다음과 같은 연립상미분방정식이 있다.
{ y 1 ′ = y 1 + 2 y 2 y 2 ′ = − y 1 − y 2 + y 3 y 3 ′ = 3 y 1 + 2 y 2 − 2 y 3 \begin{cases} y _ {1} ' = y _ {1} +2y _ {2} \\ y _ {2} ' = -y _ {1} -y _ {2} +y _ {3} \\ y _ {3} ' =3y _ {1} +2y _ {2} -2y _ {3} \end{cases} ⎩ ⎨ ⎧ y 1 ′ = y 1 + 2 y 2 y 2 ′ = − y 1 − y 2 + y 3 y 3 ′ = 3 y 1 + 2 y 2 − 2 y 3 y 1 ( x ) y _ {1} \left (x \right) y 1 ( x ) y 2 ( x ) y _ {2} \left (x \right) y 2 ( x ) y 3 ( x ) y _ {3} \left (x \right) y 3 ( x ) y → ( x ) = [ y 1 ( x ) y 2 ( x ) y 2 ( x ) ] \overrightarrow{y} \left (x \right) = \begin{bmatrix*}[r] y _ {1} \left (x \right) \\ y _ {2} \left (x \right) \\ y _ {2} \left (x \right) \end{bmatrix*} y ( x ) = y 1 ( x ) y 2 ( x ) y 2 ( x ) p 1 → \overrightarrow{p _ {1}} p 1 p 2 → \overrightarrow{p _ {2}} p 2 p 3 → \overrightarrow{p _ {3}} p 3 y → ( x ) = c 1 e k 1 x p 1 → + c 2 e k 2 x p 2 → + c 3 e k 3 x p 3 → \overrightarrow{y} \left (x \right) =c _ {1} e ^ {k _ {1} x} \overrightarrow{p _ {1}} +c _ {2} e ^ {k _ {2} x} \overrightarrow{p _ {2}} +c _ {3} e ^ {k _ {3} x} \overrightarrow{p _ {3}} y ( x ) = c 1 e k 1 x p 1 + c 2 e k 2 x p 2 + c 3 e k 3 x p 3 ∣ p 2 → ∙ p 3 → ∣ \left | \overrightarrow{p _ {2}} \bullet \overrightarrow{p _ {3}} \right | p 2 ∙ p 3 c 1 , c 2 , c 3 c _ {1}, c _ {2}, c _ {3} c 1 , c 2 , c 3 k 1 , k 2 , k 3 k _ {1}, k _ {2}, k _ {3} k 1 , k 2 , k 3 k 1 < k 2 < k 3 k _ {1} <k _ {2} <k _ {3} k 1 < k 2 < k 3
① 7 2 19 \displaystyle \frac{7} {2 \sqrt {19}} 2 19 7 내 답 ② 2 6 \displaystyle \frac{2} {\sqrt {6}} 6 2 내 답 ③ 10 144 \displaystyle \frac{10} {\sqrt {144}} 144 10 내 답