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

\(\begin{array}{l}{\bf{2x' - y' = y + 3x + }}{{\bf{e}}^{\bf{t}}}{\bf{,}}\\{\bf{3y' - 4x' = y - 15x + }}{{\bf{e}}^{{\bf{ - t}}}}\end{array}\)

One will solve the given system using the elimination method. First, one will rewrite it in operator form:

\(\begin{array}{c}{\bf{2x' - 3x - y' - y = }}{{\bf{e}}^{\bf{t}}}\\{\bf{ - 4x' + 15x + 3y' - y = }}{{\bf{e}}^{{\bf{ - t}}}}\\{\bf{(2D - 3)[x] - (D + 1)[y] = }}{{\bf{e}}^{\bf{t}}}\\{\bf{( - 4D + 15)[x] + (3D - 1)[y] = }}{{\bf{e}}^{{\bf{ - t}}}}\end{array}\)

One can eliminate \({\bf{y}}\) from the system by multiplying the first equation by \(\left( {{\bf{3 D - 1}}} \right)\) and the second by \(\left( {{\bf{D + 1}}} \right)\) and then adding those two equations together:

\(\begin{array}{c}{\bf{(3D - 1)(2D - 3)[x] - (3D - 1)(D + 1)[y] = (3D - 1)}}{{\bf{e}}^{\bf{t}}}\\{\bf{(D + 1)( - 4D + 15)[x] + (D + 1)(3D - 1)[y] = (D + 1)}}{{\bf{e}}^{{\bf{ - t}}}}\\\left( {{\bf{6}}{{\bf{D}}^{\bf{2}}}{\bf{ - 11D + 3}}} \right){\bf{[x] - }}\left( {{\bf{3}}{{\bf{D}}^{\bf{2}}}{\bf{ + 2D - 1}}} \right){\bf{[y] = 3}}{{\bf{e}}^{\bf{t}}}{\bf{ - }}{{\bf{e}}^{\bf{t}}}\\\left( {{\bf{ - 4}}{{\bf{D}}^{\bf{2}}}{\bf{ + 11D + 15}}} \right){\bf{[x] + }}\left( {{\bf{3}}{{\bf{D}}^{\bf{2}}}{\bf{ + 2D - 1}}} \right){\bf{[y] = - }}{{\bf{e}}^{{\bf{ - t}}}}{\bf{ + }}{{\bf{e}}^{{\bf{ - t}}}}\\\left( {{\bf{2}}{{\bf{D}}^{\bf{2}}}{\bf{ + 18}}} \right){\bf{[x] = 2}}{{\bf{e}}^{\bf{t}}}\\\left( {{{\bf{D}}^{\bf{2}}}{\bf{ + 9}}} \right){\bf{[x] = }}{{\bf{e}}^{\bf{t}}}\end{array}\)

First, one will find a homogeneous solution. The auxiliary equation is \({{\bf{r}}^{\bf{2}}}{\bf{ + 9 = 0}}\) and its roots are \({{\bf{r}}^{\bf{2}}}{\bf{ = - 9}} \Rightarrow {{\bf{r}}_{{\bf{1,2}}}}{\bf{ = \pm 3i}}\) and the homogeneous solution for \({\bf{x}}\) is:

\({{\bf{x}}_{\bf{h}}}{\bf{(t) = }}{{\bf{c}}_{\bf{1}}}{\bf{cos3t + }}{{\bf{c}}_{\bf{2}}}{\bf{sin3t}}\).

Now one can find a particular solution for\({\bf{x}}\).

One will apply the method of undetermined coefficients. Assume that a particular solution has a form of\({{\bf{x}}_{\bf{p}}}{\bf{(t) = A}}{{\bf{e}}^{\bf{t}}}\). Then\({\bf{x}}_{\bf{p}}^{}{\bf{'(t) = x}}_{\bf{p}}^{}{\bf{''(t) = A}}{{\bf{e}}^{\bf{t}}}\).

So, substituting it into the differential equation one will get:

\(\begin{array}{c}\left( {{{\bf{D}}^{\bf{2}}}{\bf{ + 9}}} \right)\left[ {{{\bf{x}}_{\bf{p}}}} \right]{\bf{ = x}}_{\bf{p}}^{}{\bf{''(t) + 9}}{{\bf{x}}_{\bf{p}}}{\bf{(t)}}\\{\bf{ = A}}{{\bf{e}}^{\bf{t}}}{\bf{ + 9A}}{{\bf{e}}^{\bf{t}}}{\bf{ = 10A}}{{\bf{e}}^{\bf{t}}}{\bf{ = }}{{\bf{e}}^{\bf{t}}}\\{\bf{10A = 1}}\; \Rightarrow {\bf{A = }}\frac{{\bf{1}}}{{{\bf{10}}}}\end{array}\)

So, the particular solution for \({\bf{x}}\) is \({{\bf{x}}_{\bf{p}}}{\bf{(t) = }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}\) and the general solution is:

\({\bf{x(t) = }}{{\bf{x}}_{\bf{h}}}{\bf{(t) + }}{{\bf{x}}_{\bf{p}}}{\bf{(t) = }}{{\bf{c}}_{\bf{1}}}{\bf{cos3t + }}{{\bf{c}}_{\bf{2}}}{\bf{sin3t + }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}\)

If one subtracts the equations of the initial system, one will get

\({\bf{6x' - 18x - 4y' = }}{{\bf{e}}^{\bf{t}}}{\bf{ - }}{{\bf{e}}^{{\bf{ - t}}}} \Leftrightarrow {\bf{4y' = 6x' - 18x - }}{{\bf{e}}^{\bf{t}}}{\bf{ + }}{{\bf{e}}^{{\bf{ - t}}}}\).

So, one will find the first derivative of \({\bf{x}}\) and substitute it into the previous equation to obtain\({\bf{y'}}\).

\(\begin{array}{c}{\bf{x'(t) = - 3}}{{\bf{c}}_{\bf{1}}}{\bf{sin3t + 3}}{{\bf{c}}_{\bf{2}}}{\bf{cos3t + }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}{\bf{;}}\\{\bf{y'(t) = 6x' - 18x - }}{{\bf{e}}^{\bf{t}}}{\bf{ + }}{{\bf{e}}^{{\bf{ - t}}}}\\{\bf{ = 6}}\left( {{\bf{ - 3}}{{\bf{c}}_{\bf{1}}}{\bf{sin3t + 3}}{{\bf{c}}_{\bf{2}}}{\bf{cos3t + }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}} \right){\bf{ - 18}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{cos3t + }}{{\bf{c}}_{\bf{2}}}{\bf{sin3t + }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}} \right){\bf{ - }}{{\bf{e}}^{\bf{t}}}{\bf{ + }}{{\bf{e}}^{{\bf{ - t}}}}\\{\bf{ = 18}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){\bf{cos3t - 18}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{sin3t - }}\frac{{{\bf{11}}{{\bf{e}}^{\bf{t}}}}}{{\bf{5}}}{\bf{ + }}{{\bf{e}}^{{\bf{ - t}}}}{\bf{;}}\\{\bf{y'(t) = }}\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){\bf{cos3t - }}\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{sin3t - }}\frac{{{\bf{11}}{{\bf{e}}^{\bf{t}}}}}{{{\bf{20}}}}{\bf{ + }}\frac{{{{\bf{e}}^{{\bf{ - t}}}}}}{{\bf{4}}}\end{array}\)

Integrating the previous equation, one will obtain\({\bf{y}}\left( {\bf{t}} \right)\).

\(\begin{array}{c}{\bf{y(t) = }}\int {\left( {\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){\bf{cos3t - }}\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{sin3t - }}\frac{{{\bf{11}}{{\bf{e}}^{\bf{t}}}}}{{{\bf{20}}}}{\bf{ + }}\frac{{{{\bf{e}}^{{\bf{ - t}}}}}}{{\bf{4}}}} \right)} {\bf{dt}}\\{\bf{ = }}\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right)\int {{\bf{cos}}} {\bf{3tdt - }}\frac{{\bf{9}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right)\int {{\bf{sin}}} {\bf{3tdt - }}\frac{{{\bf{11}}}}{{{\bf{20}}}}\int {{{\bf{e}}^{\bf{t}}}} {\bf{dt + }}\frac{{\bf{1}}}{{\bf{4}}}\int {{{\bf{e}}^{{\bf{ - t}}}}} {\bf{dt}}\\{\bf{ = }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){\bf{sin3t + }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{cos3t - }}\frac{{{\bf{11}}}}{{{\bf{20}}}}{{\bf{e}}^{\bf{t}}}{\bf{ - }}\frac{{\bf{1}}}{{\bf{4}}}{{\bf{e}}^{{\bf{ - t}}}}\\{\bf{ = }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{cos3t - }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right){\bf{sin3t - }}\frac{{{\bf{11}}}}{{{\bf{20}}}}{{\bf{e}}^{\bf{t}}}{\bf{ - }}\frac{{\bf{1}}}{{\bf{4}}}{{\bf{e}}^{{\bf{ - t}}}}\end{array}\)

So, the general solutions \({\bf{x}}\left( {\bf{t}} \right)\)and\({\bf{y}}\left( {\bf{t}} \right)\)for the given system are

\(\begin{array}{c}{\bf{x(t) = }}{{\bf{c}}_{\bf{1}}}{\bf{cos3t + }}{{\bf{c}}_{\bf{2}}}{\bf{sin3t + }}\frac{{{{\bf{e}}^{\bf{t}}}}}{{{\bf{10}}}}{\bf{,}}\\{\bf{y(t) = }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){\bf{cos3t - }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ - }}{{\bf{c}}_{\bf{2}}}} \right){\bf{sin3t - }}\frac{{{\bf{11}}}}{{{\bf{20}}}}{{\bf{e}}^{\bf{t}}}{\bf{ - }}\frac{{\bf{1}}}{{\bf{4}}}{{\bf{e}}^{{\bf{ - t}}}}\end{array}\)