Answer on Question #79727 – Math – Differential Equations
Question
1. d x d t = 2 x , d y d t = 4 y \frac{dx}{dt} = 2x, \frac{dy}{dt} = 4y d t d x = 2 x , d t d y = 4 y
Solution
Matrix of the system:
A = ( 2 0 0 4 ) A = \left( \begin{array}{cc} 2 & 0 \\ 0 & 4 \end{array} \right) A = ( 2 0 0 4 )
Solve the characteristic equation:
det ( A − λ E ) = ∣ 2 − λ 0 0 4 − λ ∣ = 0 \det(A - \lambda E) = \left| \begin{array}{cc} 2 - \lambda & 0 \\ 0 & 4 - \lambda \end{array} \right| = 0 det ( A − λ E ) = ∣ ∣ 2 − λ 0 0 4 − λ ∣ ∣ = 0 ( 2 − λ ) ( 4 − λ ) = 0 (2 - \lambda)(4 - \lambda) = 0 ( 2 − λ ) ( 4 − λ ) = 0 λ 1 = 2 , λ 2 = 4 \lambda_1 = 2, \quad \lambda_2 = 4 λ 1 = 2 , λ 2 = 4
Find eigenvectors of the matrix:
a) λ 1 = 2 \lambda_1 = 2 λ 1 = 2
( 2 − 2 0 0 4 − 2 ) ∼ ( 0 0 0 2 ) ∼ ( 0 − 2 0 0 ) \left( \begin{array}{cc} 2 - 2 & 0 \\ 0 & 4 - 2 \end{array} \right) \sim \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array} \right) \sim \left( \begin{array}{cc} 0 & -2 \\ 0 & 0 \end{array} \right) ( 2 − 2 0 0 4 − 2 ) ∼ ( 0 0 0 2 ) ∼ ( 0 0 − 2 0 ) { α 1 = 2 C 1 α 2 = 0 ⇒ → A 1 = ( 2 0 ) \left\{ \begin{array}{l} \alpha_1 = 2C_1 \\ \alpha_2 = 0 \end{array} \right. \Rightarrow \xrightarrow{A_1} = \left( \begin{array}{c} 2 \\ 0 \end{array} \right) { α 1 = 2 C 1 α 2 = 0 ⇒ A 1 = ( 2 0 )
b) λ 2 = 4 \lambda_2 = 4 λ 2 = 4
( 2 − 4 0 0 4 − 4 ) ∼ ( − 2 0 0 0 ) \left( \begin{array}{cc} 2 - 4 & 0 \\ 0 & 4 - 4 \end{array} \right) \sim \left( \begin{array}{cc} -2 & 0 \\ 0 & 0 \end{array} \right) ( 2 − 4 0 0 4 − 4 ) ∼ ( − 2 0 0 0 ) { α 1 = 0 α 2 = − 2 C 1 ⇒ ⟶ A 2 = ( 0 − 2 ) \left\{ \begin{array}{c} \alpha_ {1} = 0 \\ \alpha_ {2} = - 2 C _ {1} \end{array} \right. \Rightarrow \underset {A _ {2}} {\longrightarrow} = \left( \begin{array}{c} 0 \\ - 2 \end{array} \right) { α 1 = 0 α 2 = − 2 C 1 ⇒ A 2 ⟶ = ( 0 − 2 ) ( x y ) = C 1 ( 2 0 ) e 2 t + C 2 ( 0 − 2 ) e 4 t \binom {x} {y} = C _ {1} \binom {2} {0} e ^ {2 t} + C _ {2} \binom {0} {- 2} e ^ {4 t} ( y x ) = C 1 ( 0 2 ) e 2 t + C 2 ( − 2 0 ) e 4 t
Answer: ( x y ) = C 1 ( 2 0 ) e 2 t + C 2 ( 0 − 2 ) e 4 t . \binom{x}{y}=C_1\binom{2}{0}e^{2t}+C_2\binom{0}{-2}e^{4t}. ( y x ) = C 1 ( 0 2 ) e 2 t + C 2 ( − 2 0 ) e 4 t .
Question
2. d x d t = 2 x , d y d t = 2 y \frac{dx}{dt} = 2x, \frac{dy}{dt} = 2y d t d x = 2 x , d t d y = 2 y
Solution
Matrix of the system:
A = ( 2 0 0 2 ) A = \left( \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right) A = ( 2 0 0 2 )
Solve the characteristic equation:
det ( A − λ E ) = ∣ 2 − λ 0 0 2 − λ ∣ = 0 \det (A - \lambda E) = \left| \begin{array}{cc} 2 - \lambda & 0 \\ 0 & 2 - \lambda \end{array} \right| = 0 det ( A − λ E ) = ∣ ∣ 2 − λ 0 0 2 − λ ∣ ∣ = 0 ( 2 − λ ) 2 = 0 (2 - \lambda) ^ {2} = 0 ( 2 − λ ) 2 = 0 λ = 2 \lambda = 2 λ = 2
Find eigenvectors of the matrix:
( 2 − 2 0 0 2 − 2 ) ( V 11 V 21 ) = 0 \left( \begin{array}{cc} 2 - 2 & 0 \\ 0 & 2 - 2 \end{array} \right) \left( \begin{array}{c} V _ {1 1} \\ V _ {2 1} \end{array} \right) = 0 ( 2 − 2 0 0 2 − 2 ) ( V 11 V 21 ) = 0 { 0 ⋅ V 11 + 0 ⋅ V 21 = 0 0 ⋅ V 11 + 0 ⋅ V 21 = 0 \left\{ \begin{array}{l} 0 \cdot V _ {1 1} + 0 \cdot V _ {2 1} = 0 \\ 0 \cdot V _ {1 1} + 0 \cdot V _ {2 1} = 0 \end{array} \right. { 0 ⋅ V 11 + 0 ⋅ V 21 = 0 0 ⋅ V 11 + 0 ⋅ V 21 = 0 V 1 = ( 0 1 ) V _ {1} = \left( \begin{array}{c} 0 \\ 1 \end{array} \right) V 1 = ( 0 1 ) V 2 = ( 1 0 ) V _ {2} = \left( \begin{array}{c} 1 \\ 0 \end{array} \right) V 2 = ( 1 0 ) x ( t ) = C 1 e 2 t x (t) = C _ {1} e ^ {2 t} x ( t ) = C 1 e 2 t y ( t ) = C 2 e 2 t y (t) = C _ {2} e ^ {2 t} y ( t ) = C 2 e 2 t
Answer:
x ( t ) = C 1 e 2 t x (t) = C _ {1} e ^ {2 t} x ( t ) = C 1 e 2 t y ( t ) = C 2 e 2 t y (t) = C _ {2} e ^ {2 t} y ( t ) = C 2 e 2 t
Answer provided by https://www.AssignmentExpert.com
#339914 on Dec 2023