$\dot{x}=x+y+5$

$\dot{y}=x-y$

Is this system linear or nonlinear differential equations?

Kolby Castillo
2022-09-22
Answered

Is this system of differential equations linear?

$\dot{x}=x+y+5$

$\dot{y}=x-y$

Is this system linear or nonlinear differential equations?

$\dot{x}=x+y+5$

$\dot{y}=x-y$

Is this system linear or nonlinear differential equations?

You can still ask an expert for help

Nancy Phillips

Answered 2022-09-23
Author has **12** answers

Explanation:

The system is linear, yes. It can be written as

$$\left[\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}5\\ 0\end{array}\right]$$

The system is linear, yes. It can be written as

$$\left[\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}5\\ 0\end{array}\right]$$

Medenovgj

Answered 2022-09-24
Author has **1** answers

Step 1

You could write your system of equations as

$$\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}5\\ 0\end{array}\right)$$

and solve it this way:

1. Solve your homogeneous system. In this case, it would be

$${\overrightarrow{x}}_{h}=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}\\ -2{c}_{2}{e}^{-t}\end{array}\right).$$

2. Solve for your particular solution.

2.1. Rewrite your system as $\overrightarrow{x}(t{)}^{\prime}=A\overrightarrow{x}(t)+\overrightarrow{g}$ where $\overrightarrow{g}=\left(\begin{array}{c}5\\ 0\end{array}\right)$. Then guess a particular solution. In this case, ${\overrightarrow{x}}_{p}=\overrightarrow{a}.$

2.2. Take the derivative,

$${\overrightarrow{x}}_{p}=0,$$

and plug things in.

$$0=A\overrightarrow{a}+\overrightarrow{g}$$

$$A\overrightarrow{a}=-\overrightarrow{g}.$$

2.3. Solve the system.

$$\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right)=\left(\begin{array}{c}-5\\ 0\end{array}\right)$$

$$\{\begin{array}{l}{a}_{1}+{a}_{2}=-5\\ {a}_{1}-{a}_{2}=0\end{array}$$

$${a}_{1}=-2.5\text{and}{a}_{2}=-2.5$$

$$\overrightarrow{a}=\left(\begin{array}{c}-2.5\\ -2.5\end{array}\right).$$

Step 2

So, the solution is

$$\overrightarrow{x}(t)=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}-2.5\\ -2{c}_{2}{e}^{-t}-2.5\end{array}\right).$$

You could write your system of equations as

$$\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}5\\ 0\end{array}\right)$$

and solve it this way:

1. Solve your homogeneous system. In this case, it would be

$${\overrightarrow{x}}_{h}=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}\\ -2{c}_{2}{e}^{-t}\end{array}\right).$$

2. Solve for your particular solution.

2.1. Rewrite your system as $\overrightarrow{x}(t{)}^{\prime}=A\overrightarrow{x}(t)+\overrightarrow{g}$ where $\overrightarrow{g}=\left(\begin{array}{c}5\\ 0\end{array}\right)$. Then guess a particular solution. In this case, ${\overrightarrow{x}}_{p}=\overrightarrow{a}.$

2.2. Take the derivative,

$${\overrightarrow{x}}_{p}=0,$$

and plug things in.

$$0=A\overrightarrow{a}+\overrightarrow{g}$$

$$A\overrightarrow{a}=-\overrightarrow{g}.$$

2.3. Solve the system.

$$\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right)=\left(\begin{array}{c}-5\\ 0\end{array}\right)$$

$$\{\begin{array}{l}{a}_{1}+{a}_{2}=-5\\ {a}_{1}-{a}_{2}=0\end{array}$$

$${a}_{1}=-2.5\text{and}{a}_{2}=-2.5$$

$$\overrightarrow{a}=\left(\begin{array}{c}-2.5\\ -2.5\end{array}\right).$$

Step 2

So, the solution is

$$\overrightarrow{x}(t)=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}-2.5\\ -2{c}_{2}{e}^{-t}-2.5\end{array}\right).$$

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