Is this system of differential equations linear? Suppose I have a system of differential equations like below: dotx = x + y + 5, doty = x - y

Is this system of differential equations linear?
$\stackrel{˙}{x}=x+y+5$
$\stackrel{˙}{y}=x-y$
Is this system linear or nonlinear differential equations?
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Nancy Phillips
Explanation:
The system is linear, yes. It can be written as
$\left[\begin{array}{c}\stackrel{˙}{x}\\ \stackrel{˙}{y}\end{array}\right]=\left[\begin{array}{cc}1& 1\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}5\\ 0\end{array}\right]$
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Medenovgj
Step 1
You could write your system of equations as
$\left(\begin{array}{c}\stackrel{˙}{x}\\ \stackrel{˙}{y}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}5\\ 0\end{array}\right)$
and solve it this way:
1. Solve your homogeneous system. In this case, it would be
${\stackrel{\to }{x}}_{h}=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}\\ -2{c}_{2}{e}^{-t}\end{array}\right).$
2. Solve for your particular solution.
2.1. Rewrite your system as $\stackrel{\to }{x}\left(t{\right)}^{\prime }=A\stackrel{\to }{x}\left(t\right)+\stackrel{\to }{g}$ where $\stackrel{\to }{g}=\left(\begin{array}{c}5\\ 0\end{array}\right)$. Then guess a particular solution. In this case, ${\stackrel{\to }{x}}_{p}=\stackrel{\to }{a}.$
2.2. Take the derivative,
${\stackrel{\to }{x}}_{p}=0,$
and plug things in.
$0=A\stackrel{\to }{a}+\stackrel{\to }{g}$
$A\stackrel{\to }{a}=-\stackrel{\to }{g}.$
2.3. Solve the system.
$\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\end{array}\right)=\left(\begin{array}{c}-5\\ 0\end{array}\right)$
$\left\{\begin{array}{l}{a}_{1}+{a}_{2}=-5\\ {a}_{1}-{a}_{2}=0\end{array}$

$\stackrel{\to }{a}=\left(\begin{array}{c}-2.5\\ -2.5\end{array}\right).$
Step 2
So, the solution is
$\stackrel{\to }{x}\left(t\right)=\left(\begin{array}{c}{c}_{1}{e}^{t}+{c}_{2}{e}^{-t}-2.5\\ -2{c}_{2}{e}^{-t}-2.5\end{array}\right).$