= Solution
Retain the <derivative> marks and coefficients from the printed system. In matrix form it is $M\dot u+Ku=b(t)$, with
$$
M=\begin{pmatrix}3&1\\1&4\end{pmatrix},\qquad
K=\begin{pmatrix}5&-1\\-2&7\end{pmatrix}.
$$
Since $\det M=11$, multiplication by its inverse gives
$$
\dot x+2x-y=e^{-t}+e^{-3t},\qquad
\dot y-x+2y=-e^{-t}+e^{-3t}.
$$
The <linear transformation> $u=x+y$, $v=x-y$ diagonalizes this system:
$$
\dot u+u=2e^{-3t},\qquad \dot v+3v=2e^{-t}.
$$
Both start at zero. An <integrating factor> gives $u=v=e^{-t}-e^{-3t}$, so
$$
\boxed{x(t)=e^{-t}-e^{-3t},\qquad y(t)=0}.
$$
Direct substitution into the two original equations confirms the different forcing coefficients; the exact cancellation in $y$ depends on those coefficients.
Back to article page
