Question

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Answer 1

Answer on Question #60150 – Math – Differential Equations

Question

Solve the linear system of differential equations

$$\left\{ \begin{array}{l} \frac {d x}{d t} = x - 2 y \\ \frac {d y}{d t} = x - y \end{array} \right.$$

Solution

Given

$$\left\{ \begin{array}{l} \frac {d x}{d t} = x - 2 y \\ \frac {d y}{d t} = x - y \end{array} \right.$$

1) Equation $\frac{dy}{dt} = x - y$ from (1) gives

$$x = \frac {d y}{d t} + y$$

2) Differentiate (2) with respect to $t$ :

$$\frac {d x}{d t} = \frac {d ^ {2} y}{d t ^ {2}} + \frac {d y}{d t}$$

3) Using (2) and (3) substitute for $x$ and $\frac{dx}{dt}$ into the first equation $\frac{dx}{dt} = x - 2y$ of (1):

$$\frac {d ^ {2} y}{d t ^ {2}} + \frac {d y}{d t} = \frac {d y}{d t} + y - 2 y,$$ $$\frac {d ^ {2} y}{d t ^ {2}} = - y.$$

The characteristic equation is

$$a ^ {2} = - 1,$$ $$a ^ {2} + 1 = 0.$$

Its solutions are

$$a _ {1} = i, a _ {2} = - i,$$

hence

$$y (t) = C _ {1} \cos (t) + C _ {2} \sin (t)$$

4) Differentiating (4) with respect to $t$ :

$$\frac {d y}{d t} = - C _ {1} \sin (t) + C _ {2} \cos (t)$$

Substituting (4), (5) into (2)

$$x(t) = \frac{dy}{dt} + y = -C_1 \sin(t) + C_2 \cos(t) + C_1 \cos(t) + C_2 \sin(t) = \\ = (C_2 - C_1) \sin(t) + (C_1 + C_2) \cos(t).$$

5) The general solution of system (1) is

$$\left\{ \begin{array}{c} x(t) = (C_2 - C_1) \sin(t) + (C_1 + C_2) \cos(t), \\ y(t) = C_1 \cos(t) + C_2 \sin(t), \end{array} \right. \quad C_1, C_2 \text{ are arbitrary real constants}.$$

Answer:

$$\left\{ \begin{array}{c} x(t) = (C_2 - C_1) \sin(t) + (C_1 + C_2) \cos(t), \\ y(t) = C_1 \cos(t) + C_2 \sin(t), \end{array} \right. \quad C_1, C_2 \text{ are arbitrary real constants}.$$

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