The solution of the system of differential equations $\begin{array} { l }
\frac { d x } { d t } = - 6 x + 5 y \
\frac { d y } { d t } = - 5 x + 4 y \
x ( 0 ) = 1 / 3 , y ( 0 ) = 0
\end{array}$ is
A) $x = \left( e ^ { t } - 5 t e ^ { t } \right) / 3 , y = - 5 t e ^ { t } / 3$
B) $x = \left( e ^ { - t } - t e ^ { - t } \right) / 3 , y = - t e ^ { - t } / 3$
C) $x = \left( e ^ { - t } + t e ^ { - t } \right) / 3 , y = t e ^ { - t } / 3$
D) $x = \left( e ^ { - t } + 5 t e ^ { - t } \right) / 3 , y = 5 t e ^ { - t } / 3$
E) $x = \left( e ^ { - t } - 5 t e ^ { - t } \right) / 3 , y = - 5 t e ^ { - t } / 3$

Question

The solution of the system of differential equations $\begin{array} { l } 
\frac { d x } { d t } = - 6 x + 5 y \
\frac { d y } { d t } = - 5 x + 4 y \
x ( 0 ) = 1 / 3 , y ( 0 ) = 0
\end{array}$ is
A) $x = \left( e ^ { t } - 5 t e ^ { t } \right) / 3 , y = - 5 t e ^ { t } / 3$
B) $x = \left( e ^ { - t } - t e ^ { - t } \right) / 3 , y = - t e ^ { - t } / 3$
C) $x = \left( e ^ { - t } + t e ^ { - t } \right) / 3 , y = t e ^ { - t } / 3$
D) $x = \left( e ^ { - t } + 5 t e ^ { - t } \right) / 3 , y = 5 t e ^ { - t } / 3$
E) $x = \left( e ^ { - t } - 5 t e ^ { - t } \right) / 3 , y = - 5 t e ^ { - t } / 3$

Quizplus · Accepted Answer

The correct solution can be found by solving the system of differential equations using methods such as eigenvalues and eigenvectors or the method of undetermined coefficients. The given initial conditions $x(0) = 1/3, y(0) = 0$ help determine the specific solution from the general solution. The answer choice E, $x = \left( e ^ { - t } - 5 t e ^ { - t } \right) / 3 , y = - 5 t e ^ { - t } / 3$, correctly reflects the behavior of the system as it involves the exponential decay term $e^{-t}$ modified by a linear term in $t$, which is consistent with the dynamics described by the differential equations and the initial conditions.

The Solution of the System of Differential Equations dxdt=−6x+5ydydt=−5x+4yx(0)=1/3,y(0)=0\begin{array} { l } \frac { d x } { d t } = - 6 x + 5 y \\\frac { d y } { d t } = - 5 x + 4 y \\x ( 0 ) = 1 / 3 , y ( 0 ) = 0\end{array}dtdx​=−6x+5ydtdy​=−5x+4yx(0)=1/3,y(0)=0​

The Solution of the System of Differential Equations $\begin{array} { l } \frac { d x } { d t } = - 6 x + 5 y \\\frac { d y } { d t } = - 5 x + 4 y \\x ( 0 ) = 1 / 3 , y ( 0 ) = 0\end{array}$