ODE1.1

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Solve the system of ODE's: {\begin{cases}x_{t}=6x-3y\\y_{t}=2x+y\end{cases}}\,

Guess the solutions

x=Ae^{{\lambda t}},x_{t}=\lambda Ae^{{\lambda t}}\,
y=Be^{{\lambda t}},y_{t}=\lambda Be^{{\lambda t}}\,

Plug these guesses into the original equation.

\lambda Ae^{{\lambda t}}=6Ae^{{\lambda t}}-3Be^{{\lambda t}}\,
\lambda Be^{{\lambda t}}=2Ae^{{\lambda t}}+Be^{{\lambda t}}\,

Write this system of equations in matrix form.

{\begin{bmatrix}6-\lambda &-3\\2&1-\lambda \\\end{bmatrix}}{\begin{bmatrix}A\\B\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}\,

Find lambda by taking the determinant of the matrix.

(6-\lambda )(1-\lambda )+6=0\,
6-7\lambda +\lambda ^{2}+6=0\,
\lambda ^{2}-7\lambda +12=0\,
\lambda ={\frac  {7\pm {\sqrt  {49-48}}}{2}}={\frac  {7\pm 1}{2}}=3,4\,

There are two values for lambda, so there will be two solutions and the sum of these two solutions is the general solution.

If \lambda =3\,,

3A-3B=0\,
2A-2B=0\,

Add these equations to get
A-B=0\,, so A=B=C_{1}\,

One solution is x=C_{1}e^{{3t}},y=C_{1}e^{{3t}}\,

If \lambda =4\,,

2A-3B=0\,
2A-3B=0\,

These equations imply
B=(2/3)A\,, so if B=C_{2},A=(3/2)C_{2}\,

Then the general solution is

x=C_{1}e^{{3t}}+(3/2)C_{2}e^{{4t}}\,
y=C_{1}e^{{3t}}+C_{2}e^{{4t}}\,


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