Answer
a) $α(t)$ = $e^{2t}$
b) $y(t)$ = $Ce^{-2t}-e^{-3t}$
c) $y$ = $2e^{-2t}-e^{-3t}$
Work Step by Step
the equation is of the from
$y'+A(t)y$ = $B(t)$
$A(t)$ = $2$
$B(t)$ = $e^{-3t}$
by theorem 1, $α(t)$ is defined by
$α(t)$ = $e^{\int{A(t)}dt}$ = $e^{2t}$
b)
$y(t)$ = $\frac{1}{α(t)}(\int{α(t)B(t)+C})$
$y(t)$ = $e^{2t}(\int{e^{-t}dt}+C)$
$y(t)$ = $Ce^{-2t}-e^{-3t}$
c)
$y(0)$ = $1$
$1$ = $-1+C$
$C$ = $2$
$y$ = $2e^{-2t}-e^{-3t}$
