The given question is not at all clear. So, I am taking a example and solving that.
Question: Find a particular solution of:
y ′ = [ 2 − 1 − 1 1 0 − 1 1 − 1 0 ] y + [ e t 0 e − t ] \mathbf{y}^{\prime}=\left[\begin{array}{ccc}
2 & -1 & -1 \\
1 & 0 & -1 \\
1 & -1 & 0
\end{array}\right] \mathbf{y}+\left[\begin{array}{c}
e^{t} \\
0 \\
e^{-t}
\end{array}\right] y ′ = ⎣ ⎡ 2 1 1 − 1 0 − 1 − 1 − 1 0 ⎦ ⎤ y + ⎣ ⎡ e t 0 e − t ⎦ ⎤
Solution:
The characteristic polynomial of the coefficient matrix is
∣ 2 − λ − 1 − 1 1 − λ − 1 1 − 1 − λ ∣ = − λ ( λ − 1 ) 2 \left|\begin{array}{lll}
2-\lambda & -1 & -1 \\
1 & -\lambda & -1 \\
1 & -1 & -\lambda
\end{array}\right|=-\lambda(\lambda-1)^{2} ∣ ∣ 2 − λ 1 1 − 1 − λ − 1 − 1 − 1 − λ ∣ ∣ = − λ ( λ − 1 ) 2
y 1 = [ 1 1 1 ] , y 2 = [ e t e t 0 ] , and y 3 = [ e t 0 e t ] \mathbf{y}_{1}=\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],\quad \mathbf{y}_{2}=\left[\begin{array}{l}
e^{t} \\
e^{t} \\
0
\end{array}\right],\quad \text { and } \quad \mathbf{y}_{3}=\left[\begin{array}{l}
e^{t} \\
0 \\
e^{t}
\end{array}\right] y 1 = ⎣ ⎡ 1 1 1 ⎦ ⎤ , y 2 = ⎣ ⎡ e t e t 0 ⎦ ⎤ , and y 3 = ⎣ ⎡ e t 0 e t ⎦ ⎤
are linearly independent solutions.
Y = [ 1 e t e t 1 e t 0 1 0 e t ] Y=\left[\begin{array}{lll}
1 & e^{t} & e^{t} \\
1 & e^{t} & 0 \\
1 & 0 & e^{t}
\end{array}\right] Y = ⎣ ⎡ 1 1 1 e t e t 0 e t 0 e t ⎦ ⎤
y p = Y u \mathbf{y}_{p}=Y \mathbf{u} y p = Y u Y u ′ = f Y \mathbf{u}^{\prime}=\mathbf{f} Y u ′ = f
[ 1 e t e t 1 e t 0 e 0 e t ] [ u 1 ′ u 2 ′ u 3 ′ ] = [ e t 0 e − t ] \left[\begin{array}{lll}
1 & e^{t} & e^{t} \\
1 & e^{t} & 0 \\
e & 0 & e^{t}
\end{array}\right]\left[\begin{array}{l}
u_{1}^{\prime} \\
u_{2}^{\prime} \\
u_{3}^{\prime}
\end{array}\right]=\left[\begin{array}{l}
e^{t} \\
0 \\
e^{-t}
\end{array}\right] ⎣ ⎡ 1 1 e e t e t 0 e t 0 e t ⎦ ⎤ ⎣ ⎡ u 1 ′ u 2 ′ u 3 ′ ⎦ ⎤ = ⎣ ⎡ e t 0 e − t ⎦ ⎤
The determinant of Y Y Y
∣ 1 e t e t 1 e t 0 1 0 e t ∣ = − e 2 t \left|\begin{array}{lll}
1 & e^{t} & e^{t} \\
1 & e^{t} & 0 \\
1 & 0 & e^{t}
\end{array}\right|=-e^{2 t} ∣ ∣ 1 1 1 e t e t 0 e t 0 e t ∣ ∣ = − e 2 t
Thus, by Cramer's rule,
u 1 ′ = − 1 e 2 t ∣ e t e t e t 0 e t 0 e − t 0 e t ∣ = − e 3 t − e t e 2 t = e − t − e t u 2 ′ = − 1 e 2 t ∣ 1 e t e t 1 0 0 1 e − t e t ∣ = − 1 − e 2 t e 2 t = 1 − e − 2 t \begin{aligned}
u_{1}^{\prime} &=-\frac{1}{e^{2 t}}\left|\begin{array}{lll}
e^{t} & e^{t} & e^{t} \\
0 & e^{t} & 0 \\
e^{-t} & 0 & e^{t}
\end{array}\right| &=-\frac{e^{3 t}-e^{t}}{e^{2 t}}=e^{-t}-e^{t} \\
u_{2}^{\prime} &=-\frac{1}{e^{2 t}}\left|\begin{array}{ccc}
1 & e^{t} & e^{t} \\
1 & 0 & 0 \\
1 & e^{-t} & e^{t}
\end{array}\right| &=-\frac{1-e^{2 t}}{e^{2 t}}=1-e^{-2 t}
\end{aligned} u 1 ′ u 2 ′ = − e 2 t 1 ∣ ∣ e t 0 e − t e t e t 0 e t 0 e t ∣ ∣ = − e 2 t 1 ∣ ∣ 1 1 1 e t 0 e − t e t 0 e t ∣ ∣ = − e 2 t e 3 t − e t = e − t − e t = − e 2 t 1 − e 2 t = 1 − e − 2 t
u 3 ′ = − 1 e 2 t ∣ 1 e t e t 1 e t 0 1 0 e − t ∣ = e 2 t e 2 t = 1 u_{3}^{\prime}=-\frac{1}{e^{2 t}}\left|\begin{array}{ccc}
1 & e^{t} & e^{t} \\
1 & e^{t} & 0 \\
1 & 0 & e^{-t}
\end{array}\right|=\frac{e^{2 t}}{e^{2 t}} \quad=1 u 3 ′ = − e 2 t 1 ∣ ∣ 1 1 1 e t e t 0 e t 0 e − t ∣ ∣ = e 2 t e 2 t = 1
Therefore
u ′ = [ e − t − e t 1 − e − 2 t 1 ] \mathbf{u}^{\prime}=\left[\begin{array}{l}
e^{-t}-e^{t} \\
1-e^{-2 t} \\
1
\end{array}\right] u ′ = ⎣ ⎡ e − t − e t 1 − e − 2 t 1 ⎦ ⎤
Integrating and taking the constants of integration to be zero yields
u = [ − e t − e − t e − 2 t 2 + t t ] y p = Y u = [ 1 e t e t 1 e t 0 1 0 e t ] [ − e t − e − t e − 2 t 2 + t t ] = [ e t ( 2 t − 1 ) − e − t 2 e t ( t − 1 ) − e − t 2 e t ( t − 1 ) − e − t ] \mathbf{u}=\left[\begin{array}{l}
-e^{t}-e^{-t} \\
\frac{e^{-2 t}}{2}+t \\
t
\end{array}\right] \\
\mathbf{y}_{p}=Y \mathbf{u}=\left[\begin{array}{lll}
1 & e^{t} & e^{t} \\
1 & e^{t} & 0 \\
1 & 0 & e^{t}
\end{array}\right]\left[\begin{array}{l}
-e^{t}-e^{-t} \\
\frac{e^{-2 t}}{2}+t \\
t
\end{array}\right]=\left[\begin{array}{c}
e^{t}(2 t-1)-\frac{e^{-t}}{2} \\
e^{t}(t-1)-\frac{e^{-t}}{2} \\
e^{t}(t-1)-e^{-t}
\end{array}\right] u = ⎣ ⎡ − e t − e − t 2 e − 2 t + t t ⎦ ⎤ y p = Y u = ⎣ ⎡ 1 1 1 e t e t 0 e t 0 e t ⎦ ⎤ ⎣ ⎡ − e t − e − t 2 e − 2 t + t t ⎦ ⎤ = ⎣ ⎡ e t ( 2 t − 1 ) − 2 e − t e t ( t − 1 ) − 2 e − t e t ( t − 1 ) − e − t ⎦ ⎤
#340153 on Dec 2023
Comments