Q2RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

\(\begin{array}{l}{\bf{x' = x + 2y }}\\{\bf{y' =  - 4x - 3y}}\end{array}\)

 

Step-by-Step Solution

Answer

The solution to the given system is:

 

\(\begin{array}{c}{\bf{x(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t, }}\\{\bf{y(t) = }}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}\end{array}\)

Step 1: Substituting the equation into the second equation

 

The given equations are:

 

\(\begin{array}{l}{\bf{x' = x + 2y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right){\bf{ }}\\{\bf{y' =  - 4x - 3y}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 2 \right)\end{array}\)

 

From the first equation, we have;

 

\(\begin{array}{c}{\bf{y = }}\frac{{{\bf{x' - x}}}}{{\bf{2}}}\;\;\;\\{\bf{y' = }}\frac{{{\bf{x'' - x'}}}}{{\bf{2}}}\end{array}\)

 

Substituting the equations into the second equation of the given system one will get;

 

\(\begin{array}{c}\frac{{{\bf{x}}''{\bf{ - x}}'}}{{\bf{2}}}{\bf{ =  - 4x - }}\frac{{\bf{3}}}{{\bf{2}}}\left( {{\bf{x}}'{\bf{ - x}}} \right)\\{\bf{x}}''{\bf{ - x}}'{\bf{ =  - 8x - 3}}\left( {{\bf{x}}'{\bf{ - x}}} \right)\\{\bf{x}}''{\bf{ + 2x}}'{\bf{ + 5x = 0}}\end{array}\)

Step 2: Finding the general solution for \({\bf{x}}\)

The auxiliary equation corresponding to the previous homogeneous differential equation is \({{\bf{r}}^{\bf{2}}}{\bf{ + 2r + 5 = 0}}\)and its roots are:

 

\({{\bf{r}}_{{\bf{1,2}}}}{\bf{ = }}\frac{{{\bf{ - 2 \pm }}\sqrt {{\bf{4 - 20}}} }}{{\bf{2}}} \Rightarrow {{\bf{r}}_{{\bf{1,2}}}}{\bf{ =  - 1 \pm 2i}}\)

 

So, the general solution for \({\bf{x}}\) is\({\bf{x(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}.\)

Step 3: Substitute the value of \({\bf{x}}\) into \({\bf{y}}\)

One will now find the first derivative of \({\bf{x}}\) and substitute it into \({\bf{y = }}\frac{{{\bf{x}}'{\bf{ - x}}}}{{\bf{2}}}\) to obtain a solution for\({\bf{y}}\).

 

\(\begin{array}{c}{\bf{x'(t) = }}\left( {{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}} \right){\bf{'}}\\{\bf{ =  - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - 2}}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t + 2}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t}}\\{\bf{2y(t) = x'(t) - x(t)}}\\{\bf{ =  - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - 2}}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t + 2}}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}\\{\bf{ = 2}}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - 2}}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}\\ \Rightarrow {\bf{y(t) = }}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}\end{array}\)

 

Therefore, the solution to the given system is:

 

\(\begin{array}{c}{\bf{x(t) = }}{{\bf{c}}_{\bf{1}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t + }}{{\bf{c}}_{\bf{2}}}{{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t, }}\\{\bf{y(t) = }}\left( {{{\bf{c}}_{\bf{2}}}{\bf{ - }}{{\bf{c}}_{\bf{1}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{cos2t - }}\left( {{{\bf{c}}_{\bf{1}}}{\bf{ + }}{{\bf{c}}_{\bf{2}}}} \right){{\bf{e}}^{{\bf{ - t}}}}{\bf{sin2t}}\end{array}\)