The functions
y 1 ( t ) = e 3 t \displaystyle {y}_{{1}}{\left({t}\right)}={e}^{{{3}{t}}} y 1 ( t ) = e 3 t and
y 2 ( t ) = e − 3 t \displaystyle {y}_{{2}}{\left({t}\right)}={e}^{{-{3}{t}}} y 2 ( t ) = e − 3 t are known to be solutions of
y ′ ′ + A y ′ + B y = 0 \displaystyle {y}{''}+{A}{y}'+{B}{y}={0} y ′′ + A y ′ + B y = 0 , where
A \displaystyle {A} A and
B \displaystyle {B} B are constants. Determine
A \displaystyle {A} A and
B \displaystyle {B} B . [Hint: Find the characteristic equation.]
A \displaystyle {A} A =
B \displaystyle {B} B =
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