- #1
karush
Gold Member
MHB
- 3,269
- 5
$\tiny{31.6}$
Solve the initial value problem
$Y'=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|Y
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|,
\quad Y(0)=\left|\begin{array}{rr} 1 \\1 \end{array}\right| $
ok so we have the form $y'=AY+G$
rewrite as
$$\displaystyle
\left|\begin{array}{rr}y_1^\prime \\y_2^\prime \end{array}\right|
=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|
\left|\begin{array}{rr}y_1 \\y_2\end{array}\right|
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|$$
ok so the next thing to do is find eigenvalues of A so
$\left| \begin{array}{cc}
-\lambda+2&1\\-1&-\lambda+2\end{array}
\right|
=\left(-\lambda+2\right)^{2}+1$
so roots are
$\lambda_{1}=2 + i, \qquad \lambda_{2}=2 - i$so far ? hopefully
Solve the initial value problem
$Y'=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|Y
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|,
\quad Y(0)=\left|\begin{array}{rr} 1 \\1 \end{array}\right| $
ok so we have the form $y'=AY+G$
rewrite as
$$\displaystyle
\left|\begin{array}{rr}y_1^\prime \\y_2^\prime \end{array}\right|
=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|
\left|\begin{array}{rr}y_1 \\y_2\end{array}\right|
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|$$
ok so the next thing to do is find eigenvalues of A so
$\left| \begin{array}{cc}
-\lambda+2&1\\-1&-\lambda+2\end{array}
\right|
=\left(-\lambda+2\right)^{2}+1$
so roots are
$\lambda_{1}=2 + i, \qquad \lambda_{2}=2 - i$so far ? hopefully