2) Solve the following system
ππ₯
ππ‘
= π₯ + π¦
ππ¦
ππ‘
= β2π₯ β π¦Β
"\\dfrac{dy}{dt}=-2x-y"
The first equation of the system is equivalent toΒ
"y=\\frac{dx}{dt}-x"ThenΒ
"\\frac{dy}{dt}=\\frac{d^2x}{dt^2}-\\frac{dx}{dt}"Put these in the second equation of the system. We have the following differential equation:
which is equivalent to
Its characteristic equationΒ "k^2+1=0."Β
Therefore,Β "x(t)=C_1\\sin t+C_2\\cos t".
ThenΒ
"\\frac{dx}{dt}=C_1\\cos t-C_2\\sin t"Β and
"=C_1\\cos t-C_2\\sin t-C_1\\sin t-C_2\\cos t"
Β .Consequently, the system has the general solution:
"\\begin{cases}\nx(t)=C_1\\sin t+C_2\\cos t\\\\\ny(t)=(-C_1-C_2)\\sin t+(C_1-C_2))\\cos t\n\\end{cases}"
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