Find the solution of the following couple of different differential equations x^'=4x-y and y^'=-4x+4y when t=0 ,x(0)=1 and y(0)=1
"\\begin{cases}\nx^\\prime =4x-y\n\\\\\ny^\\prime=-4x+4y\n\\end{cases}" and "x(0)=y(0)=1" .
From the first equation we get "x^{\\prime \\prime}=4x^\\prime -y^\\prime =4x^\\prime -\\left(-4x+4y \\right)=4x^\\prime +4x-4y=4x^\\prime +4x-4(4x-x^\\prime )=4x^\\prime +4x-16x+4x^\\prime"
We have second order differential equation "x^{\\prime \\prime}-8x^\\prime +12x=0" .
The characteristic equation is "\\lambda ^2-8\\lambda +12=0" .
The roots are "\\lambda _1=2,\\" "\\lambda _2=6" .
The general solution is "x(t)=c_1e^{2t}+c_2e^{6t}" .
Then we can find "y=4x-x^\\prime =4\\left( c_1e^{2t}+c_2e^{6t} \\right) -\\left(2c_1e^{2t}+6c_2e^{6t}\\right)= 2c_1e^{2t}-2c_2e^{6t}" .
Using "x(0)=y(0)=1" , we get a system:
"\\begin{cases}\n1=c_1+c_2\n\\\\\n1=2c_1-2c_2\n\\end{cases}" "\\begin{cases}\n2=2c_1+2c_2\n\\\\\n1=2c_1-2c_2\n\\end{cases}" "\\begin{cases}\nc_1=\\tfrac{3}{4}\n\\\\\nc_2=\\tfrac{1}{4}\n\\end{cases}"
Answer: "x(t)=\\tfrac{1}{4}\\left(3e^{2t}+e^{6t}\\right), \\ \\ y(t)=\\tfrac{1}{2}\\left(3e^{2t}-e^{6t}\\right)" .
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