Answer to Question #203681 in Differential Equations for rahimah

Question #203681

2) Solve the following system

𝑑π‘₯

𝑑𝑑

= π‘₯ + 𝑦

𝑑𝑦

𝑑𝑑

= βˆ’2π‘₯ βˆ’ 𝑦 


1
Expert's answer
2021-06-08T09:36:34-0400
"\\dfrac{dx}{dt}=x+y"

"\\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:


"\\frac{d^2x}{dt^2}-\\frac{dx}{dt}=-2x-\\frac{dx}{dt}+x"


which is equivalent to


"\\frac{d^2x}{dt^2}+x=0"



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


"y=\\frac{dx}{dt}-x"

"=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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