Question

simultaneous differential equations

1. dx/dt=2x
dy/dt=4y


2. dx/dt=2x
dy/dt=2y

Answer 1

Answer on Question #79727 – Math – Differential Equations

Question

1. $\frac{dx}{dt} = 2x, \frac{dy}{dt} = 4y$

Solution

Matrix of the system:

$$A = \left( \begin{array}{cc} 2 & 0 \\ 0 & 4 \end{array} \right)$$

Solve the characteristic equation:

$$\det(A - \lambda E) = \left| \begin{array}{cc} 2 - \lambda & 0 \\ 0 & 4 - \lambda \end{array} \right| = 0$$ $$(2 - \lambda)(4 - \lambda) = 0$$ $$\lambda_1 = 2, \quad \lambda_2 = 4$$

Find eigenvectors of the matrix:

a) $\lambda_1 = 2$

$$\left( \begin{array}{cc} 2 - 2 & 0 \\ 0 & 4 - 2 \end{array} \right) \sim \left( \begin{array}{cc} 0 & 0 \\ 0 & 2 \end{array} \right) \sim \left( \begin{array}{cc} 0 & -2 \\ 0 & 0 \end{array} \right)$$ $$\left\{ \begin{array}{l} \alpha_1 = 2C_1 \\ \alpha_2 = 0 \end{array} \right. \Rightarrow \xrightarrow{A_1} = \left( \begin{array}{c} 2 \\ 0 \end{array} \right)$$

b) $\lambda_2 = 4$

$$\left( \begin{array}{cc} 2 - 4 & 0 \\ 0 & 4 - 4 \end{array} \right) \sim \left( \begin{array}{cc} -2 & 0 \\ 0 & 0 \end{array} \right)$$ $$\left\{ \begin{array}{c} \alpha_ {1} = 0 \\ \alpha_ {2} = - 2 C _ {1} \end{array} \right. \Rightarrow \underset {A _ {2}} {\longrightarrow} = \left( \begin{array}{c} 0 \\ - 2 \end{array} \right)$$ $$\binom {x} {y} = C _ {1} \binom {2} {0} e ^ {2 t} + C _ {2} \binom {0} {- 2} e ^ {4 t}$$

Answer: $\binom{x}{y}=C_1\binom{2}{0}e^{2t}+C_2\binom{0}{-2}e^{4t}.$

Question

2. $\frac{dx}{dt} = 2x, \frac{dy}{dt} = 2y$

Solution

Matrix of the system:

$$A = \left( \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right)$$

Solve the characteristic equation:

$$\det (A - \lambda E) = \left| \begin{array}{cc} 2 - \lambda & 0 \\ 0 & 2 - \lambda \end{array} \right| = 0$$ $$(2 - \lambda) ^ {2} = 0$$ $$\lambda = 2$$

Find eigenvectors of the matrix:

$$\left( \begin{array}{cc} 2 - 2 & 0 \\ 0 & 2 - 2 \end{array} \right) \left( \begin{array}{c} V _ {1 1} \\ V _ {2 1} \end{array} \right) = 0$$ $$\left\{ \begin{array}{l} 0 \cdot V _ {1 1} + 0 \cdot V _ {2 1} = 0 \\ 0 \cdot V _ {1 1} + 0 \cdot V _ {2 1} = 0 \end{array} \right.$$ $$V _ {1} = \left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$ $$V _ {2} = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$$ $$x (t) = C _ {1} e ^ {2 t}$$ $$y (t) = C _ {2} e ^ {2 t}$$

Answer:

$$x (t) = C _ {1} e ^ {2 t}$$ $$y (t) = C _ {2} e ^ {2 t}$$

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