{\bf{ - y''' - y' + y' = 0}}\\{\bf{ - y''' = 0 }}\\{\bf{y''' = 0}}\end{array}\)

Integrating the previous equation three times one will get:

\(\begin{array}{c}{\bf{y''(t) = }}\int {{\bf{y'''}}} {\bf{(t)dt = }}{{\bf{c}}_{\bf{1}}}\\{\bf{y'(t) = }}\int {{\bf{y''}}} {\bf{(t)dt = }}{{\bf{c}}_{\bf{1}}}{\bf{t + }}{{\bf{c}}_{\bf{2}}}\\{\bf{y(t) = }}\int {{\bf{y'}}} {\bf{(t)dt = }}{{\bf{c}}_{\bf{1}}}\frac{{{{\bf{t}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}{\bf{t + }}{{\bf{c}}_{\bf{3}}}\end{array}\)

One can rename the constant \({\bf{ }}\frac{{{{\bf{c}}_{\bf{1}}}}}{{\bf{2}}}\)to\({{\bf{c}}_{\bf{1}}}\) since those constants are arbitrary, so one has that the general solution for \({\bf{y}}\left( {\bf{t}} \right)\)is\({\bf{y(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}{\bf{t + }}{{\bf{c}}_{\bf{3}}}\)

It remains to find a general solution for \({\bf{x}}{\bf{.}}\)Now find it from \({\bf{x' = - y'' - y'}}{\bf{.}}\)One will first find the second derivative of \({\bf{y}}\left( {\bf{t}} \right)\)and then substitute it into the previous equation.

\(\begin{array}{c}{\bf{y'(t) = }}\left( {{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}{\bf{t + }}{{\bf{c}}_{\bf{3}}}} \right){\bf{' = 2}}{{\bf{c}}_{\bf{1}}}{\bf{t + }}{{\bf{c}}_{\bf{2}}}\\{\bf{y''(t) = 2}}{{\bf{c}}_{\bf{1}}}\\{\bf{x'(t) = - y'' - y}}\\{\bf{ = - 2}}{{\bf{c}}_{\bf{1}}}{\bf{ - }}\left( {{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}{\bf{t + }}{{\bf{c}}_{\bf{3}}}} \right)\\{\bf{ = - }}{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}{\bf{t - }}\left( {{{\bf{c}}_{\bf{3}}}{\bf{ + 2}}{{\bf{c}}_{\bf{1}}}} \right)\end{array}\)

Integrating the previous equation, one will get the general solution for\({\bf{x}}\):

\(\begin{array}{c}{\bf{x(t) = }}\int {{\bf{x'}}} {\bf{dt = }}\int {\left( {{\bf{ - }}{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}{\bf{t - }}\left( {{{\bf{c}}_{\bf{3}}}{\bf{ + 2}}{{\bf{c}}_{\bf{1}}}} \right)} \right)} {\bf{dt }}\\{\bf{ = - }}{{\bf{c}}_{\bf{1}}}\frac{{{{\bf{t}}^{\bf{3}}}}}{{\bf{3}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}\frac{{{{\bf{t}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - }}\left( {{{\bf{c}}_{\bf{3}}}{\bf{ + 2}}{{\bf{c}}_{\bf{1}}}} \right){\bf{t + }}{{\bf{c}}_{\bf{4}}}\end{array}\)

Therefore, the solution for the given system is:

\(\begin{array}{c}{\bf{x(t) = - }}\frac{{{{\bf{c}}_{\bf{1}}}}}{{\bf{3}}}{{\bf{t}}^{\bf{3}}}{\bf{ - }}\frac{{{{\bf{c}}_{\bf{2}}}}}{{\bf{2}}}{{\bf{t}}^{\bf{2}}}{\bf{ - }}\left( {{{\bf{c}}_{\bf{3}}}{\bf{ + 2}}{{\bf{c}}_{\bf{1}}}} \right){\bf{t + }}{{\bf{c}}_{\bf{4}}}\\{\bf{y(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{t}}^{\bf{2}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}{\bf{t + }}{{\bf{c}}_{\bf{3}}}\end{array}\)